Class 12 Medical Physics - Nuclei

Find the binding energy of an $\alpha$ -particle from the following data:
Mass of helium nucleus $=4.001265$ a.m.u; mass of proton $=1.007277$ a.m.u. Mass of neutron $=1.00866$ a.m.u (take $1$ a.m.u $=931.5$ MeV)

$\alpha$ particle

$\rightarrow _2^4He$

Given,

Mass of helium nucleus

$=4.001265$ a.m.u

Mass of proton

$=1.007277$ a.m.u

Mass of neutron

$=1.00866$ a.m.u

We have Binding Energy,

$E=\triangle m c^2$

So,

$\triangle m=$ mass of Helium

$-2$ (mass of neutron + proton)

$=4.001265-2(1.007277+1.00866) \\=4.001265-4.031874 \\=-0.030609u$

$E=\triangle mc^2 \\=0.030609\times 931.5MeV \\=28.5122835MeV$

The binding energies per nucleon for deutron $(_{1}H^{2})$ and helium $(_{2}He^{4})$ are $1.1$ MeV and $7.0$ MeV respectively. Calculate the energy released, when two deutrons fuse to form a helium nucleus $(_{2}He^{4})$ .

$_1H^2+ \; _1H^2\rightarrow \; _2He^4$

So, Energy released = Energy of

$_2He^4$

$-2$ (Energy of

$_1H^2$ )

$=7.0-2\times 1.1 \\=4.8MeV$

The binding energy per nucleon for $_{6}C^{12}$ nucleons is $7.68$ MeV and that for $_{6}C^{13}$ is $7.47$ MeV. Calculate the energy required to remove a neutron from $_{6}C^{13}$ nucleons.

Given data:
Binding energy per nucleon of

$_{6}C^{13}=7.47$ MeV
Binding energy per nucleon of

$_{6}C^{12}=7.68$ MeV
Binding energy of neutron

$=?$
Solution:
We can write the reaction as:

$_{6}C^{13}\rightarrow _{6}C^{12} + _{0}n^1$
Total binding energy of

$_{6}C^{13}=7.47\times 13=97.11$ MeV
Total binding energy of

$_{6}C^{12}=7.68\times 12=92.16$ MeV
Total binding energy of the reactant

$=$ Total binding energy of the product

$\therefore 97.11$ MeV

$=92.16$ MeV

$+$ Binding energy of a neutron.

$\therefore$ Binding energy of a neutron

$=97.11-92.16$

$=4.95$ MeV.

What is (a) mass defect, and (b) binding energy in Oxygen $^{16}_{8}{O}$ , whose nuclear mass is $15.995$ amu. $(m_p=1.0078$ amu $;$ $m_n=1.0087$ amu $)$ .

Oxygen

$^{16}_8O$ has

$8$ protons and

$8$ neutrons.

Nuclear mass of

$^{16}_8O$ ,

$M_{N} = 15.995$ amu

(a) Mass defect,

$\Delta M = 8m_p + 8m_n - M_N$

$\therefore$

$\Delta M = 8(1.0078) + 8(1.0087) - 15.995 = 0.137$ amu

(b) Binding energy,

$E =\Delta M \times 931.5$ MeV

$\implies$

$E = 0.137\times 931.5 = 127.62$ MeV

A neutron breaks into a proton and an electron. Calculate the energy produced in this reaction in MeV. Mass of an electron $=9\times {10}^{-31}kg$ , mass of proton $=1.6725\times {10}^{-27}kg$ , mass of neutron $=1.6747\times {10}^{-27}kg$ and speed of light $=3\times {10}^{8}m{s}^{-1}$

Mass defect of the process is given by

$\Delta m=\left[ 1.6747\times { 10 }^{ -27 }-\left( 1.6725\times { 10 }^{ -27 }+9\times { 10 }^{ -31 } \right) \right] 0.0013\times { 10 }^{ -27 }kg$
According to mass-energy relationship
Energy released

$E=\Delta mc^2$

$E=\left( 0.0013\times { 10 }^{ -27 } \right) { \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }=\cfrac { 1.17\times { 10 }^{ -3 } }{ 1.6\times { 10 }^{ -19 } } =0.73\times { 10 }^{ }eV=0.73MeV$

Calculate the energy released if $U^{238}$ -nucleus emits an $\alpha-$ particle.
OR
Calculate the energy released in MeV in the following nuclear reaction
$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Q$
Given Atomic mass of $^{238}U = 238.05079 \,u$
Atomic mass of $^{234}Th = 234.04363 \,u$
Atomic mass of alpha particle $= 4.00260 \,u$
$1 \,u = 931.5 \,MeV / c^2$
Is the decay spontaneous ? Give reasons.

The process is

$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He + Q$
The energy released

$(\alpha - particle)$

$Q = (M_U - M_{TH} - M_{He}) c^2$

$= (238.05079 - 234.04363 - 4.00260) u \times c^2 = (0.00456 \,u) \times c^2$

$= 0.00456 \times \left ( \dfrac{931. - MeV}{c^2} \right ). c^2 = 4.25 \,MeV$
Yes, the decay is spontaneous (since Q is positive).

What are the different types of nuclear reactions?

There are two types of nuclear reactions:
(i) Nuclear fission
(ii) Nuclear fusion.

Answer the following question:
Identify the nature of the radioactive radiations emitted in each step of the decay process given below:
$^{A}_{Z}X \rightarrow ^{A-4}_{Z-1}W$

If atomic number decreases by 2 and mass number decreases by 4 an alpha particle is emitted out. So,

$^{a}_{z}X \overset{\alpha}{\longrightarrow} ^{A-4}_{Z-2}Y$
If a

$\beta$ -is emitted out, the atomic number increases by 1, while mass number remains uncharged. So,

$^{a-4}_{z-2}Y \overset{\beta}{\longrightarrow} ^{A-4}_{Z-1}W$

Answer the following question:
A heavy nucleus splits into two lighter nuclei. Which one of the two-parent nucleus or the daughter nuclei has more binding energy per nucleon ?

Parent nucleus has lower binding energy per nucleon compared to that of the daughter nuclei. When a heavy nucleus splits into two lighter nuclei, nucleons get more tightly bound.

Give reason for the release of $21.6\ MeV$ energy in the fusion of deuterium reaction.

$\underset{\text{(deuterium)}}{^{2}_{1}H}\ +\ \underset{\text{(deuterium)}}{^{2}_{1}H}\ \rightarrow\ \underset{\text{(helium isotope)}}{^{3}_{2}He}\ +\ ^{1}_{0}n\ +\ 3.3\ MeV$

$\underset{\text{(helium)}}{^{3}_{2}H}\ +\ \underset{\text{(deuterium)}}{^{2}_{1}H}\ \rightarrow\ \underset{\text{(helium)}}{^{4}_{2}He}\ +\ \underset{\text{(proton)}}{^{1}_{0}H}\ +\ 18.3\ MeV$

When two deuterium nuclei fuse,

$3.3 MeV$ energy is released and the nucleus of helium isotope is formed. In this process again this helium isotope gets fused with one deuterium nucleus to form a helium nucleus and

$18.3\ MeV$ is released. Thus, in all three deuterium nuclei fuse to form a helium nucleus with a release of

$21.6\ MeV$ .

Write one nuclear fusion reaction.

$\underset{\text{(deuterium)}}{^{2}_{1}H}\ +\ \underset{\text{(deuterium)}}{^{2}_{1}H}\ \rightarrow\ \underset{\text{(helium isotope)}}{^{3}_{2}He}\ +\ ^{1}_{0}n\ +\ 3.3\ MeV$

$\underset{\text{(helium)}}{^{3}_{2}H}\ +\ \underset{\text{(deuterium)}}{^{2}_{1}H}\ \rightarrow\ \underset{\text{(helium)}}{^{4}_{2}He}\ +\ \underset{\text{(proton)}}{^{1}_{0}H}\ +\ 18.3\ MeV$

Why is a very high temperature required for the process of nuclear fusion? State the approximate temperature required.

When two nuclei approach each other, due to their positive charge, the electrostatic force of repulsion becomes too strong between them that they do not fuse. Hence, at ordinary temperature and pressure nuclear fusion is not possible. To make the fusion possible, a high temperature of approximately

$107 K$ and high pressure is required. Due to thermal agitations both nuclei acquire sufficient kinetic energy at such a high temperature so as to overcome the force of repulsion between them when they approach each other and so they get fused.

Which of the isotope of Uranium is easily fissionable? Give reason.

$^{235}_{92}U$ is more easily fissionable than the isotope

$^{238}_{92}U$ . This is because the fission of

$^{238}_{92}U$ nucleus is possible only by the fast neutrons, while the fission of

$^{235}_{92}U$ nucleus can be even by the slow neutrons.

The isotope $_{ 5 }^{ 12 }{ B }$ having a mass 12.014 u undergoes $\beta$ -decay to $_{ 6}^{ 12 }{ C }$ , $_{ 6 }^{ 12 }{ C }$ has an excited state of the nucleus ( $_{ 5 }^{ 12 }{ C^* }$ ) at 4.041 MeV above its ground state. If $_{ 5 }^{ 12 }{ B }$ decays to $_{ 6 }^{ 12 }{ C^* }$ , the maximum kinetic energy of the $\beta$ -particle in units of MeV is
(1 u = 931.5 MeV/ ${ c }^{ 2 }$ , where c is the speed of light in vacuum).

$_{ 5 }^{ 12 }B\rightarrow _{ 6 }^{ 12 }{ C }+{ \beta }^{ - }+{ \gamma }^{ - }$

$Q=(m_{ 5 }^{ 12 }{ B }-m_{ 6 }^{ 12 }{ C }){ c }^{ 2 }\\ \quad =(m_{ 5 }^{ 12 }{ B }-(m_{ 6 }^{ 12 }{ C }+\Delta m)){ c }^{ 2 }\\ \quad =(m_{ 5 }^{ 12 }{ B }-m_{ 6 }^{ 12 }{ C }){ c }^{ 2 }-\Delta m{ c }^{ 2 }\\ \quad =0.014\times 931.5-4.041\\ \quad =9\ MeV$

The binding energy of $_{ 17 }^{ }{ { Cl }^{ 35 } }$ nucleus is $298$ MeV. Find its atomic mass. The mass of hydrogen atom $\left( _{ 1 }^{}{ { H }^{ 1 } } \right)$ is $1.008143$ a.m.u. Given $1$ a.m.u. $=931$ MeV.

The

$_{ 17 }^{ }{ { Cl }^{ 35 } }$ atom has

$17$ proton and

$18$ neutrons in its nucleus.
Mass of

$17\ _{ 1 }^{}{ { H }^{ 1 } }$ atom has

$17$ protons and

$18$ neutrons in its nucleus
Mass of

$17$ proton

$=17\times 1.008143$ amu

$=17.138431$ amu
Mass of

$18$ neutrons

$=18\times 1.008986=18.161748$ a.m.u
Total

$=35.300179$ a.m.u
Mass defect

$\Delta m=298/931=0.320085$ a.m.u
The atomic mass of

$_{ 17 }^{ }{ { Cl }^{ 35 } }$ would be the sum of equivalent of the binding energy of the nucleus. Hence atomic mass of

$_{ 17 }^{ }{ { Cl }^{ 35 } }=35.300179-0.320085=34.980094$ a.m.u

(a) Show that the mass M of an atom is given approximately by $M_{app}$ = $Am_p$ ,where A is the mass number and $m_p$ is the proton mass. For (b) $^1 H$ , (c) $^{31} P$ , (d) $^{120} Sn$ , (e) $^{197}Au$ , and (f) $^{239} Pu$ , use Table 42-1 to find the percentage deviation between $M_{app}$ and M:
Percentage deviation = $\frac {M_{app}M}{M}100$
(g) Is a value of $M_{app}$ accurate enough to be used in a calculation ofa nuclear binding energy?

1706937_1780864_ans_88dd358e54fe4d61af3bd607349a16e0.png

Consider the fission of $^{238}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final stable end products,after the beta decay of the primary fission fragments,are $^{140}Ce$ and $^{99}Ru$ .
(a) What is the total of the beta-decay events in the two beta-decay chains?
(b) Calculate $Q$ for this fission process. The relevant atomic and particle masses are
$^{238}U$ $238.05079 \ u$ $^{140}Ce$ $139.90543 \ u$
$n$ $1.00866 \ u$ $^{99}Ru$ $98.90594 \ u$

1779583_1781597_ans_ccc6e48d24ab4c7fb667d778e826c279.jpg

A $^{238}U$ nucleus emits a 4.196 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual $^{234}Th$ nucleus into account.

1776856_1781567_ans_5e746ec06fc3412fb5c7421cb5d8c87f.jpg

(a) Calculate the disintegration energy $Q$ for the fission of the molybdenum isotope $^{98}Mo$ into two equal parts. The masses you will need are $97.905 41 \ u$ for $^{98}Mo$ and $48.950 02 \ u$ for $^{49}Sc$ .(b) If $Q$ turns out to be positive, discuss why this process does not occur spontaneously.

1974673_1780832_ans_fb3f638a67594fd6b33b1462a81ca6f7.jpg

Verify the three $Q$ values reported for the reactions given in the above figure. The needed atomic and particle masses are
$^1H$ $1.007 825 \ u$ $^4He$ $4.002 603 \ u$
$^2H$ $2.014 102 \ u$ $e^{\pm}$ $0.000 548 6 \ u$
$^3He$ $3.016 029 \ u$

(Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.)

1797939_1781820_ans_0fcf49f071d04a069ee847179fe64c3a.jpg

Coal burns according to the reaction $C + O_2 \rightarrow CO_2$ . The heat of combustion is $3.3 \times 10^7 \ J/kg$ of atomic carbon consumed.
(a) Express this in terms of energy per carbon atom.
(b) Express it in terms of energy per kilogram of the initial reactants, carbon and oxygen.
(c) Suppose that the Sun (mass $= 2.0 \times 10^{30} \ kg$ ) were made of carbon and oxygen in combustible proportions and that it continued to radiate energy at its present rate of $3.9 \times 10^{26} \ W$ . How long would the Sun last?

1796053_1781832_ans_97c9494c2d5c4ceca8a6749e3f48ac52.jpg

We have seen that $Q$ for the overall proton-proton fusion cycle is $26.7 \ MeV$ . How can you relate this number to the $Q$ values for the reactions that make up this cycle, as displayed in the above figure?

1794515_1781818_ans_a6b0a39a7d504c66a7bea0d9a3dfbe86.jpg

Calculate the Q values f the fusion reaction $^4He +^4 He = ^8Be$
In such a fusion energetically favourable? the atomic mass of $^8 Be$ is 8.0053 u and that of $^4He$ is 4.0026 u.

$^4H +^4 H \rightarrow ^8Be$

$M(^2H) = 4.0026 u$

$M(^8 Be ) = 8.0053 u$

$Q value =[ 2 M(^2H) -M(^8Be)] = ( 2 \times 4.0026 - 8.0053 ) u$

$= 0.0001 \times 931 Mev$

$=0.0001\times 931\times 10^{6}$

$= 93.1 Kev$

Lithium (Z = 3) has two stable isotopes $^6Li$ and $^7Li$ . When neutrons are bombarded on lithium sample, electrons and $\alpha$ - particles are ejected. Write down the nuclear processes taking place.

The nuclear process taking place is,

${^6_3}L_i +n \rightarrow {^7_3}L_{i}$

${^7_3}L_{i} +n\rightarrow {^8_3}L_i \rightarrow {^8_4}Be+\bar{v}+e^{-}$

${ ^8_3}Be \rightarrow ^4_2He + ^4_2 He$

Find the energy liberated in the reaction
$^{223}Ra \rightarrow ^{209}Pb + ^{14}C$
the atomic masses needed are as follows
$^{223}Ra \quad ^{209}Pb \quad ^{14}C$
$22.018u \quad 208.981 u \quad 14.003 u$

$^{223}R_a = 223.018 u, ^{209}Pb= 208.981 u; ^{14}C = 14.003 u$

$^{223}R_a \rightarrow ^{209}Pb + ^{14}C$

$\triangle m = mass ^{223}R_a - mass ( ^{209}Pb +^{14}C )$

$\rightarrow = 223.018 -(208.981 +14.003) = 0.034$

$energy = \triangle M \times u = 0.034 \times 931 = 31.65 MeV$

Potassium -40 can decay in three modes. it can decay by $\beta^-$ emission $\beta^+$ emission or electron capture .(a) write the equations showing the end products.(b) Find the Q-values in each of the three cases. Atomic masses of $^{40}_{18}Ar, \ ^{40}_{19}K$ and $^{40}_{20}Ca$ are 39.9624 u , 39.9640 u and 39.9626 u respectively.

1870528_1724399_ans_85924a1f8356447794a5b68c60a7e0ab.jpg

How much quantity of radioactive substance will remain in decayed after 5 half-lives period ?

Initial quantity

$\rightarrow A_o$

After

$1^{st}$ half life

$\rightarrow A_o/2$

After

$2^{nd}$ half life

$\rightarrow (A_o/2)/2=A_o/4$

After

$3^{rd}$ half life

$\rightarrow (A_o/4)2=A_o/8$

After

$4^{th}$ half life

$\rightarrow (A_o/8)/2=A_o/16$

After

$5^{th}$ half life

$\rightarrow (A_o/16)/2=A_o/32$ .

1200340_1399802_ans_ed0a18d635584bec8e9a90fc65520621.jpg

Hydrogen bomb is based on the principle of

Hydrogen Bomb is based on the principle of nuclear fusion. Nuclear fusion is a reaction in which two or more atomic nuclei are combined to form one or more different atomic nuclei and subatomic particles. The difference in mass between the reactants and products is manifested as either the release or absorption of energy.

The mass defect for the nucleus of $_{2}He^{4}$ is $0.0303 a.m.u$ . The approximate value of binding energy per nucleon in $MeV$ will be $(1 \mathrm{a}.\mathrm{m}.\mathrm{u}=931\mathrm{M}\mathrm{e}\vee)$

Energy released

$=$ mass defect

$\times \ c^{2}$
Binding energy

$=0.0303\times 931$
(Since 1 a.m.u

$=931$ MeV)
Binding energy per nucleon

$=\dfrac{0.0303\times 931}{4}$

$=7\ MeV$

In plants and animals, the ratio of C $^{14}$ to C $^{12}$ is _______.

$1:1.35\times10^{-12}$

Animals including humans consume plants a lot thus tend to have the same ratio of carbon 14 to carbon 12 atoms. The ratio of carbon 14 to carbon 12 in living organisms is

$1:1.35\times10^{-12}$ .

The energy released during a nuclear reaction is called _______ energy

The energy released during a nuclear reaction is called binding energy.

Binding energy is the minimum energy required to disassemble a system of particles into separate parts.

When an $\alpha$ -particle is ejected, the atomic number of the atom decreases by ___________ .

When an

$\alpha$ particles...

When an alpha particle is emitted from a nucleus, the nucleus loses two protons and two neutrons. This means the atomic mass number decreases by

$4$ and the atomic number decreases by

$2$ (it loses two protons)

If the binding energy per nucleon for Li $^7$ is 5.6 MeV, the total binding energy of a lithium nucleus is _______ MeV.

For

$_{ 3 }{ Li }^{ 7 }$ nucleus

Mass defect,

$\triangle M=0.0424$

$\because 1u=931.5MeV/{ c }^{ 2 }$

$\therefore \triangle M=0.042\times 931.5MeV/{ c }^{ 2 }$

$=39.1MeV/{ c }^{ 2 }$

Binding energy,

${ E }_{ b }=\triangle M{ c }^{ 2 }$

$=\left( 39.1\times \cfrac { MeV }{ { c }^{ 2 } } \right) { c }^{ 2 }$

$=39.1MeV$

The mass of $_{2}He^{4}$ nnucleus is $0.03$ amu less than the sum of the masses of $2$ protons and $2$ neutrons. What is the binding energy per nucleon?

Here,

$\triangle m=0.03u$ (given)

So,

$E=\triangle mc^2 \\=0.03\times 931.5 \\=27.945MeV$

Find out the binding energy per nucleon of an $\alpha$ -particle in MeV. It is given that the masses of $\alpha$ -particle, proton and neutron are respectively $4.00150$ a.m.u., $1.00728$ a.m.u. and $1.00867$ a.m.u.

Binding Energy

$=\triangle mc^2$

$\triangle m=$ mass of

$\alpha$ particle

$-$ mass of total number of proton and neutron

$=4.00150-2(1.00728+1.00867) \\=4.00150-4.0319 \\=-0.0304u$

$E=\triangle mc^2$

$=0.0304\times 931MeV \\=28.30MeV$

So, Binding energy per nucleon

$=\cfrac{E}{no.\;of\;nucleon} \\=\cfrac{28.30}{4}=7.07MeV$

Calculate the binding energy of an alpha-particle. Given mass of proton $=1.0073$ a.m.u., mass of neutron $=1.0087$ a.m.u., mass of $\alpha$ -particle $=4.0015$ a.m.u.

Binding Energy

$=\triangle mc^2$

$\triangle m=$ mass of

$\alpha$ particle

$-$ mass of total number of proton and neutron

$=4.00150-2(1.00728+1.00867) \\=4.00150-4.0319 \\=-0.0304u$

$E=\triangle mc^2$

$=0.0304\times 931MeV \\=28.30MeV$

The total energy of an electron in the first excited state of the hydrogen atom is $-3.4$ eV. Find the out its (i) kinetic energy and (ii) potential energy in this state.

We know that

$\dfrac{|P.E|}{2} = |T.E| = K.E$

So,

$\dfrac{|P.E|}{2} =3.4 = K.E$

So, kinetic energy

$K.E = 3.4 \ eV$

Potential energy

$P.E =- 2\times |T.E| = -2\times (3.4) = -6.8 \ eV$

$\alpha$ -particle is emitted from a $_{88}^{226}Re$ nucleus. If the energy of $\alpha$ -particle is 4.662 MeV, then what is the total free energy in this decay?

$_{88}Ra^{226}$

$Z=88$

$A=226$

$E_{\alpha}=4.662MeV$

$E_{\alpha}=\dfrac{A-4}{A}Q$

$Q=\left(\dfrac{A-4}{A}\right)E_{\alpha}$

$Q=\left(\dfrac{226}{226-4}\right)4.662MeV$

$Q=\dfrac{226}{222}\times 4.662MeV$

$Q=4.746MeV$

The masses pf $^{11}C and ^{11}B$ are respectively $11.0114 u and 11.0093 u$ . find the maximum energy a positron can have in the $\beta^+$ decay of $^{11}C to ^{11} B$

Given,
mass of

${^{11}}C= 11.014 u ;$

mass of

${^{11}}B= 11.0093 u$

Energy liberated in the

$\beta ^{+}$ decay (Q) is given by,

$Q=[m({^{11}}C)-m({^{11}}B)-2m_e]c^{2}$

$=(11.0114u-11.0093u-2\times 0.0005486u)c^{2}$

$=0.0010028\times 931MeV$

$=0.9336MeV=933.6keV$

For maximum KE of the positron, energy of neutrino can be taken as zero.

$\therefore$ Maximum KE of the positron

$=933.6KeV$

How much energy is released in the following reaction :
$^7Li + p \rightarrow \alpha + \alpha$ tomic mass of $^Li = 7.0160 u$ and that of $^He = 4.0026 i=u$

$Li^7 + p \rightarrow \alpha +\alpha +E ;$

$Li^7= 7.016 u$

$\alpha = ^4He = 4.0026u ; p = 1.007276 u$

$E =Li^7 +P -2 \alpha = (7.016 +1.007276)u -( 2 \times 4.0026)u = 0.018076 u.$

$\rightarrow 0.018076 \times 931 = 16.828 = 16.83 MeV$

In the reaction $\mathrm{p}+^{19} \mathrm{F} \rightarrow \alpha+^{16} \mathrm{O},$ the masses are
$\begin{array}{ll}m(\mathrm{p})=1.007825 \mathrm{u}, & m(\alpha)=4.002603 \mathrm{u} \\m(\mathrm{F})=18.998405 \mathrm{u}, & m(\mathrm{O})=15.994915 \mathrm{u}\end{array}\\$
Calculate the $Q$ of the reaction from these data.

The mass change is

$\Delta M=(4.002603 \mathrm{u}+15.994915 \mathrm{u})-(1.007825 \mathrm{u}+18.998405 \mathrm{u})=-0.008712 \mathrm{u}\\$

Using Eq.

$37-50$ and Eq.

$37-46,$ this leads to

$Q=-\Delta M c^{2}=-(-0.008712 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u})=8.12 \mathrm{MeV}\\$

Which one of the following cannot emit radiation and why ? Excited nucleus, excited electron.

Excited electron cannot emit radiation. This is because energy of electronic energy levels is in the range of

$eV$ only not in

$MeV.$

$\gamma$ -radiation has energy in

$MeV.$

The nuclide $^{14} C$ contains (a) how many protons and (b) how many neutrons?

(a) 6 protons, since

$Z = 6$ for carbon
(b)8 neutrons, since

$A – Z = 14 – 6 = 8$ .

What is the binding energy per nucleon of the europium isotope $^{152}_{63} Eu$ ? Here are some atomic masses and the neutron mass.
$^{152} _{63}Eu$ 151.921 742 u $^1H$ 1.007 825 u
n 1.008 665 u

1711570_1780896_ans_6e747e33cadc47d1935f5223b7afe4d1.png

In the reactions given below:
(a) $^{11}_{6}C \rightarrow ^{z}_{y}B + x + v$
(b) $^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{a}Ne + ^{c}_{b}He$
Find the values of x, y, z and a, b, c.

$(a). x = \beta^{+} / ^{0}_{1}e,$ y = 5, z = 11 $$

$,(b) . x = 10, y = 2, z = 4$

$(c). a = 10, b = 2, c = 4$

A nucleus of mass number A, has a mass defect $\Delta m.$ Give the formula, for the binding energy per nucleon, of this nucleus.

BE per nucleon,

$B_n = \dfrac{Total \,binding \,energy}{Number \,of \,nucleons} = \dfrac{\Delta mc^2}{A}$

Where c is the speed of light in vaccum.

The mass of H atom is less than sum of the masses of a proton and electron. Why is this?

There is something called the strong nuclear force (strong interaction) that binds protons and neutrons (and other particles) together. This force overall reduces the energy of the system - energy released as heat. The binding energy (not of electrons to protons but of protons to neutrons) becomes greater as elements become heavier as you increase atomic mass until you reach iron, at which point it begins to decrease. This is why fusing hydrogen atoms together releases energy and splitting uranium does the same.

so from the einstein's equation the binding force is responsible,

Distinguish between isotopes and isobars. Give one example for each of the species.

$Isotopes$	$Isobars$
1. The nuclides having the same atomic number (Z) 2. But different atomic masses (A) are called isotopes. Example: $^{1}_{1}H, ^{2}_{1}H, ^{3}_{1}H$	1. The nuclides having the same atomic mass (A) but 2. Different atomic numbers (Z) are called isobars. Examples: $^{3}_{1}H, ^{3}_{2}He$

What other name is given to nuclear fusion? Give reason.

Nuclear fusion is also called as thermo nuclear reaction because nuclear fusion takes place at very high temperature.

Write the nuclear equations for:
${\beta}^{+}$ - decay of $_{ 43 }^{ 97 }{ Tc }$

$_{ 43 }^{ 97 }{ Tc }\rightarrow _{ 42 }^{ 97 }{ Mo }+_{ +1 }^{ 0 }{ e }+\overline { v }$

Write the nuclear equations for:
$\alpha$ - decay of $_{ 88 }^{ 226 }{ Ra }$

$\quad _{ 88 }^{ 226 }{ Ra }\rightarrow _{ 86 }^{ 222 }{ Rn }+_{ 2 }^{ 4 }{ He }$

Write the nuclear equations for:
${\beta}^{-}$ - decay of $_{ 83 }^{ 210 }{ Bi }$

$\quad _{ 83 }^{ 210 }{ Bi }\rightarrow _{ 84 }^{ 240 }{ Po }+_{ -1 }^{ 0 }{ e }+\overline { v }$

What is radioactivity?

The act of emitting radiation spontaneously, as its name implies, is known as radioactivity. An atomic nucleus that is unstable for some reason "wants" to give up some energy in order to shift to a more stable configuration achieves this.
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting ionizing radiation. As each nucleus disintegrates, it generates a charged particle with enough kinetic energy to penetrate solid matter.

Fill in the blanks.
The equation $E=mc^2$ is derived by the scientist _________.

Albert Einstein

Albert Einstein formulated the equation

$E=mc^{2}$ in 1905

Define binding energy of electron.

1311427_1138860_ans_836a4abe32784cd78d98ce7e515b818f.jpg

Four physical quantities are given in Column I and their order the values in Column II. Match the options appropriately :

A. Thermal energy for air molecule at room temperature = 0.02eV

B. Binding energy of heavy nuclei per nucleon = 7 MeV

C. X-ray photon energy = 10KeV.

D. Photon energy of visible light = 2 eV

When the number of nucleons in nucleus increase, the binding energy per nucleon first _ and then _ with increase of mass number.

When the number of in nucleus increases, the binding energy per nucleon first increases because high binding energy is required to separate a stable nucleus into its constituent protons and neutrons. When the mass number is increasing the binding energy per nucleon gradually decrease. So there is misufficient binding energy to hold the nucleus together.

The mass defect per nucleon is called ___________.

The mass defect per nucleon

$\dfrac{\Delta m}{A}=P$ , is called the packing fraction of the nucleus.

For greater stability, a nucleus should have greater value of binding energy per nucleon. Why?

Binding energy is the energy that holds a nucleus together and is equal to the mass defect of the nucleus. For stable nucleus, we can see that native prefers even number of protons and an even number of neutrons. The odd combination of protons and neutrons in the nucleus are very less and are only found in lighter elements. For elements with large number of protons in the nucleus, the coulomb electrostatic force of repulsion becomes significant and the number of neutrons must be greater to compensate this repulsion effect.

Therefore the basic point is that the binding energy of a nucleon is directly proportional to the stability of the nucleus,. Since stability is what determines whether the nucleus will decay, which ultimately is affected by the binding energy.

State any two differences between chemical reactions and nuclear reactions.

Nuclear reactions involve a change in an atom's nucleus, usually producing a different element. Chemical reactions on the other hand, involve only a rearrangement of electrons and do not involve changes in the nuclei.

Fill in the Blanks
In nuclear reactions we have the conservation of ___________.

Energy

Define specific binding energy.

The specific binding energy is defined as the binding energy per nucleon. The highest specific binding energy is

$8.8 MeV$ , and occurs for nickel.

A $\mu-meson$ (charge $-e$ , mass $=207m$ , where $m$ is the mass of the electron) can be captured by a proton to form a hydrogen-like "mesic" atom. Calculate the radius of the first Bohr orbit, the binding energy, and the wavelength of the line in the Lyman series for such an atom. The mass of the proton is $1836$ times the mass of the electron. The radius of the first Bohr orbit and the binding energy of Hydrogen are $0.529\mathring{A}$ and $13.6\space eV$ , respectively. Take $R = 109678\space cm^{-1}$

Bohrs radius is given by

$r=\dfrac { { n }^{ 2 }{ h }^{ 2 } }{ 4{ \pi }^{ 2 }mkz{ e }^{ 2 } } \\ e=-1.6\times { 10 }^{ 19 }C\\ n=1\\ h=6.63\times { 10 }^{ -34 }㎡㎏/s\\ m=207\times 9.11\times { 10 }^{ -31 }kg\\ z=1\\ g=9\times { 10 }^{ 9 }\\ { r }_{ 1 }=\dfrac { { \left( 1 \right) }^{ 2 }{ \left( 6.63\times { 10 }^{ -34 } \right) }^{ 2 }\times 7\times 7 }{ 4\times 22\times 22\times 207\times 9.11\times { 10 }^{ -31 }\times 9\times { 10 }^{ 9 }\times 1\times 1.6\times { 10 }^{ 19 }\times 1.6\times { 10 }^{ 19 } } \\ =2.55\times { 10 }^{ -13 }m\\ =0.00255\quad$

Binding energy is the energy of electron in first order is given by

${ E }_{ 1 }=\dfrac { 2{ \pi }^{ 2 }m{ k }^{ 2 }{ z }^{ 2 }{ e }^{ 4 } }{ { n }^{ 2 }{ h }^{ 2 } } \\$

For

$n=1\\ m=207\times 9.11\times { 10 }^{ -31 }kg\\ z=1$

For Hydrogen Binding energy is

${ E }_{ 1 }^{ ' }=\dfrac { -2{ \pi }^{ 2 }m{ k }^{ 2 }{ z }^{ 2 }{ e }^{ 4 } }{ { n }^{ 2 }{ h }^{ 2 } } =13.6ev$

${ E }_{ 1 }=207\times 13.6\\ =2.8keV$

For lyman series for hydrogen atom

$R=\dfrac { 2{ \pi }^{ 2 }m{ k }^{ 2 }{ e }^{ 4 } }{ c{ h }^{ 3 } } =109678{ ㎝ }^{ -1 }=10967800{ m }^{ -1 }\\ { R }^{ ' }=\dfrac { 2{ \pi }^{ 2 }{ m }_{ c }{ k }^{ 2 }{ e }^{ 4 } }{ c{ h }^{ 3 } } =207\times R\\ =207\times 10967800\\ =22703346{ m }^{ -1 }$

Radius of First Bohrs orbit=0.00255

Binding Energy =2.8keV

Wavelength of Lymon =

$={ R }^{ ' }\left[ \dfrac { 1 }{ { 1 }^{ 2 } } -\dfrac { 1 }{ { n }^{ 2 } } \right]$

In Column I consider each process just before and just after it occurs. Initial system is isolated from all other bodies. Consider all product particles (even those having rest mass zero) in the system. Match the system in Column I with the result they produce in Column II

$A$

$_{ 92 }^{ 238 }{ U }\rightarrow _{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }$

#Proton is conserved.

#Momentum is always conserved in absence of external force.

#Since reaction is spontaneous so mass is converted into energy.

#Charge is always conserved

So,

$(A)\rightarrow(2),(3),(4)$

$B$

$_{ 92 }^{ 238 }{ U }\rightarrow _{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }$

#Proton is conserved.

#Momentum is always conserved in absence of external force.

#Since reaction is spontaneous so mass is converted into energy.

#Charge is always conserved

So,

$(B)\rightarrow(2),(3),(4)$

$C$

$_{ 92 }^{ 238 }{ U }\rightarrow _{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }$

#Proton is conserved.

#Momentum is always conserved in absence of external force.

#Since reaction is spontaneous so mass is converted into energy.

#Charge is always conserved

So,

$(C)\rightarrow(2),(3),(4)$

$D$ In

$\beta^-$ decay proton increases by +1

Here, other atom is not given so

$(2)$ will not be the answer.

So,

$(D)\rightarrow(1),(3),(4)$

Match column I with column II

$A$ Stability of nucleus

$\alpha \; \cfrac{1}{Neutron-Proton\;ratio} \\ \alpha \;packing \; fraction \\ \alpha \;Mass \; defect$

So,

$(A)\rightarrow(3),(4),(5)$

$B$ Any substance decays when Binding energy decrease by a certain value.

So,

$(B)\rightarrow(2)$

$C$ Negative energy is directly proportional to stability of substance

So,

$(C)\rightarrow(1)$

$D$ Stopping potential is always negative.

So,

$(D)\rightarrow(1)$

In Column I some of the nuclear reactions are given. Match this with the energy involved in these reactions in Column II :

Energy released =

$\triangle m{ c }^{ 2 }$

$A$

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 1 }^{ 3 }{ H }+_{ 1 }^{ 1 }{ H }+{ E }_{ 1 }\\ \triangle m=2m(_{ 1 }^{ 2 }{ H })-m(_{ 1 }^{ 3 }{ H })-m(_{ 1 }^{ 1 }{ H })\\ =2\times 2.014-3.016-1.008\\ =0.004amu\\ { E }_{ 1 }=\triangle m{ c }^{ 2 }=0.004\times 931=3.7MeV\approx 4MeV$

So,

$(A)\rightarrow (3)$

$B$

$_{ 1 }^{ 3 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ H }+{ E }_{ 2 }\\ \triangle m=m(_{ 1 }^{ 3 }{ H })+m(_{ 1 }^{ 2 }{ H })-m(_{ 2 }^{ 4 }{ He })-m(_{ 0 }^{ 1 }{ H })\\ =3.016+2.014-4.002-1.008=0.02amu\\ { E }_{ 2 }=\triangle m{ c }^{ 2 }=0.02\times 931=18.62MeV$

So,

$(B)\rightarrow (2)$

$C$

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 2 }^{ 3 }{ He }+_{ 0 }^{ 1 }{ H }+{ E }_{ 3 }\\ \triangle m=2m(_{ 1 }^{ 2 }{ H })-m(_{ 2 }^{ 3 }{ He })-m(_{ 0 }^{ 1 }{ H })\\ =2\times 2.014-3.016-1.008\\ =0.004amu\\ { E }_{ 3 }=0.004\times 931=3.3MeV$

So,

$(C)\rightarrow (1)$

$D$

$_{ 2 }^{ 3 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 1 }^{ 1 }{ H }+{ E }_{ 4 }\\ \triangle m=m(_{ 2 }^{ 3 }{ H })+m(_{ 1 }^{ 2 }{ H })-m(_{ 2 }^{ 4 }{ He })-m(_{ 1 }^{ 1 }{ H })\\ =3.016+2.014-4.002-1.008\\ =0.02amu\\ { E }_{ 4 }=\triangle m{ c }^{ 2 }=0.02\times 931=18.62MeV$

So,

$(D)\rightarrow (2)$

Draw a plot of BE/A versus mass number A for 2 $\leq$ A $\leq$ Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei.

From the plot, it can be clearly seen that after an atomic mass of nearly 56, which corresponds to Iron, the binding energy starts decreasing (i.e. Iron is the most tightly bound nucleus). Before Iron the binding energy is rapidly increasing mostly. Thus in general we can say that if two light nuclei fuse to form a heavy nuclei, they will release energy and hence will reach a more stable state.

Write the relation for binding energy (B.E) (in MeV) of a nucleus of mass $\displaystyle _{Z}^{A}{M}$ , atomic number (Z) and mass number (A) in terms of the masses of its constituents - neutrons and protons.

The binding energy in (MeV) is

$E_B=(Zm_p+Nm_n-M_A^Z)c^2$
Where

$m_p=$ mass of proton ,

$m_n=$ mass of neutron,

$c=$ velocity of light.
Now

$N=A-Z$ thus,

$E_B=[Zm_p+(A-Z)m_n-M_A^Z]c^2=[Z(m_p-m_n)+Am_n-M_A^Z)c^2$

The fission properties of $^{239}_{94}Pu$ are very similar to those of $^{235}_{92}U$ . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure $^{239}_{94}Pu$ undergo fission?

Number of atoms in 1 kg of pure Pu=

$\dfrac{6.023\times 10^{23}}{239}\times 1000=2.52\times 10^{24}$

As average energy released in fission is

$180 MeV$ , the total energy released is

$\Rightarrow2.52\times 10^{24}\times 180MeV$

$\Rightarrow4.53\times 10^{26}MeV$ .

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :
$^2_1H\, +\, ^2_1\, \rightarrow\, ^3_2He\, +\, n\, +\, 3.27\, MeV$ .

Number of atoms in 2 kg deuterium =

$\dfrac{6.023\times 10^{23}\times 2000}{2}=6.023\times 10^{26}$

Energy released when 2 atoms fuse =

$3.27MeV$

Thus, total energy released =

$\dfrac{3.27}{2}\times 6.023\times 10^{26}MeV$

$=15.75\times 10^{13}J$ .

Energy consumed by bulb per second=

$100J$ .

Thus, time for which the bulb glows =

$\dfrac{15.75\times 10^{13}}{100}s$

$=15.75\times 10^{11}s=4.99\times 10^7 years$ .

(i) What characteristic property of nuclear force explains the constancy of binding energy per nucleon $\left({BE}/{A}\right)$ in the range of mass number $'A'$ lying $30 < A < 170$ ?
(ii) Show that the density of nucleus over a wide range of nuclei is constant-independent of mass number $A$ .

(i)

The approximate constancy of

$\dfrac{BE}{A}$ over most of the range is saturation property of nuclear force.

In heavy nuclei, nuclear size > range of the nuclear force.

So a nuclear sense approximately a constant number of neighbors and thus the nuclear

$\dfrac{BE}{A}$ levels off at high A. This saturation of the nuclear force.

(ii)

The mass of the nucleus of the atom of the mass number A

A= A a.m.u.

= A

$\times 1.660565 \times 10^{-27} kg$

The volume of the nucleus is

$\dfrac{4 \pi r^{3}}{3}= \dfrac {4 \pi}{3} (R_0 A^{-3})^{3}= \dfrac {4 \pi R_0^{3} A}{3}$

$\therefore Volume = \dfrac {4 \pi}{3}(1.1 \times 10^{-15})^{3} \times A m^{3}$

Now the density of the nucleus=

$\dfrac {m}{ V}$ ; where m is mass and V is the volume of the nucleus.

$\rho = \dfrac {A \times 1.660565 \times 10^{-27}}{\dfrac{4 \pi}{3}(1.1 \times 10^{-15})^{3} \times A}= 2.97 \times 10^{17} {Kg} m^{-3}$

Thus, we can see the density of the nuclei is independent of mass number and it is constant for all nuclei .

Write symbolically the nuclear $\beta^+$ decay process of $_{6}^{11}C$ . Is the decayed product X an isotope or isobar of ( $_{6}^{11}C$ )? Given the mass values m ( $_{6}^{11}C$ ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

$\beta^+$ decay:

$_{116}C \rightarrow _{115} X+e^+ + \nu$

$e^+=$ Positron

$\nu=$ Neutrino

As mass number of

$C$ and

$X$ are the same,

$C$ and

$X$ are isobars.

Energy released

$Q=(m_c-m_x-m_e)c^2$

$=[(m_c+116m_e)-(m_x+115m_e)]c^2$

$=[m_c-m_x]c^2$

$=[11.011434-11.009305]c^2$

We have

$c^2=931.5 MeV/amu$

$=0.002129 \times 931.5 MeV$

$=1.983$

$MeV$

The mass of deutron $(_1H^2)$ nucleus is $2.013553$ a.m.u. If the masses of proton and neutron are $1.007275$ a.m.u. and $1.008665$ a.m.u. respectively. Calculate the mass defect, the packing fraction, binding energy and binding energy per nucleon.

Mass defect (

$\triangle m$ )

$\\=mass\;of\;nucleus-(mass\;of\;neutron+mass\;of\;proton)$

$\triangle m=2.013553-(1.007275+1.008665) \\=-0.002387amu$

Binding Energy = Mass defect X 931 MeV

$=-0.002387\times 931\\=-2.222MeV$

Binding Energy per nucleon

$=-1.111MeV$

Obtain the binding energy of the nuclei $_{26}Fe^{56}$ and $_{83}Bi^{209}$ in units of MeV from the following data: mass of hydrogen atom $=1.007825$ a.m.u., mass of neutron $=1.08665$ a.m.u.; mass of $_{26}Fe^{56}$ atom $=55.934939$ a.m.u. and mass of $_{83}Bi^{209}$ atom $=208.980388$ a.m.u. Also calculate binding energy per nucleon in the two cases.

(i)

$_{26}Fe^{56}$ nucleus contains 26 protons .

Number of neutrons

$=(56-26) = 30$ neutrons

Now,

Mass of 26 protons

$= 26$ X

$1.007825 = 26.20345 u$

Mass of 30 neutrons

$= 30$ X

$1.008665= 30.25995 u$

Total mass of 56 nucleons

$= 56.46340 u$

Mass of

$_{26}Fe{56}$ nucleus

$= 55.934939 u$

Therefore,

Mass defect,

$\triangle m$

$= 56.46340-55.934939 = 0.528461 u$

Total Binding Energy

$= 0.528461$ X

$931.5 Mev = 492.26 MeV$

Average binding energy per nucleon

$= \frac{492.26}{56}=8.790 MeV$

(ii)

$_{83}Bi^{209}$ nucleus contains 83 protons

Number of neutrons

$=209-83=126 neutrons$

Now,

Mass of 83 protons

$= 83$ X

$1.007825= 83.649475 amu$

Mass of 126 neutrons

$= 126 X 1.008665= 127.091790 amu$

Therefore, total mass of nucleons

$= 83.649475+127.091790=210.741260amu$

Given, mass of nucleus

$=208.980388 amu$

Now, mass defect

$\triangle m$

$=210.741260-208.980388=1.760872$

Total binding energy

$=1.760872$ X

$931.5 = 1640.26 MeV$

Therefore, average binding energy per nucleon =

$\frac{1640.26}{209}=7.848 Mev$

Find out the binding energy per nucleon of an $\alpha$ -particle in MeV. It is given that the masses of $\alpha$ -particles, proton and neutron are respectively $4.00150$ a.m.u., $1.00728$ a.m.u. and $1.00867$ a.m.u.

Binding Energy

$E=\triangle m{ c }^{ 2 }$

$\triangle m=mass\;of\;\alpha\;particle-(mass\;of\;proton+mass\;of\;neutron) \\ \triangle m=4.00150-2(1.00728+1.00867) \\=4.00150-4.0319\\=-0.0304amu$

$E=\triangle m{ c }^{ 2 }$

$E=-0.0304\times 931 \\E=28.30MeV$

(a) Draw the plot of blinding energy per nucleon (BE/A)
as a function of mass number. Write two important conclusion that can be drawn regarding the nature of nuclear source.
(b) Use this graph to explain the release of energy in both the process of nuclear fusion and fission.
(c) Write the basic nuclear process of neutron undergoing $\beta$ -decay. why is the detection of neutrinos found very difficult?

(a)Conclusions drawn from the graph:

(i) The nuclear force is very attractive and strong to produce binding energy of the order of MeV per nucleon.

(ii)The nuclear force is a short range force which can be shown by its value being constant in the range 30<A<170.

(b)In nuclear fission a heavier nucleus breaks into lighter nuclei to release high binding energy per nucleon. In nuclear fusion, the binding energy of the fused heavier nucleon is more than the binding energy of the lighter nuclei which combined to form the heavier nuclei. Energy is also released in nuclear fusion.

(c)the basic nuclear process of neutron undergoing

$\beta$ decay is:

$n\rightarrow p+\vec e+\vec \nu$

Neutrinos have high penetrating power which makes their detection difficult.

The binding energy per nucleon of $_{2}He^{4}$ is $7.075$ MeV. What is its total binding energy?

Binding energy per nucleon

$=7.075MeV$

No. of nucleon

$=4$

So, Total Binding Energy

$=7.075\times 4 =28.3MeV$

_____________ is the energy released during nuclear reactions.

Kinetic energy

In a nuclear reaction, the total energy is conserved. The "missing " rest mass must therefore reappear as kinetic energy released in the reaction, its source is the nuclear binding energy.

The mass of $_{3}Li^{7}$ nucleus is $0.042$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

We have Mass number = 7 and Atomic number= 3

If mass,

$m = 1 u$ and coulomb force

$c= 3$ X

$10^8 ms^{-1}$ , then

$E=931 MeV$ that is,

$1u=931 MeV$

Binding Energy =

$0.042$ X

$931=39.10 MeV$

Therefore, BInding Energy per Nucleon

$= \dfrac{ Binding Energy}{Mass Number}=\dfrac{39.10}{7}= 5.58= 5.6MeV$

Calculate the (i) mass defect, (ii) binding energy and (iii) the binding energy per nucleon for a $_{6}C^{12}$ nucleus. Nuclear mass of $_6C^{12}=12.000000$ a.m.u., mass of hydrogen nucleus $=1.007825$ a.m.u. and mass of neutron $=1.008665$ a.m.u.

Given,

Mass of one proton = 1.007825 a.m.u

Mass of one neutron= 1.008665 a.m.u

Nuclear mass of

$_6C^{12}$ = 12 a.m.u

(i)

$_6C^{12}$ has 6 proton, 6 electron and 6 neutron.

Mass of nucleus = Mass of 6 proton + Mass of 6 neutron

$=(6 \times 1.007825) + (6 \times 1.008665)$

$=12.09894\, u$

Mass defect

$(\Delta m)$

$=12.09849u-12u=0.098931 \, u$

(ii) Binding Energy = Mass Defect

$\times$

$931.5 \, MeV$

$=0.09849u \times 931.5 MeV/u=92.15\, MeV$

(iii) Binding Energy per Nucleon= Binding Energy / Number of Nucleons

Number of nucleon = Mass Number = 12

So, Binding Energy per nucleon

$=\frac{92.15}{12}=7.68 \, MeV$

The binding energy per nucleon for $C^{12}$ is $7.68$ MeV and that for $C^{13}$ is $7.47$ MeV. What is the energy required to remove a neutron from $C^{13}$ ?

Hint:- Binding energy is defined as the minimum energy required to separate the protons and neutrons of an atomic nucleus. It is equivalent to the mass defect.

Step 1: Note the given values

Initial binding energy per nucleon when an atom is

$C^{12}=7.68Mev$

Final binding energy per nucleon when atom is

$C^{13}=7.47Mev$

Step 2: Calculate energy required to remove a neutron

$C^{13} + energy \rightarrow C^{12}+n$

So energy released or required to remove a neutron will be difference of initial and final energy.

$E=E_i-E_f$

= 13 × 7.47 −12×7.68

= 97.11 −92.16

= 4.95 MeV

With the help of one example, explain how the neutron to proton ratio changes during alpha decay of a nucleus.

Alpha decay involves the ejection of helium nucleus which is just two protons and two neutrons. This means that the atomic mass decreases by 4 and atomic number decreases by 2. Thia decay us typical radioisotopes which have nuclei that ate too large.

Explain nuclear fusion with an equation.

Nuclear fusion is a reaction in which two or more nuclei come closer enough to form one or more different atomic nuclei and subatomic particles.

$_{ 1 }^{ 3 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+_{ 0 }^{ 1 }{ n }+Energy$

The mass of deutron $(_1H^2)$ nucleus is $2.013553$ a.m.u. If the masses of proton and neutron are $1.007275$ a.m.u. and $1.008665$ a.m.u. respectively, calculate the mass defect, the packing fraction , binding energy and binding energu per nucleon.

Mass defect

$=$ Mass of atom

$-$ Mass of neutron and proton

Mass defect

$=2.013553-(1.007275+1.008665) \\=0.002387$

Binding Energy

$=0.002387\times 931=2.22MeV$

Binding Energy per nucleon

$=\cfrac{2.22}{2}=1.11MeV$

Obtain the binding energy of a nitrogen nucleus $(_7N^{14})$ from the following data: $m_H=1.00783$ a.m.u., $m_n=1.0087$ a.m.u.; $m_N=14.00307$ a.m.u.

$m_{H}$ = 1.00783 amu

$m_{n}$ = 1.0087 amu

$m_{N}$ = 14.0030u amu

$_{7}N^{14}$ , these are 7 protons and 7 neutrons.

$\therefore$ mass defect,

$\Delta =( 7m_{H}+7m_{n}) - m_{N}$

=(7×1.00783)+(7×1.00807)-14.00307=0.11243 amu

Binding energy of nitrogen nucleus

$\Delta$ ×931 MeV =0.11243×931

=104.67

What is the significance of binding energy per nucleon of a nucleus?

Binding energy per nucleon: It is average energy required to remove a nucleon from the nucleus to infinite distance. Higher the average (binding energy / nucleon), greater is the stability of the nucleus.

In our Nature, where is the nuclear fusion reaction taking place continuously?

In our nature, sun is main star which generates its energy by nuclear fusion of hydrogen nuclei into helium. Fusion of hydrogen atoms at sun occurs at sun at a solar core temperature of 14 million kelvin.

What is the signeficance of binding energy per nucleon of a nucleus of a radioactive element?

Binding energy per nucleon basically tells the stability of a nucleus. Larger the binding energy per nucleon, the greater the work that must be done to remove the nucleon from the nucleus, the more stable the nucleus.

Write one balanced reaction representing nuclear fusion.

Nuclear fusion reaction:

$_1H^2 + _1H^2 \rightarrow _1H^3 + _1H^1 + 4.0 MeV$ energy

$_1H^3 + _1H^2 \rightarrow _2He^4 + _0n^1 + 17.6 MeV$ energy

Which reaction is responsible for solar energy? Name the major component of solar energy that reaches us.

Thermo-nuclear fusion (Proton-proton chain ) reaction is responsible for solar energy.

The major component of solar energy that reaches us is infra-red radiation (heat radiation)

Define mass defect of a nucleus. Binding energy of ${ _{ 8 }^{ }{ O } }^{ 16 }$ is $127.5MeV$ . Write the value of its 'binding energy per nucleon'. Write the value of $1eV$ energy in joule.

Mass defect is the amount by which the mass of an atomic nucleus differs from the sum of the masses of its constituent particles, being the mass equivalent of the energy released in the formation of the nucleus.

Nuclear binding energy per nucleon for the given atom=

$\dfrac{BE}{A}$

$=\dfrac{127.5MeV}{16}$

$=7.96MeV$

Value of

$1eV$ is equal to

$1.6\times 10^{-19}J$ .

The atomic mass of Uranium $^{238}_{92}U$ is 238-0508 u, while that of Thorium $^{234}_{90}Th$ is 234.0436 u, and that of Helium $^4_2He$ is 4.0026 u. Alpha decay converts $^{238}_{92}U$ into $^{234}_{90}Th$ as shown below:
$^{238}_{92}U \rightarrow ^{234}_{90}Th + ^4_2He + energy$
Determine the energy released in this reaction.

Atomic mas of Uranium

$=$ 238.0508 u
Atomic mass of Thorium

$=$ 234.0436 u
Atomic mass of Helium

$=$ 4.0026 u
Mass of reactants

$=$ 238.0508 u
Mass of products

$=$ 234.0436

$+$ 4.0026
----------------------
238.0462 u
---------------------

$\Delta m = 238.0508$

$- 238.0462$
------------------
0.0046 u
------------------
Energy Released

$= 0.0046 \times 931$

$= 4.28 Mev$

In a certain star, three alpha particles undergo fusion in a single reaction to form $^{12}_{6}C$ nucleus. Calculate the energy released in this reaction in $MeV$ .
Given: $m(^{4}_{2}He) = 4.002604\ u$ and $m(^{12}_{6}C) = 12.000000\ u$

Mass of

$3$ alpha particles

$= 3\times 4.002604\ u$

$= 12.007812\ u$ .
Mass of 1

$_{6}C^{12} atom = 12.007812\ u$
Mass Defect

$= \triangle m = 12.007812 - 12.000000 = 0.007812$
Energy released

$= 0.007812\times 931.5 = 7.2768\ MeV$ .

(a) What do you mean by $Q$ value of a nuclear reaction?
(b) Write down the expression for $Q$ value in the class of $\alpha$ decay.

a) The Q-value of the reaction is defined as the difference between the sum of the masses of the initial reactants and the sum of the masses of the final products, in energy units (usually in MeV). It is the measure of amount of energy released by that reaction.
b) The Q-value of the decay,

$Q_{\alpha}$ is the difference of the mass of the parent (

$m_P$ ) and the combined mass of the daughter ((

$m_P$ ) and the

$\alpha$ -particle (

$m_{\alpha}$ ), multiplied by

$c^2$ :

$Q_{\alpha} = (m_P - m_D - m_{\alpha})c^2$ .

What are isotopes and isobars?

$Isobars$ : Isobars are atoms of different elements having the same atomic mass but different atomic number. e.g. -

$^{40}S, \ ^{40}Cl, \ ^{40}Ar$ .

$Isotopes$ : Isotopes are atoms with the same number of protons but that have a different number of neutrons. e.g. - Carbon -14, Iodine - 131 etc.

Write any five differences between Nuclear Fission and Nuclear Fusion.

Nuclear Fission	Nuclear Fusion
Fission is the splitting of a large atom into two or more smaller ones.	Fusion is the fusing of two or more lighter atoms into a larger one.
Fission reaction does not normally occur in nature.	Fusion occurs in stars, such as the sun.
Fission produces many highly radioactive particles.	Few radioactive particles are produced by fusion reaction, but if a fission "trigger" is used, radioactive particles will result from that.
Critical mass of the substance and high-speed neutrons are required.	High density, high temperature environment is required.
Takes little energy to split two atoms in a fission reaction	Extremely high energy is required to bring two or more protons close enough that nuclear forces overcome their electrostatic repulsion.

Calculate the binding energy and binding energy per nucleon (in MeV) of a nitrogen nucleus $(^{14}_7N)$ from the following data :
Mass of proton = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen nucleus = 14.00307 u

Nuclear reaction :

$7p+7n\rightarrow ^{14}_7 N$

Mass defect

$\Delta M = 7m_p + 7m_n - m_{nucleus}$

$\therefore$

$\Delta M = 7(1.00783)+7(1.00867) - 14.00307$ amu

$\implies$

$\Delta M = 0.11243$ amu

Thus binding energy

$E = \Delta M \times 931.5$ MeV

$\therefore$

$E = 0.11243\times 931.5 = 104.73$ MeV

Binding energy per nucleon

$= \dfrac{104.73}{14} = 7.48$ MeV

Calculate mass defect and binding energy per nucleon of $^{20}_{10}Ne,$ given
Mass of $^{20}_{10}Ne=19.992397$ u
Mass of $^1_1H=1.007825$ u
Mass of $^1_0n=1.008665$ u.

The required nuclear reaction is :

$10 \ ^1_0n + 10 \ ^1_1H \ \rightarrow \ ^{20}_{10}Ne$

Mass defect

$\Delta M =10 (m)_{^1_0n} + 10 (m)_{^1_1H} -(m)_{ ^{20}_{10} Ne}$

$\therefore$

$\Delta M =10 (1.008665) + 10 (1.007825) - 19.992397 = 0.172503 \ u$

Binding energy

$E= \Delta M \times 931.5 \ MeV$

$\therefore$

$E = 0.172503 \times 931.5 = 160.69 \ MeV$

Binding energy per nucleon

$u = \dfrac{E}{20} = \dfrac{160.69}{20} = 8.03 \ MeV$

What is the name given to the atoms whose nuclei contain the same number of proton but different number of neutrons?

Two atoms are said to be isotopes if their nucleii have same number of protons but different number of neutrons.

If speed of light is doubled then how much binding energy of nucleus will become?

Let the mass defect of nucleus be

$\Delta M$ and speed of light initially be

$c$ .
Binding energy of nucleus

$E_B = \Delta M\times c^2$

$\implies \ E_B\propto c^2$
Given :

$c' = 2c$

$\implies \ \dfrac{E'_B}{E_B} = \dfrac{(c')^2}{c^2} = \dfrac{(2c)^2}{c^2} = 4$

$\implies \ E'_B = 4E_B$
Thus binding energy increases by

$4$ times.

The three stable isotopes of neon: $_{ 10 }^{ 20 }{ Ne },_{ 10 }^{ 21 }{ Ne }$ and $_{ 10 }^{ 22 }{ Ne }$ have respective abundances of $90.51$ %, $0.27$ % and $9.22$ %. The atomic masses of the three isotopes are $19.99u,20.99u$ and $21.99u$ respectively. Obtain the average atomic mass of neon.

$A=\cfrac { 90.51\times 19.99+0.27\times 20.99+9.22\times 21.99 }{ 100 } =20.18mu$

What is meant by binding energy? If the masses of proton, neutron and alpha $(\alpha)$ particles are $1.00728\ amu, 1.00867\ amu$ and $4.00150\ amu$ respectively, then find the binding energy per nucleon of alpha particle. $[1\ amu = 931\ MeV]$ .

Binding energy of nucleus is defined as the energy that holds the components of nucleus together or is also defined as the amount of energy required to break the nucleus completely into its constituent particles.
Mathematically, binding energy is equal to the product of mass defect of nucleus and the square of speed of light.
Binding energy

$E_B = \Delta M \times c^2 = \Delta M \ (in \ amu )\times 931 \ MeV$
Alpha particle has 2 protons and 2 neutrons. So alpha particle has 4 nucleons.
Mass defect of alpha particle

$\Delta M = 2m_p + 2m_n - m_{He} = 2(1.00728)+2(1.00867) - 4.00150 = 0.0304 \ MeV$
Thus binding energy

$E_B = \Delta M\times 931 = 0.0304\times 931 = 28.3 \ MeV$
Binding energy per nucleon

$= \dfrac{E_B}{4} = \dfrac{28.3}{4} = 7.0756 \ MeV$

The binding energies per nucleon for deuterium and helium are $1.1$ MeV and $7.0$ MeV respectively. How much amount energy, in joules, will be liberated when $10^6$ deuterons take part in the reaction.

If two deuterium nuclei react to form a single helium nucleus

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }\rightarrow _{ 2 }^{ 4 }{ H }+Q$

Energy released

$Q=4\times 7\times -2\times 2\times 1.1=23.6MeV$

${ 10 }^{ 6 }$ deuterium atom will have

$\frac { { 10 }^{ 6 } }{ 2 }$ helium atom as a result.

Total energy liberated is

$E=\frac { { 10 }^{ 6 } }{ 2 } \times 23.6\\ =11.8\times { 10 }^{ 12 }eV\\ 1eV=1.6\times { 10 }^{ -19 }J\\ 11.8\times { 10 }^{ 12 }eV=11.8\times { 10 }^{ 12 }\times 1.6\times { 10 }^{ -19 }J\\ =1.89\times { 10 }^{ -6 }J$

(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number $A$ .
(b) A radioactive isotope has a half-life of $10$ years. How long will it take for the activity to reduce to $3.125$ %?

A. The value of binding energy per nucleon gives a measure of the stability of that nucleus greater is the binding energy per nucleon of a nucleus , more stable is the nucleus.

The above graph shown, binding energy per nucleon drawn against mass number A.

The saturation effect of nuclear forces is the property responsible for the approximate constancy of binding energy in range 30<A<17030<A<170

It is clear from the curve that binding energy per nucleon of the fused nucle is more than there of the light nuclei taking part nuclear fusion .

Hence energy gets released in the process.

In a nuclear fission , the sum of the masses of the final products is less than the sum of the masses of the reactant component

$N = N_{o} e ^{- \lambda t}$

But

$N = \dfrac{N_{o}}{32}$

$\dfrac{N_{o}}{32} = N_{o} e^{-\lambda t}$

$2^{-5} = e^{-\lambda t}$

5 ln2 = \lambda t$$

but given half life so

$\lambda = \dfrac{ln2}{10}$

so t = 50 Years

Four nuclei of an element undergo fusion to form a heavier nucleus, with release of energy. Which of the two - the parent or the daughter nucleus - would have higher binding energy per nucleon?

Since the unstable parent nuclei fuse to form a heavier stable daughter nuclei in a nuclear fusion reaction releasing some energy. So, daughter nuclei is more stable than parent nuclei. Thus daughter nucleus has more binding energy per nucleon than parent nucleus.

What is meant by the terms half-life of a radioactive substance and binding energy of a nucleus?

half life of a radioactive substance is defined as the time by which half of the initially presented radioactive substance has decayed.
Half-life is calculated by

$T_{1/2} = \dfrac{\ln 2}{\lambda}$
Binding energy of nucleus is defined as the minimum amount of energy required to break the nucleus into its constituent neutrons and protons.
Binding energy is calculated by

$E_B = \Delta M \ c^2$
where

$\Delta M$ is the mass defect of nucleus.

The binding energy per nucleon for $_{6}C^{12}$ is $7.68\ MeV$ and that for $_{6}C_{13}$ is $7.47\ MeV$ . Calculate the energy required to remove a neutron from $_{6}C_{13}$ .

Any atom has a nucleus around which electrons are revolving.

Every nucleon consists of at least one proton. Many nuclei have some neutrons in them along with protons.

Protons and neutrons are together called as nucleons.

Binding energy (BE) is the amount of energy required break the nucleus and separate all the nucleons.

BE per nucleon is given by

$BE/A= \cfrac{BE}{A}$ Here

$A$ is the number of nucleons in the atom.

$_6C_{13}$ means that there are 6 protons, 13 nucleons and 13-6=7 neutrons

Energy required to remove one neutron will be equal to BE per nucleon=7.47MeV

Write one balanced equation each to show.
(i) Nuclear fission
(ii) Nuclear fusion
(iii) Emission of $\beta^-$ (i.e., a negative beta particle).

Balanced Nuclear fission reaction :

$^1_0n \ + \ ^{235}_{92}U \ \rightarrow \ ^{144}_{56}Ba \ + \ ^{89}_{36}Kr \ + \ 3 \ ^1_0n$
Balanced nuclear fusion reaction :

$^2_1H \ + \ ^3H_1 \ \rightarrow \ ^4_2He \ + \ ^1_0n$
Balanced

$\beta^{-}$ emission reaction :

$^{14}_6C \ \rightarrow \ ^{14}_7N \ + \ ^0_{-1}e \ + \ \bar{\nu}$
where

$^{0}_{-1}e$ is the negative beta particle and

$\bar{\nu}$ is the anti-neutrino.

Name the isotope of Uranium which is easily fissionable.

Out of two isotopes of Uranium

$_{92}U^{235}$ and

$_{92}U^{238}$ given,

$_{92}U^{235}$ is easily fissionable.

Write a nuclear reaction when Uranium $238$ emits an alpha particle to form a Thorium(Th) nucleus.

Nuclear reaction of Uranium-238 emitting an alpha particle to form Thorium is given by :

$_{92}U^{238} \ \rightarrow \ \ _{90}Th^{234} \ + \ _2He^4$
where

$_2He^4$ is an alpha particle.

Differentiate between half life and average life of a radioactive substance.

The difference between half life and average life is given below :-

Half life	Average life
(i) The time required for half of the radioactive substance to decay is called as half life of that substance. (ii) It is denoted by $t _{\frac{1}{2}}$ (iii) $t_{\frac{1}{2}} =$ half life $= \left(\dfrac{\ln 2}{\lambda} \right)$ where $\lambda$ is decay constant.	(i) The expected life for the radioactive substance within a sample is called as average life of that substance. (ii) It is denoted by $\tau$ . (iii) $\tau =$ average life $= \dfrac{1}{\lambda}$

Write the nuclear equations for:
$\alpha$ - decay of $_{ 92 }^{ 242 }{ Pu }$

$_{ 92 }^{ 242 }{ Pu }\rightarrow _{ 92 }^{ 238 }{ U }+_{ 2 }^{ 4 }{ He }$

The kinetic energy of an $\alpha -$ particle which flies out of the nucleus of a $Ra^{226}$ atom in radioactive disintegration is $4.78\ M\ ev$ . Find the total energy evolved during the escape of the $\alpha -$ particle.

mass of alpha particle

$m_a=4m$ , where

$m$ is mass of a proton.

mass of Radium nucleus is

$m_u=226m$ after emission the residue will have mass

$m_r=226-4=222m$

We know

$KE=momentum^2/2m$

so for alpha particle, it should be like this

$E=p^2_a/2m_a$

or momentum of alpha particle is

$p_a=\sqrt[2]{2m_aE}=\sqrt[2]{8mE}$

But during emission the residue nucleus should get same momentum in opposite direction to make net momentum as zero

before and after emission so it should also be

$p_r=\sqrt[2]{8mE}$

If kinetic energy of residue is

$E_r$ then

$E_r=p^2_r/2m_r=\dfrac{8mE}{2\times 222m}=0.018E$

So net energy released in the emission is the sum of above two i.e.

$E+0.018E=1.018E$

Put

$E=4.78Mev$ we get total energy as

$4.866Mev$

The figure indicates the potential energy curve of a diatomic molecule. What will be the value of binding energy?

$\dfrac{d{P.E}}{dR}=$ Force on molecule

$\dfrac{d{P.E}}{dR}=0$ at

$R=2\mathring { A }$

$F=0$

So the atom is in equilibrium at

$R=2\mathring { A }$

$E$ at

$r=2=-8J$

Binding energy to break the atoms=

$8J$

How much energy will be released when $10\ kg$ of $U^{235}$ is completely converts into energy :

$10\ kg$ of

$U^{235}$

fission of

$1\ atm$ of

$U^{235}=202.5\ me\ V$

$=19.54\ TJ/mol$

$=83.14\ TJ/kg$

$\therefore U^{235}$ would field

$831.4\ TJ$

In a old rock, ratio of nuclei of uranium and lead is $1:1$ . Half life of uranium is $4.5 \times 10^n$ yrs. Let initially it contains only uranium nuclei. How old is the rock?

1343595_1112107_ans_bc64f44aa1d0463aa3eb8305edd54b9f.jpg

When a $U^{238}$ nucleus originally at rest, decay by emitting an alpha particle having a speed $1.5\times 10^{7}\ m^{-1}$ . Then calculate the kinetic energy of the residual nucleus.

Mass of $\alpha$ particle, ${{m}_{\alpha}}=4\,u$

Mass of nucleus after fission, ${{m}_{n}}=234\,u$

From conservation of momentum

Final momentum = initial momentum

${{m}_{\alpha}}{{v}_{\alpha}}+{{m}_{n}}{{v}_{n}}=0$

$\Rightarrow 4\times 1.5\times {{10}^{7}}+234{{v}_{n}}=0$

$\Rightarrow {{v}_{n}}=2.56\times {{10}^{5}}\,m{{s}^{-1}}$

Hence, velocity of nucleus is $2.56\times {{10}^{5}}\,m{{s}^{-1}}$

Calculate the energy released in $meV$ in the following nuclear reaction:

$_{ 92 }^{ 298 }{ U\longrightarrow _{ 90 }^{ 234 }{ TH }+_{ 2 }^{ 4 }{ He }+Q }$

[Mass of $_{ 92 }^{ 298 }U=238.05079\ u$ Mass of $_{ 90 }^{ 234 }{ TH }=234.043630\ u]$

Given:

Mass of

$_{ 92 }^{ 298 }U=238.05079\ u$ , Mass of

$_{ 90 }^{ 234 }{ TH}=234.043630\ u$

Mass Defect = Mass of uranium - Mass of thorium - Mass of helium

$\text{ Mass defect} = 238.05079 \ u - 234.043630\ u - 4.002600 \ u = 0.004564 \ u$

Now,

$1 \ u \ \text{releases } 931.5 \ MeV/c^2 \text{energy}$

$\Rightarrow$ Energy released by

$0.004564 \ u$ =

$(0.00456 \times 931.5 ) MeV/c^2 = 4.24764 \ MeV/c^2$

Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, $2\ \leq \ A\ \leq 240$ . How do you explain the constancy fo biding energy per nucleon in the range $30\ \leq \ A\ \leq 170$ using the property that nuclear force is short-ranged?

1311769_1139169_ans_8198b35cd77a47ac89dc978c014a1b29.jpg

Find the binding of the nucleus of lithium isotope $_3Li^7$ and hence find the binding energy per nucleon in it $(M_{_3Li^7} = 7.014353 \ \text{amu}, M_{_1H^1} = 1.007826$ , mass of neutron = $1.00867 \ u$ )

Mass defect,

$\Delta m=3H+(7-3)n-Li=3H+4n-Li=3\times 1.007826u +4\times 1.00867u-7.014353u=0.043805u$

so binding energy

$BE=\Delta m \times 931.5Mev=0.043805\times 931.5Mev=40.8Mev$

So binding energy per nucleon will be

$\dfrac{40.8Mev}{\text{mass number}}=\dfrac{40.8}{7}=5.829Mev/nucleon$

Find the binding energy of the nucleus of lithium isotope $_{3}{Li}^{7}$ and hence find the binding energy per nucleon in it.\left({ M }_{ _{ 3 }{ Li }^{ 7 } }=7.014353amu, { M }_{ _{ 1 }{ H }^{ 1 } }=1.007826, mass of netron=1.00867\ u\right)

1310248_1138258_ans_b1d9aedde24b4bcba18aab929ede96e4.jpg

Which process is responsible in Hydrogen bomb ?
If 20 N force is applied on a wall which doesn't move the wall than work done by wall is............

Nuclear fusion is responsible in Hydrogen Bomb involving joining of hydrogen nuclei to form Helium and releasing destructive amount of energy.

As there is no displacement of wall, work done(dot product of force and displacement) by the wall is

$0$ (zero).

1133101_1159034_ans_41da93e908ba4a2ca145bdb6567a7fad.jpg

matter is neither created nor destroyed a chemical reaction More simply the mass of products is equal to the mass of the reaction in a chemical reaction.

The law of conservation of mass states that the net change in mass of the reactants and products before and after a chemical reaction is zero

$.$ This means mass can neither be created nor destroyed in other words

$,$ the total mass in a chemical reaction remains constant

$.$

Obtain the binding energy (in $MeV$ ) of a nitrogen nucleus $(^{14}_{7}N)$ , given $m(^{14}_{7}N)=14.00307\ u$

${ m }_{ p }=1.007825amu;\quad { m }_{ n }=1.008665amu$

The nitrogen nucleus contains

$7$ protons and

$7$ neutrons

The binding energy of nitrogen nucleus

$=\left[ \left\{ 7\times 1.007825+7\times 1.008665 \right\} -14.00307 \right] \times 931.5=\left[ \left\{ 7.054775+7.060655 \right\} -14.00307 \right] \times 931.5=104.66MeV$

Define binding energy of a satellite. Obtain an expression for the binding energy of a satellite revolving around the earth at a certain altitude.

The minimum amount of energy required for a satellite to escape from earth's gravitational influence is called as Binding Energy of a satellite. r is a radius of orbit i.e. Radius of earth + height at which satellite is orbiting the earth

Let the mass of satellite is m and r is the radius of the circular orbit. the satellite moves in the circumference of circular orbit around the earth.

now, at equilibrium condition,

centripetal force = gravitational force

mV²/r = GMm/ r2

mV² = GMm/ r

½ mV² = GMm/ 2r

KE = GMm/ 2r

now, potential energy between satellite and the earth is given by

PE = - GMm/ r

[ here negative sign indicates force acts between satellite and earth is attractive]

TE = KE+PE

TE = GMm/ 2r + - GMm/ r

TE = - GMm/2r

E_r = - GMm/ r

here negative sign indicates that the satellite is bound to the earth by attractive force and cannot leave it on its own. To move the satellite to infinity, so, the binding energy of a satellite revolving in a circular orbit around the earth is

Explain binding energy curve.Write its importance .What is gamma decay? Explain with example.

$Binding\,\,energy\,curve: -$ The fact that there is a peak in the binding energy curve in he region of stability near iron means that either the breakup of heavier nuclei

$($ fission

$)$ or the combining of lighter nuclei

$($ fusion

$)$ will yield nuclei which are more tightly bound

$($ less mass per nucleon

$).$

Importance of Binding energy curve

$:-$ Nuclear binding energy is defined as the energy required to break up a nucleus into its individual nucleons

$i.e.,$ protons and neutrons

$.$ The more important than the total binding energy is the energy per nucleon against mass number

$.$

$gamma\,decay:-$ One of the three main types of radioactive decay is known as gamma decay

$\left( {\gamma - decay} \right).$

During gamma decay

$,$ the energy of the parent atom is changed by the emission of a photon

$.$ The resulting energy of the daughter atom is lower than the parent atom

$.$ A photon is a massless particle with a very small wavelength

$.$

Find the binding energy of a H-atom in the state n=2

$n_1 = 1 , \ n_2 = \infty$

The binding energy is given by:

$E = \dfrac{-13.6}{n_1^2} - \dfrac{-13.6}{n_2^2}$

$= 13.6 \Big( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \Big)$

$= 13.6 (1/\infty - 1/4)$

$= -13.6/4 = -3.4 \ eV$

The mass of the nucleus $_7N^{14}$ is $14.00000amu$ . Calculate the binding energy per nucleon of $_7N^{14}$ in MeV. Mass of proton $=1.00759amu$ , mass of neutron $=1.00898amu$ and $1 amu=931 MeV.$

The mass defect of nucleus is

$\Delta m=(Zm_p+Zm_n-mass \,of \,nucleus)$

$\Delta m=7[1.00759+1.300898]-14=0.11500a.m.u$
The energy equivalent of this mass is E=107.98MeV
Therefore the binding energy per nucleon is 7.713MeV

A sample has $1000$ nuclear samples find after how much time will they remain $250$ .

given,

$\dfrac{1000}{2}=500$ nuclear samples after the first

$t_{\frac{1}{2}}$

$\dfrac{500}{2}=250$ nuclear samples after another

$t_{\frac{1}{2}}$

considering

$100$ years for decay.

$\therefore 2\times t_{\frac{1}{2}}=100\rightarrow t_{\frac{1}{2}}=\dfrac{100}{2}=50 years$

Explain why, in a nuclear reactor, the chain reaction stops if the control rods are fully inserted into the graphite.

The control rods absorb all the neutrons, stopping the nuclear chain reaction

Assume that the mass of a nucleus is approximately given by $M = Am_p$ , where $A$ is the mass number. Estimate the density of matter in $kgm^{-3}$ inside a nucleus. What is the specific gravity of nuclear matter?

$M = Am_p , f= M/V , m_p = 1.007276 u$

$R =R_gA^{1/3} = 1.1 \times 10^{-15} A^{1/3}, u= 1.6605402 \times 10^{-27} kg$

$= \dfrac { A\times 1.007276 \times 1.6605402 \times 10^{-27} }{\dfrac{4}{3} \times 3.14 \times R^3} = 0.300159 \times 10^{18}= 3 \times 10^{17} kg/m^3$

The specific gravity of the nuclear matter

$\dfrac{Density \ of \ matter}{Density \ of \ water}$

$\therefore Specific \ gravity=\dfrac{3\times 10^{17}}{10^{3}}=3\times 10^{14} kg/m^{3}$

What are isotopes?

The atoms of the same element, having same atomic number

$Z$ , but different mass number

$A$ , are called isotopes.

Define binding energy and obtain an expression for binding energy of a satellite revolving in a circular orbit round the earth.

Binding Energy can be defined as the minimum energy required to be supplied to it order to free the satellite from the gravitational influence of the planet. (i.e; in order to take the satellite from the orbit to as point if infinity.)
Consider a satellite revolving around a planet with speed V at a distance r.

$KE=\dfrac{1}{2} m_sV^2$

$P.E= \dfrac{GM_pm_s}{r}$

$E=\dfrac{GM_pm_s}{r^2}$ =

$\dfrac{msV^2}{r}$

$\Rightarrow m_3V^2$ =

$\dfrac{4M_pm_s}{r}$

$KE=\dfrac{1}{2}$

$\dfrac{Gm_sM_p}{r}$

$T.E=K.E+P.E=\dfrac{1}{2}$

$\dfrac{GM_pm_s}{r}-\dfrac{GM_pm_s}{r}$

$T.E=\dfrac{-GM_pm_s}{2r}$

$B.E+T.E=0$ (at infinite

$TE=0$ )

$B.E=\dfrac{GP_pm_s}{2r}=0$

$B.E=\dfrac{GM_pm_s}{2r}$
Hence Binding energy of satellite G of man

$m_s$ revolving around a planet of man

$M_p$ in a radius r
is

$=\dfrac{GM_pm_s}{2r}$

1212296_1555769_ans_d2b7fae094f143b0b11f8d8b7a0647e0.png

If the nucleon of a nucleous are separated far aprat from each others the sum of mass of all these nucleons is larger than the mass of nucleus where does this mass difference come.

When nucleons of nucleus are separated apart from each other then the sum of the nucleons is greater than the mass of nucleus.This is direct mass defeat of nucleus. Some mass has been converted into energy which holds the nucleous together.

Write the nuclear equations for:
${\beta}^{+}$ - decay of $_{ 6 }^{ 11 }{ C }$

$_{ 6 }^{ 11 }{ C }\rightarrow _{ 5 }^{ 11 }{ B }+_{ +1 }^{ 0 }{ e }+\overline { v }$

Write nuclear reaction equations for
Electron capture of $_{ 54 }^{ 120 }{ Xe }$

$_{ 54 }^{ 120 }{ Xe }+_{ -1 }^{ 0 }{ e }\rightarrow _{ 53 }^{ 120 }{ I }+\overline { v }$

Find whether alpha decay or any of the beta decay are allowed for $_{ 89 }^{ 236 }{ Ac }$ .

Given masses are $M(_{ 89 }^{ 236 }{ Ac })=226.028356$ a.m.u; $M(_{ 87 }^{ 222 }{ Fr })=222.017415$ a.m.u; $M(_{ 90 }^{ 236 }{ Th }) =226.017388$ a.m.u; $M(_{ 88 }^{ 226 }{ Ra })=226.025406$ a.m.u; $M(_{ 2 }^{ 4 }{ He })=4.002603$ a.m.u

Our first step will be to write the reaction, then to find the disintegration energy

$Q$ . If

$Q> 0$ , the decay is allowed

$\alpha \quad decay:_{ 89 }^{ 226 }{ Ac }\rightarrow _{ 87 }^{ 222 }{ Fr }+\alpha$

$Q=\left[ M\left( _{ 89 }^{ 226 }{ Ac } \right) -M\left( _{ 87 }^{ 222 }{ Fr } \right) -M\left( _{ }^{ 4 }{ He } \right) \right] { c }^{ 2 }=5.50MeV$ (

$\alpha$ decay is allowed)

$\beta \quad decay:_{ 89 }^{ 226 }{ Ac }\rightarrow _{ 90 }^{ 226 }{ Th }+{ \beta }^{ - }+\bar { v } \quad$

$Q=\left[ M\left( _{ 89 }^{ 226 }{ Ac } \right) -M\left( _{ 90 }^{ 226 }{ Th } \right) \right] { c }^{ 2 }=1.12MeV$ (

$\beta$ decay is allowed)

${ \beta }^{ - }\quad decay:_{ 89 }^{ 226 }{ Ac }\rightarrow _{ 88 }^{ 226 }{ Ra }+{ \beta }^{ - }+v$

$Q=\left[ M\left( _{ 89 }^{ 226 }{ Ac } \right) -M\left( _{ 88 }^{ 226 }{ Ra } \right) \right] { c }^{ 2 }=-0.38MeV$

Electron capture:

$:_{ 89 }^{ 226 }{ Ac }+{ e }^{ - }\rightarrow _{ 88 }^{ 226 }{ Ra }+v$

$Q=\left[ M\left( _{ 89 }^{ 226 }{ Ac } \right) -M\left( _{ 88 }^{ 226 }{ Ra } \right) \right] { c }^{ 2 }=0.64MeV$ (electron capture is alowed)

${ _{ 92 }^{ }{ U } }^{ 238 }$ changes to ${ _{ 85 }^{ }{ At } }^{ 210 }$ by a series of $\alpha$ and $\beta$ decays. Find number of $\alpha$ - decays undergone (an integer)

One alpha decay decreases atomic number by 2 and atomic mass by 4

One beta decay decreases atomic number by 1

$x$ and

$y$ are number of

$\alpha$ decays and

$\beta$ decays respectively

$92-2x+y=85$

$\Rightarrow2x-y=7$
similarly

$238-4x=210$

solving the above two equations we get

$x=7$ and

$y=7$

Identify $X$ in the following nuclear reactions:
$\ ^{1}H+\ ^{0}Be\rightarrow X+n$

None of the reactions given includes a beta decay so the number of protons, the number of neutrons, and the number of electrons are each conserved. Atomic numbers (numbers of protons and numbers of electrons) and molar masses (combined number of protons and neutrons) can be founded in Appendix

$F$ of the text.

$\ ^{1}H$ has

$1$ proton,

$1$ electron, and

$0$ neutrons

and

$\ ^{9}Be$ has

$4$ protons,

$4$ electrons and

$9-4=5$ neutrons, so

$X$ has

$1+4=5$ protons

$1+4=5$ electrons and

$0+5-1=4$ neutrons. One of the neutrons is freed in the reaction.

$X$ must be boron with a molar mass of

$5+4=9\ g/mol:\ ^{9}B$ .

What is the minimum energy that is required to break a nucleus of $^{12} \mathrm{C}$ ( of mass $11.99671 \mathrm{u}$ ) into three nuclei of $^{4} \mathrm{He}$ (of mass $4.00151 \mathrm{u} \text { each }) ?$

From Eq. 28-37, we have

$\begin{aligned}Q &=-\Delta M c^{2}=-[3(4.00151 \mathrm{u})-11.99671 \mathrm{u}] c^{2}=-(0.00782 \mathrm{u})(931.5 \mathrm{MeV} / \mathrm{u}) \\&=-7.28 \mathrm{Mev}\end{aligned}\\$

Thus, it takes a minimum of

$7.28 \mathrm{MeV}$ supplied to the system to cause this reaction. We note that the masses given in this problem are strictly for the nuclei involved; they are not the "atomic" masses that are quoted in several of the other problems in this chapter.

Draw the graph showing the variation of binding energy per nucleon with mass numbers. Give the reason for the decrease of binding energy per nucleon for nuclei with higher mass number.

The graph of the binding energy per nucleon versus mass number A is shown in figure.

The decrease of the binding energy per nucleon for nuclei with high mass number is due to increased coulomb repulsion between protons inside the nucleus.

What is the binding energy per nucleon of $^{262} Bh$ ? The mass of the atom is 262.1231 u.

The binding energy of a nucleus is the difference is given by:

$\Delta E_{be} = \sum(mc^2) - Mc^2$

where

$\sum(mc)^2$ is the tootal mass energy of the individual protons neutrons, the number of protons is Z and the number of neutrons is A-Z, the mass of the proton is

$m_H$ and the mass of the neutrons is

$m_n$ , so we can write:

$\sum(mc)^2 = (Zm_H + (A-Z)m_n)c^2$

substitute into the above equation to get:

$\Delta E_{be} = (Zm_H + (A-Z)m_n - M)c^2$

for americium

$^{262}_{107}Rf, M = 262.1231u$ , and using

$m_H = 1.007825u$ and

$m_m = 1.008655u$ , we get:

$\Delta E_{be} = ((107)(1.007825u) + (262-107)(1.008665u) - 262.1231u)c^2 = (2.05725u)c^2$

but

$1 u = 931.494013 MeV/c^2$ , so:

$\Delta E_{be} = (2.05725 \times 931.494013 MeV/c^2)c^2 = 1916.316058MeV$

the binding energy per nucleon equals this energy divided by the number of nucleons A, that is

$$\Delta E_{be,n} = \dfrac{\Delta E_{be}}{A}=\dfrac{1916.316058MeV}{262}=7.3142MeV$$

What is the binding energy per nucleon of the rutherfordium isotope $^{259} _{104}Rf$ ? Here are some atomic masses and the neutron mass.
$^{259}_{104}Rf$ 259.105 63 u $^1H$ 1.007 825 u
n 1.008 665 u

The binding energy of a nucleus is the difference is given by:

$\Delta E_{be} = \sum(mc^2) - Mc^2$

where

$\sum(mc)^2$ is the tootal mass energy of the individual protons neutrons, the number of protons is Z and the number of neutrons is A-Z, the mass of the proton is

$m_H$ and the mass of the neutrons is

$m_n$ , so we can write:

$\sum(mc)^2 = (Zm_H + (A-Z)m_n)c^2$

substitute into the above equation to get:

$\Delta E_{be} = (Zm_H + (A-Z)m_n - M)c^2$

for americium

$^{259}_{104}Rf, M = 259.10563u$ , and using

$m_H = 1.007825u$ and

$m_m = 1.008655u$ , we get:

$\Delta E_{be} = ((104)(1.007825u) + (259-104)(1.008665u) - 259.10563u)c^2 = (2.051245u)c^2$

but

$1 u = 931.494013 MeV/c^2$ , so:

$\Delta E_{be} = (2.051245 \times 931.494013 MeV/c^2)c^2 = 1910.722437MeV$

the binding energy per nucleon equals this energy divided by the number of nucleons A, that is

$\Delta E_{be,n} = \dfrac{\Delta E_{be}}{A}=\dfrac{1910.722437MeV}{259}=7.3773MeV$

The first of the following two reactions can occur, but
the second cannot.
Explain.
$K^0_S \rightarrow \pi^+ + \pi^-$ (can occur)

$\wedge^0 \rightarrow \pi^+ + \pi^-$ (cannot occur)

Baryon number conservation allows the first and forbids the second

Compare and contrast the nature of $\alpha-, \beta-$ and $\gamma-$ radiations.

Comparison of Properties of

$\alpha-, \beta-,$ and

$\gamma-$ rays.

	Property	$\alpha$ -particle	$\beta$ -particle	$\gamma$ $-rays
1.	Nature	Nucleus of Helium	Very fast moving electron $(e^-)$	electromagnetic wave of wavelength $= 10^{-2}$
2.	Charge	$+2e$	$-e$	No charge
3.	Rest mass	$6.6 \times 10^{-27} kg$	$9.1 \times 10^{-31} kg$	zero
4.	Velocity	$1.4 \times 10^7 \,m/s$ to $2.2 \times 10^7 \,m/s$	$0.3 \,c$ to $0.98 \,c$	$c = 3 \times 10^8 \,m/s$
5.	Ionising Power	high, 100 times that of $\beta$ -particle	100 times more than $\gamma$ -rays	very small
6.	Penetrating Power	very small	high, 100 times more than $\alpha$ -particles	very high, 100 times more than $\beta$ -particles

Answer the following question
In the following nuclear reaction
$n + ^{235}_{92}U \rightarrow ^{144}_{Z}Ba + ^{A}_{36}X + 3n,$
Assign the values of Z and A.

$n + ^{235}_{92}U \rightarrow ^{144}_{Z}Ba + ^{A}_{36}X + 3n,$
From law of conservation of atomic number

$0 + 92 = Z + 36$

$\Rightarrow Z = 92 - 36 = 56$
From law of conservation of mass number,

$1 + 235 = 144 + A + 3 \times 1$

$A = 236 - 147 = 89$

Calculate the binding energy per $^{40}_{20}CA$ nucleon nucleus.
[Given: $m\left ( ^{40}_{20}Ca \right ) = 39.962589 \,u$
$m_n$ (mass of a neutron) $= 1.008665 \,u$
$m_p$ (mass of a proton) $= 1.007825 \,u$
$1 \,u = 931 \,MeV / c^2]$

Total Binding energy of

$^{40}_{20}Ca$ nucleus

$= 20m_p + 20m_n - M (^{40}_{20}Ca)$

$= 20 \times 1.007825 + 20 \times 1.008665 - 39.962589$

$= 0.367211 \,u = 0.367211 \times 931 \,MeV = 341.87 \,MeV$

$\therefore$ Binding energy per nucleus

$= \dfrac{341.87}{40} MeV / nucleon$

$= 8.55 \,MeV / nucleon$

Give one example of Isotopes.

Examples: Hydrogen has three isotopes namely protium

$^{1}_{1}H$ , deuterium

$^{2}_{1}H$ and tritium

$^{3}_{1}H$ .

Name the constituents of the nucleus of an atom.

Nucleus consists of protons and neutrons.
1702185_1809213_ans_7a8a2f2b8be34bdfaec902c615dc9c46.png

A radioactive source emits three type of radiations. Name the radiation of zero mass.

Gamma radiation, unlike alpha or beta, does not consist of any particles, instead consisting of a photon of energy being emitted from an unstable nucleus. Having no mass or charge, gamma radiation can travel much farther through air than alpha or beta, losing (on average) half its energy for every 500 feet.

Before the neutrino hypothesis, the beta decay process was thought to be the transition $n\rightarrow p+{e}^{-}$ . If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range.

Neutron was at rest before

$\beta$ decay from neutron. Hence energy of neutron

$={E}_{n}={m}_{n}{c}^{2}$ and momentum of neutron

${p}_{n}=0$ as its velocit is zero
By the law of conservation of momentum

${p}_{n}={p}_{p}+{p}_{e}$ (beta)

$O={p}_{p}+{p}_{e}$
Let

${p}_{e}={p}_{p}$ then

$\Rightarrow$

$\left| { p }_{ p } \right| =\left| { p }_{ e } \right| =p(eV)$
Energy of proton

$={ E }_{ p }=\sqrt { \left( { m }_{ p }^{ 2 }{ c }^{ 4 }+{ p }_{ p }^{ 2 }{ c }^{ 2 } \right) }$
Energy of electron

$\left( \beta \right) ={ E }_{ e }=\sqrt { \left( { m }_{ p }^{ 2 }{ c }^{ 4 }+{ p }_{ e }^{ 2 }{ c }^{ 2 } \right) } \left( \because \left| { p }_{ e } \right| ={ p }_{ p } \right)$
From conservation

$\therefore { E }_{ p }=\sqrt { \left( { m }_{ p }^{ 2 }{ c }^{ 4 }+{ p }_{ p }^{ 2 }{ c }^{ 2 } \right) } ;{ E }_{ e }=\sqrt { \left( { m }_{ }^{ 2 }{ c }^{ 4 }+{ p }_{ }^{ 2 }{ c }^{ 2 } \right) }$
By the law of conservation energy

${ \left( { m }_{ p }^{ 2 }{ c }^{ 4 }+{ p }_{ p }^{ 2 }{ c }^{ 2 } \right) }^{ 1/2 }+{ \left( { m }_{ e }^{ 2 }{ c }^{ 4 }+{ p }_{ }^{ 2 }{ c }^{ 2 } \right) }^{ 1/2 }={ m }_{ n }{ c }^{ 2 }$

$\because { m }_{ p }{ c }^{ 2 }=936MeV$

${ m }_{ n }{ c }^{ 2 }=938MeV;{ m }_{ e }{ c }^{ 2 }=0.5MeV$
As the energy difference in neutron and proton is very small,

$pc$ will be small,

$pc$ will be small

$pc<<{ m }_{ p }{ c }^{ 2 }\quad$ while

$pc$ may be greater than

${m}_{e}{c}^{2}$ so by neglecting

${ \left( { m }_{ p }{ c }^{ 2 } \right) }^{ 2 }={ \left( 0.5 \right) }^{ 2 }$ (given)

$\\Rightarrow { m }_{ p }{ c }^{ 2 }+\cfrac { { p }_{ }^{ 2 }{ c }^{ 2 } }{ 2{ m }_{ p }^{ 2 }{ c }^{ 4 } } ={ m }_{ n }{ c }^{ 2 }-pc$

$\Rightarrow { m }_{ p }{ c }^{ 2 }+\cfrac { { p }_{ }^{ 2 }{ c }^{ 2 } }{ 2{ m }_{ p }^{ 2 }{ c }^{ 4 } } +pc={ m }_{ n }{ c }^{ 2 }$
Again

$pc<<{ m }_{ p }{ c }^{ 2 }$ so neglecting

$\cfrac { { p }_{ }^{ 2 }{ c }^{ 2 } }{ 2{ m }_{ p }^{ 2 }{ c }^{ 4 } }$ we get

$Pc={ m }_{ n }{ c }^{ 2 }-{ m }_{ p }{ c }^{ 2 }=938MeV-936MeV=2MeV$ is the momentum

$E=m{ c }^{ 2 }\quad$

$\because { E }^{ 2 }={ m }^{ 2 }{ c }^{ 4 }$

$E$ is the energy of either proton or neutron then

${ E }_{ p }=\sqrt { \left( { m }_{ p }^{ 2 }{ c }^{ 4 }+{ p }_{ }^{ 2 }{ c }^{ 2 } \right) } =\sqrt { \left( { 936) }^{ 2 }+2^{ 2 } \right) } =936MeV$

${ E }_{ e }=\sqrt { \left( { m }_{ e }^{ 2 }{ c }^{ 4 }+{ p }_{ }^{ 2 }{ c }^{ 2 } \right) } =\sqrt { \left( { a) }^{ 2 }+2^{ 2 } \right) } =2.06MeV$

A radioactive source emits three types of radiations. Name the radiation which travels with the speed of light.

Beta radiation is more penetrating than alpha radiation. It can pass through the skin, but it is absorbed by a few centimetres of body tissue or a few millimetres of aluminium.Gamma radiation is the most penetrating of the three radiations. It can easily penetrate body tissue.

A radioactive source emits three types of radiations. Name the radiation which is most penetrating.

A nuclide 1 is said to be the mirror isobar of nuclide 2 if ${Z}_{1}={N}_{2}$ and ${Z}_{2}={N}_{1}$
Which nuclide out of the two mirror isobars have greater binding energy and why?

As the neutrons in

${ _{ 12 }^{ }{ Mg } }^{ 23 }$ are

$11$ and in

${ _{ 11 }^{ }{ Na } }^{ 23 }$ are

$12$ so, the number of neutrons in

$Na$ is larger than

$Mg$ and hence nuclear short range attractive forces in

$Na$ will e larger than
So,

${ _{ 11 }^{ }{ Na } }^{ 23 }$ has more binidng energy than

${ _{ 12 }^{ }{ Mg } }^{ 23 }$

The mass of a H - atom is less than the sum of the masses of a proton and electron. Why is this ?

During formation of an atom with neucleon and electrons, it need the energy to blind the nucleons together in nucleus. This energy comes from Einstein mass - energy relation :

$E = \Delta mC^z$
Where

$\Delta m = \left[Zm_p + \left[ A- Z\right]m_n\right] - M$
So the mass of a H atom is

$m_p + m_e - \dfrac{B.E.}{C^2}$
Where

$B.E. = 13.6 eV$

State what is meant by the term isotopes. Use the terms proton number and nucleon number in your explanation.

Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

What happens to the atomic number of the nucleus when an $\alpha$ particle is emitted?

$\Large\underline{\textbf{Concept used:}}$

$\large {\textbf{Alpha Decay}}$

Alpha decay is a radioactive decay pathway, in which a radioactive nucleus emits an alpha particle (Helium nucleus) and decays into a new nucleus.

Example:

$\large {}_{92}^{234}U \rightarrow {}_{90}^{230}Th +{}_{2}^{4} \alpha$

The general equation for alpha decay is

$\large {}_{Z}^{A}X \rightarrow {}_{Z-2}^{A-4}Y +{}_{2}^{4} \alpha$

where,

$X$ is the parent nucleus and

$Y$ is the daughter nucleus.

$\Large\underline{\textbf{Answer/Explanation}}$

We already know that, that in alpha decay, the parent nucleus emits an

$alpha \ particle$ , which is essentially a

$\textbf{Helium nucleus( two protons and two neutrons)}$

Mass number of

$\alpha$ particle=4

Atomic number of

$\alpha$ particle=2

Hence,

$\textbf{the mass number of parent nucleus reduces by 4 and atomic number also reduces by 2.}$

When an $\alpha$ particle captures one electron, what does it change to?

When an

$\alpha$ particle captures one electron it becomes a singly ionized helium

$He^+$

When an $\alpha$ particle captures two electrons, what does it change to?

When an

$\alpha$ particle captures two electrons it changes to neutral helium atom.

A radioactive source emits three type of radiations. Name the radiation which has the lowest penetrating power.

Alpha particles have the least penetration power and can be stopped by a thick sheet of paper or even a layer of clothes.

A certain nucleus $A$ (mass number $238$ and atomic number $92$ ) is radioactive and becomes a nucleus $B$ (mass number $234$ and atomic number $90$ ) by the emission of a particle. Name the particle emitted.

Emission of alpha particle reduces the proton by $2$ and neutron by $2$ , so mass number is reduced by 4 and atomic number by $2$ .
New mass number is: $238 - 4 = 234$ , and atomic number is: $92 - 2 = 90$
$The emitted particle is alpha.$

What happens to the position of an element in the periodic table when its nucleus emits an $\alpha$ particle? Give reason for your answer.

When an

$\alpha$ particle is emitted, the daughter element occupies two places to the left of the parent element in the periodic table

Reason: If a parent nucleus

$X$ becomes a new daughter nucleus

$Y$ as a result of α decay, then the change can be expressed in the form of reaction as follows:

$\overset{^{A}_{Z}X}{\text{Parent nucleus }} \rightarrow \overset{^{A-4}_{Z-2}Y}{\text{ Daughter nucleus }} + \overset{^{4}_{2}He}{\text{ alpha-particle }}$

Thus due to emission of an alpha particle, atomic number

$Z$ decreases by

$2$ units and therefore it shifts two places to the left of the parent element in the periodic table.

What happens to the mass number of the nucleus when an $\alpha$ particle is emitted?

When an

$\alpha$ particle is emitted mass number decreases by

$4$ .

$Z$ decreases by

$2$ units and therefore it shifts two places to the left of the parent element in the periodic table.

A certain nucleus $A$ (mass number $238$ and atomic number $92$ ) is radioactive and becomes a nucleus $B$ (mass number $234$ and atomic number $90$ ) by the emission of a particle.
State the change in the form of a reaction.

${\Large\underline{\textbf{Concept Used:}}}$

${\large{\textbf{Balancing nuclear reactions}}}$

Nuclear reactions follow conservation laws like any other physical/chemical change. The rules to balance a nuclear equation are:

$\bullet$ The sum of mass numbers of reactants should be equal to that of the products.

$\bullet$ The sum of charges of the reactants should be equal to the sum of charges of products.

${\Large\underline{\textbf{Step 1: Identify the reactants and products and write the reaction}}}$

Reactants:

${}_{92}^{238}A$

Products:

${}_{90}^{234}B$ and

${}_{x}^{a}X$

${}_{92}^{238}A \rightarrow{}_{90}^{234}B+{}_{x}^{a}X$

${\Large\underline{\textbf{Step 2: Balance the equation and deduce X}}}$

${{\textbf{Applying conservation of mass:}}}$

The Sum of mass numbers of reactants and that of products should be constant

$238=234+a$

$\Rightarrow a=4$

${{\textbf{Applying conservation of charge:}}}$

The Sum of atomic numbers of reactants and that of products should be constant

$92=90+x$

$\Rightarrow x=2$

Hence, the emitted particle is,

${}_{2}^{4}He$ or

$\alpha$ particle

${\Large\underline{\textbf{Step 3: Write the complete nuclear equation}}}$

${}_{92}^{238}A \rightarrow{}_{90}^{234}B+{}_{2}^{4}\alpha$

What changes occur in a nucleus of a radioactive element when it emits an alpha particle? Give one example in support of your answer.

The following are the changes which occur when an atom emits

When alpha particle emits, the atomic number decreases by

$2$ units and mass number decreases by

$4$ units
Example:

$^{238}_{92}U\ \rightarrow\ ^{234}_{90}Th\ +\ ^{4}_{2}He$

The diagram in figure shows a radioactive source $S$ placed in a thick lead walled container. The radiations given out are allowed to pass through a magnetic field. The magnetic field (shown as $x$ ) acts perpendicular to the plane of paper inwards. Arrows shows the paths of the radiation $A$ , $B$ and $C$ . Explain clearly how you arrive at naming the radiations labelled $A$ , $B$ and $C$ .

The radiation labelled as

$A$ is gamma radiation as they have no charge and thus under the action of magnetic field they go undeflected. The radiation labelled as

$B$ is alpha radiation since its mass is large and it would be deflected less in comparison to beta radiation. Fleming’s left hand rule determines the direction of deflection. As the alpha and beta have opposite charges, hence the direction of deflection of alpha and beta radiations are also opposite and thus the radiation labelled

$C$ is

$\beta$ radiation.

Name two isotopes of uranium which are fissionable.

$^{238}_{92}U$ and

$^{235}_{92}U$ are the isotopes which are fissionable.

State one similarity in the process of nuclear fission and fusion

Both nuclear fission and fusion require enormous amount of energy and both can cause ill disease which may result in ultimate death of a person.

Why is a nuclear fusion reaction called a thermo nuclear reaction?

Because this occurs at a very high temperature.

What is meant by binding energy per nucleon ? How is it related with the stability of nucleus ?

According to Einstein, this mass

$(\Delta m)$ turns into energy. This energy is called binding energy. This energy binds all nucleons of the nucleus in the form of nucleus

$\Delta m$ means that when the protons and neutrons together from the nucleus then

$Am$ mass disappears and its equivalent energy is released. By this energy the protons and neutrons are bound to the nucleus. It is obvious that breaking the proton and neutron of the nucleus will require the same outer energy. Thus it becomes clear that when the protons and neutrons from the nucleus then the some energy is released, which is called the binding energy. It is also clear from this fact that if the same energy is given to the nucleus then all nucleons will be free, Thus binding energy is the amount of energy that release all of the nucleons when giving it to the nucleus. " So, the binding energy of nucleus.

$\Delta E=\Delta mc^{2}=[\left( Zm_{p}+(A-Z)m_{n}\right)-m_u]c^{2}$ ........(2)

If the nucleons number divided in binding energy of nucleus then we get the binding energy per nucleon.

Binding energy demonstrates the stability of the atom.

$\therefore$ Binding energy per nucleon

$=\dfrac{\Delta E}{A}$

A student says that a heavy form (isotope) of hydrogen decomposes by $\alpha$ -decay. How would you react ?

$\alpha-$ decay, the daughter nucleus has atomic number less by

$2$ . For hydrogen isotopes this is not possible.

Explain nuclear mass defect.

We have read that the entire mass of the atom and positive charge is concentrated in the nucleus and nucleus is composed by the protons and neutrons. We also know how much protons and neutrons occur in the nucleus therefore, the expected mass of a nucleus can be determined by the calculation. The actual mass of nucleus can also be determined by mass spectrograph. It is found that the actual mass of any nucleus is always less than the expected mass obtained by its nucleons calculation. This difference of mass is called mass defect. Similarly,

Mass defect= Mass of nucleus obtained by the calculation - Actual mass of nucleus of

$\Delta m=m_{c}-m_{a}$

Where calculated mass is represented by

$m_{c}$ and actual mass of represented by

$m_{a}$ .

$\therefore \Delta m=$ [Mass of protons+mass of neutrons]-Actual mass of nucleus

$\Delta m=[Zm_{p}+(A-Z)m_{n}]-m$ ..... (1)

Where

$Z$ is atomic number,

$A$ mass number,

$m_{p}$ of proton,

$m_{n}$ mass of neutron and

$m$ is the actual mass.

Calculate Q-value for the reaction
$^{235}_{92}U + ^{1}_{0}n \rightarrow ^{140}_{54} Xe + ^{94}_{34}Sr + 2^{1}_{0}n + Q$
Given: Mass of $^{235}_{92}U = 235.0435\,u$
Mass of $^{140}_{54}Xe = 139.9054 \,u$
Mass of $^{94}_{34}Sr = 93.9063 \,u$
Mass of $^{1}_{0}n = 1.00867 \,u$

Mass defect

$\Delta m=[M(_{92}U^{235})+M(_0n^1)-M(_{54}Xe^{140})-M(_{38}Sr^{94})-M(2_0n^1)]$

$\Delta m=[(235.0435+1.00867)-(139.9054+93.9063)+2\times 1.00867]$

$\Delta m=[(236.05217)-(235.82904)]u$

$\Delta m=0.22313u$

$Q=\Delta mc^2$

$Q=0.22313\times 931\dfrac{MeV}{c^2}c^2 [\because 1\ u=931\dfrac{MeV}{c^2}]$

$Q=207.73MeV$

Which out of iron or lead, is easier to take out an nucleon from ?

It is easier to take out a nucleon from lead as compared to iron because the binding energy per nucleon of iron is more than that of lead.

If the nucleons of a nucleus are divided, then total mass increases. Where does this mass come from ?

1888091_1836763_ans_552ec8d867d0434eaf4bf602d1c726b8.jpg

What is meant by critical mass in nuclear chain reaction ?

Whether the mass of a fissionable material can sustain a chain reaction or not, depends on it multiplication factor. This in turn depends on the size of the material. The size of the fissionable material for which the multiplication factor

$k=1$ is called critical size and its mass is called critical mass. The chain reaction in this case remains steady or sustained.

Calculate the binding energy of an alpha particle given its mass to be 4.00151 u.

Mass of proton (

$m_p$ ) = 1.00728u

Mass of neutron (

$m_n$ )= 1.00866u

(take u= 939.565

$MeV/c^2$ )

The alpha particle has two proton and two neutron.

The combined mass of nuclei = net mass of proton + net mass of neutron

$= (2\times 1.00728) + ( 2\times 1.00866)$

$=4.03188 \ \ amu$

Binding energy - mass defect

$= 4.03188- 4.00151$

$= 0.03037 \times u$

$= 0.03037(939.565)$

$= 28.53 \ MeV$

Two stable isotopes of lithium $_{ 3 }^{ 6 }{ Li }$ and $_{ 3 }^{ 7 }{ Li }$ have respective abundance of $7.5$ % and $92.5$ %. These isotopes have masses $6.01512u$ and $7.01600u$ respectively. Find the atomic mass of lithium.

The average atomic weight of lithium

$A=\cfrac { 6.01512\times 7.5+7.01600\times 92.5 }{ 100 } =6.941amu$

What is $\alpha-$ decay? What type energy spectrum do $\alpha-$ particles have?

$\alpha-$ particle is the nucleus of helium which is double ionized particle.

$\ _{Z}X^{A}\rightarrow z-\ _{2}Y^{A-4}+\ _{2}He^{4}$
It is clear that the emission of

$\alpha-$ particle reduces the atomic number of the original nucleus from two and the mass number is

$4$ that means the atomic number of the product nucleus decreases by

$2$ and the mass number is four, it is called

$\alpha-$ decay.
Exposure from the nucleus of

$\alpha$ particles can be interpreted on the basis of quantum mechanics rather than coherent principles. According to that

$\alpha-$ particles occurs only in the nucleus and by the tunnel effect it comes out of then nucleus. The spectrum of energy of

$\alpha-$ particles is discrete and linear discrete spectrum demonstrates that nuclear power level is available in the nucleus as well as nuclear.

Calculate binding energy for $_{26}^{56}Fe$ nucleus [Given: mass of $_{26}^{56}Fe=55.9349 u,$ mass of neutron $=1.00867u,$ mass of proton $=1.00783u$ and $1\ u=931MeV/c^2$ ]

$\Delta m=[(30m_n+2m_p)-M(_{26}Fe^{56})]$

$\Delta m=(30\times 1.00867+26\times 1.00783)u-55.9349u$

$\Delta m=(30\times 2601+26.20358)u-55.9349u$

$\Delta m=56.46368-55.9349u$

$\Delta m=0.52878u$

$E_B=0.52878\times 931\ MeV$

$E_B=492.29MeV\approx 492MeV$

If the nucleuns of a nucleus are divided, then total mass increases. Where does this mass come from?

This mass is received from the binding energy of the nucleus.

Write the nuclear equations for:
${\beta}^{-}$ - decay of $_{ 15 }^{ 32 }{ P }$

$\quad _{ 15 }^{ 32 }{ P }\rightarrow _{ 16 }^{ 32 }{ S }+_{ -1 }^{ 0 }{ e }+\overline { v }$

$C^{14}$ disintegrates by $\beta-emission$ with a reaction energy (Q value) of 0.155 MeV. A $\beta-particle$ with an energy of 0.025 MeV is emitted in a direction at $135^o$ to the direction of motion of the recoil nucleus. Determine the momenta of the three particles $(\beta^-=\overline v, ^{14}N)$ involve in this disintegration in MeV/c units (where c is speed of light in vacuum). $(M_0=0.511 MeV/c^2)$

We know that the

$Q$ value of the reaction

${ E }_{ 0 }={ E }_{ v }+{ E }_{ \beta }\Rightarrow { E }_{ v }={ E }_{ 0 }-{ E }_{ \beta }=0.155-0.025=0.130MeV$
Hence, the momentum of the neutrino

$\quad { p }_{ v }={ E }_{ v }/c=0.13MeV/c$
Momentum of a

$\beta$ particle can be obtained from the relation

${ p }_{ \beta }=\sqrt { \left( 2{ M }_{ e }{ E }_{ \beta } \right) } =\sqrt { 2\times 0.511\times 0.025 } MeV/c=0.158MeV/c$
Using conservation of momentum of the system, we have

${ p }_{ \beta }\sin { { 45 }^{ o } } ={ p }_{ v }\sin { \theta } \Rightarrow \sin { \theta } =\left( { p }_{ \beta }/{ p }_{ v } \right) \sin { { 45 }^{ o } } =0.859;$

${ p }_{ R }={ p }_{ \beta }\cos { { 45 }^{ o } } +{ p }_{ v }\cos { \theta } =0.158(1/\sqrt { 2 } )+0.13\sqrt { \left[ 1-{ \left( 0.859 \right) }^{ 2 } \right] } =0.179MeV/c$

A nuclear reaction is given as
$p+^{15}N\rightarrow _Z^AX+n$
Find, A, Z and identify the nucleus X.

The nuclear reaction will be

$_1^1p+_7^{15}N=_Z^AX+_0^1n$

Now,

$1+15=A+1 \Rightarrow A=15$ and

$1+7=Z+0 \Rightarrow Z=8$

The nucleus would be isotope of oxygen i.e

$_8^{15}O$

Find the Q value of the reaction
$N^{14}+\alpha \rightarrow O^{17}+p$
The masses of $N^{14}, He^4, p,$ and $O^{17}$ are, respectively, 14.00307 u, 4.00260 u, 1.00783 u, and 16.99913 u. Find the total kinetic energy of the products if the striking $\alpha-$ particle has the minimum kinetic energy required to initiate the reaction:

${ N }^{ 14 }+\alpha \rightarrow { O }^{ 17 }+p$

Energy = E =

$\triangle mc^2$

$\triangle m=14.00307+4.00260-1.00783-16.99913 \\=-0.00129 \\ E=\triangle mc^2 \\ =-0.00129\times 931\; MeV \\=-1.20099\; MeV$

So, 1.20099 MeV energy is required to initiate the reaction.

Find whether $\alpha$ - decay or any of the $\beta-$ decay are allowed for $_{89}^{226}Ac$ . Given masses are:
$M(_{89}^{226}Ac)=226.028356 amu$
$M(_{87}^{222}Fr)=222.017415 amu$
$M(_{90}^{226}Th)=226.017388 amu$
$M(_{88}^{226}Ra)=226.025406 amu$
$M(_2^4He)=4.002603 amu$

$M\left( _{ 87 }^{ 222 }{ Fr } \right) \rightarrow \alpha -decay\\ _{ 87 }^{ 222 }{ Fr }+_{ 2 }{ He }^{ 4 }\rightarrow _{ 89 }^{ 226 }{ Ac }\\ M\left( _{ 90 }^{ 226 }{ Th } \right) \rightarrow \beta -decay\\ _{ 90 }^{ 226 }{ Th }+_{ -1 }^{ o }{ \beta }\rightarrow _{ 89 }^{ 226 }{ Ac }\\ M\left( _{ 88 }^{ 226 }{ Ra } \right) \rightarrow \beta -decay\\ _{ 88 }^{ 226 }{ Ra }+_{ -1 }^{ o }{ \beta }\rightarrow _{ 89 }^{ 226 }{ Ac }$

The nuclear reaction $n+_5^{10}B\rightarrow _3^7Li+_2^4He$ is observed to occur even when very slow-moving neutrons $(M_n=1.0087 amu)$ strike a boron atom at rest. For a particular reaction in which $K_n=0$ , the helium $(M_{He}=4.0026 amu)$ is observed to have a speed of $9.30\times 10^6 ms^{-1}$ . Determine (a) the kinetic energy of the lithium $(M_{Li}=7.0160 amu)$ and (b) the Q value of the reaction.

Energy released by the reaction

$=Q=\triangle m{ c }^{ 2 }$

$Q= \triangle m=1.0087+10.811-7.0160-4.0026 \\=0.8011 \\ Q=\triangle m{ c }^{ 2 } =0.8011\times 931 \\=745.82MeV \\=754.82\times 1.6\times 10^{-13} \\=1193.3\times 10^{-13}J$

K.E. of lithium

$=\cfrac{1}{2}mv^2 =\cfrac{1}{2}\times 7.0160\times (9.30\times 10^6)^2 \\=303.40\times 10^{12}J$

A nuclear reaction is given as
$p+^{15}N\rightarrow _Z^AX+n$
Find the Q value of the reaction

The nuclear reaction will be

$_1^1p+_7^{15}N=_Z^AX+_0^1n$

Now,

$1+15=A+1 \Rightarrow A=15$ and

$1+7=Z+0 \Rightarrow Z=8$

The nucleus would be isotope of oxygen i.e

$_8^{15}O$

The atomic mass of proton

$=1.00727$ amu ; atomic mass of neutron

$=1.00866$ amu ; atomic mass of

$^{15}O=15.9994$ amu and atomic mass of

$^{15}N=15.000108$ amu

Sum of masses of reactants

$=1.00727+15.000108=16.0074$ amu

Sum of masses of products

$=15.9994+1.00866=17.0081$ amu

The Q-value

$=$ Mass difference

$\Delta M=17.0081-16.0074=1.0007$ amu

$=1.0007\times 931 MeV=931.65 MeV$

Calculate the ground state Q value of the induced fission reaction in the equation
$n+_{92}^{235}U\rightarrow _{92}^{236}U^*\rightarrow _{40}^{99}Ze+_{52}^{134}Te+3n$
if the neutron is thermal. A thermal neutron is in the thermal equilibrium with its environment; it has an average kinetic energy given by (3/2) kT. Given: $$m(n)=1.0087 amu, M(^{235}U)=235.0439 amu, M(^{99}Zr)=98.916 amu M(^{134}Te)=133.9115 amu$$

Q value is equal to mass-energy equivalence

$Q=\triangle m{ c }^{ 2 }\\ \triangle m={ m }_{ (n) }+{ m }_{ (^{ 235 }{ U }) }-{ m }_{ (^{ 99 }{ Ze }) }-{ m }_{ (^{ 134 }{ Te }) }-3{ m }_{ (n) }\\ ={ m }_{ (^{ 235 }{ U }) }-{ m }_{ (^{ 99 }{ Ze }) }-{ m }_{ (^{ 134 }{ Te }) }-2{ m }_{ (n) }\\ =235.0439-98.916-133.9115-2\times 1.0087\\ =0.199amu\\ Q=0.199\times 931MeV\\ Q=185.269MeV$

The binding energy of $_{17}^{35}Cl$ nucleus is 298 MeV. Find its atomic mass. Given, mass of a proton $(m_p)=1.007825 amu$ , mass of neutron $(m_n)=1.008665 amu$

Here proton number

$Z=17$ and neutron number,

$N=A-Z=35-17=18$

Binding energy,

$BE=\Delta m c^2$

Mass defect ,

$\Delta m=298 MeV/c^2=298/931.5 amu=0.3199 amu$ as

$(1 amu=931.5 MeV/c^2)$

If M be atomic mass,

$\Delta m=(17m_p+18m_n)-M$

$M=(17\times 1.007825 +18\times 1.008665)-0.3199=34.9597 amu$

What is meant by binding energy of nucleus? Determine the binding energy per nucleon of an $\alpha$ particle if the masses of proton, neutron and $\alpha$ - particle are $1.00728amu$ , $1.00867amu$ and $4.00150amu$ respectively.
(Given: $1amu=931MeV$ )

Nuclear binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts. These component parts are neutrons and protons, which are collectively called nucleons.

Binding energy for

$\alpha$ -particle ia given by

$\Delta m.c^2$ .

$\Delta m=4.00150-2.(1.00728+1.00867)$

$\Delta m=0.0304amu$

Total binding energy is

$=0.0304*931$

$=28.3$

So binding energy per nucleon is

$\frac{28.3}{4}=7.0756$ .

Calculate the binding energy per nucleon for $_{10}^{20}Ne, _{26}^{56}Fe$ and $_{92}^{238}U$ . Given that mass of neutron is 1.008665 amu, mass of proton is 1.007825 amu, mass of $_{10}^{20}Ne$ is 19.9924 amu, mass of $_{26}^{56}Fe$ is 55.93492 amu, $_{92}^{238}U$ is 238.050783 amu

For

$_{10}^{20}Ne$ : mass defect ,

$\Delta m=[10m_p+(20-10)m_n]+M_{Ne}$

$\Delta m=[10\times 1.007825+10\times 1.008665]-19.9924=0.1725 amu$

Binding energy per nucleon

$BE=\dfrac{\Delta m c^2}{A}=\dfrac{0.1725 c^2}{20}=0.0086c^2 amu=0.0086\times 931.5=8.03 MeV$ as

$(1 amu=931.5 MeV/c^2)$

For

$_{26}^{56}Fe$ : mass defect ,

$\Delta m=[26m_p+(56-26)m_n]+M_{Fe}$

$\Delta m=[26\times 1.007825+30\times 1.008665]-55.93492=0.5285 amu$

Binding energy per nucleon

$BE=\dfrac{\Delta m c^2}{A}=\dfrac{0.5285 c^2}{56}=0.0086c^2 amu=0.0094\times 931.5=8.76 MeV$ as

$(1 amu=931.5 MeV/c^2)$

For

$_{92}^{238}U$ : mass defect ,

$\Delta m=[92m_p+(238-92)m_n]+M_{Ne}$

$\Delta m=[92\times 1.007825+146\times 1.008665]-238.050783=1.934 amu$

Binding energy per nucleon

$BE=\dfrac{\Delta m c^2}{A}=\dfrac{1.934 c^2}{238}=0.0086c^2 amu=0.0081\times 931.5=7.57 MeV$ as

$(1 amu=931.5 MeV/c^2)$

Find the binding energy of an $\alpha-$ particle from the following data.
Mass of the helium nucleus $=4.001265 amu$
Mass of proton $=1.007277 amu$
Mass of neutron $=1.00866 amu$
(take $1 amu=931.4813 MeV)$

The alpha particle has two proton and two neutron.

The combined mass of nuclei

$=$ net mass of proton

$+$ net mass of neutron

$=(2\times 1.007277)+(2\times 1.00866)=4.031874$ amu

Binding energy

$=$ mass defect

$=4.031874-4.001265=0.0306 amu=0.0306(931.48)=28.5 MeV$

Find whether $\alpha-decay$ or any of the $\beta-decay$ are allowed for $_{89}^{226}Ac$

Our first step will be to write the reaction, then find the disintegration energy

$Q$ .

$Q > 0$ , the decay is allowed.

$\alpha-decay: _{89}^{226}Ac\rightarrow _{87}^{222}Fr+\alpha$

$Q=[M(_{89}^{226}Ac)-M(_{87}^{222}Fr)-M(^4He)]c^2$

$=5.50 MeV$ (

$\alpha$ decay is allowed)

$\beta^-decay: _{89}^{226}Ac\rightarrow _{90}^{226}Th+\beta^-+\overline v$

$Q=[M(_{89}^{226}Ac)-M(_{90}^{226}Th)]c^2$

$=1.12 MeV$ (

$\beta^-$ decay is allowed)

$\beta^+decay: _{89}^{226}Ac\rightarrow _{88}^{226}Ra+\beta^++v$

$Q=[M(_{89}^{226}Ac)=M(_{88}^{226}Ra)-2m_e]c^2$

$=-0.38 MeV$ (

$\beta^+$ decay is not allowed)

Electron capture:

$_{89}^{226}Ac+e^-\rightarrow _{88}^{226}Ra+v$

$Q=[M(_{89}^{226}Ac=M(_{88}^{226}Ra)]c^2$

$=0.64 MeV$ (Electron capture is allowed)

Write the decay equations and expressions for the disintegration energy Q of the following decay: $\beta^-$ decay, $\beta^+$ decay, electron capture

In all the cases discussed here, we neglect any neutrino mass.

$\beta^- decay: M_{nucl}(_Z^AX)=M_{nucl})_{Z+1}^AD)+m_e+Q/c^2$
where,

$M_{nucl}$ indicates the nuclear mass.

In order to convert it to atomic mass, we add

$Zm_e$ on both sides:

$M_{nucl}(_Z^AX)+Zm_e=M_{nucl}(_{Z+1}^AD)+(Z+1)m_e+Q/c^2$

Since atomic binding energies are less than nuclear binding energies, they are neglected on the two sides of the equation.

$[M(_Z^AX)=M(_{Z+1}^AD)]c^2$

$Q$ for this equation

$=[M(_Z^AX)=M(_{Z+1}^AD)]c^2$

$\beta^+decay: M_{nucl}(_Z^AX)=M_{nucl}(_{Z-1}^AD)+m_e+Q/c^2$

In this case, only (Z-1)m is needed for the daughter atomic mass which gives us a remaining mass of

$2m_e$ .

$[M(_Z^AX)-M(_{Z-1}^AD)-2m_e]c^2$

$Q$ for this equation

$=[M(_Z^AX)-M(_{Z-1}^AD)-2m_e]c^2$

We add

$(Z-1)m_e$ to each side above and obtain:

$M_{nucl}(_Z^AX)+Zm_e=M_{nucl}(_{Z-1}^AD)+(Z-1)m_e+Q/c^2$

In this case, electron masses are balanced, i.e, cancelled out in equation:

$M(_Z^AX)-M(_{Z-1}^AD)+Q/c^2$

$Q$ for this equation

$=[M(_Z^AX)-M(_{Z-1}^AD)]c^2$

What is the minimum photon energy required to remove the least bound neutron of $_{20}^{40}Ca$ and $_{18}^{40}Ar$ . The necessary atomic masses (in u) are given below:
$M(^{40}Ca)=39.062591 u$
$M(^{39}Ca)=38.970719 u$
$M(^{40}Ca)=39.962383 u$
$M(^{39}Ar)=38.964314 u$
$m_n=1.008665 u$

Total mass (neutron plus product nucleus) after the removal is expected to be more than the initial nucleus mass.
This mass difference when expressed in energy units (MeV) is the binding energy of the least bound neutron.
The mass after break-up

$<(_{30}^{39}Ca)=38.970719 u$

$m_n=1.08665 u$
Total mass

$=39.979384 u$
The mass difference is,

$\Delta m=[M(^{39}Ca)+m_n]-m(^{40}Ca)$

$=(39.979384-39.962591)\\ =0.016793 u$

$=(0.016793)\times (931.5)MeV\\ =15.64 MeV$

So, the binding energy of least bound neutron is

$15.64\ MeV$ .
Similarly, for

$_{20}^{39}Ar$ ,

$\Delta m=[M(_{20}^{39}Ar)+m_n]-M(_{18}^{40}Ar)]$

$=(38.964314+1.008665)-39.962383 u$

$=0.010596 u=(0.010596)\times (931.5) MeV$

$=9.870\ MeV$

The inert elements are relatively unreactive because their outer shells of electrons are full. Large energies are involved in gaining or losing electrons which is therefore unlikely. An analogous behaviour takes place in the nucleus. Experimental evidence does indicate the existence of 'closed nuclear shell' when the number of protons or neutrons is

$2, 8, 20, 28, 50, 82$ or

$126$ although the concepts of individual nucleon 'orbits' and filled shells inside the nucleus is hard to believe.
The above mentioned number of protons or neutron are called magic numbers. The elements with magic number of protons have unusually high number of stable isotope. For example, aluminium with 13 protons has just one stable isotope

$^{27}Al$ , but tin with

$Z=50$ (a magic number) has 10 stable isotopes ranging from

$N=62$ to

$N=74$ , whereas neighboring indium with

$Z=49$ and antimony with

$Z=51$ have only two.
Thus, the magic numbers are associated with extra binding energies, implying higher stability.
Thus, in the given example it takes about 6 MeV more energy to remove a paired neutron from a pair that completes a closed nuclear shell (a magic number) than from a pair that does not complete a shell.

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energy of the nuclei $^{27}_{13}Al$ from the following data:
$m(^{26}_{13}Al)\, =\, 25.986895\, u$
$m(^{27}_{13}Al)\, =\, 26.981541\, u$

The reaction for the neutron separation from the aluminum atom is

$^{27}_{13} Al + E \rightarrow \ ^{26}_{13} Al + \ ^{1}_0 n$

We know that, the separation energy can be calculated as:

$E = (m (^{26}_{13} Ca) + m (^1_0n) - m (^{27}_{13} Ca)) c^2$

$= (25.986895 + 1.008665 - 260981541) c^2$

$= (0.014019) u \times c^2$

On the other hand

$1u = 931.5 MeV/c^2$

So, the energy of neutron separation will be:

$E = (0.014019) \times 931.5$

$= 13.059 MeV$

Therefore to remove a neutron from the nucleus

$13.059 MeV$ of energy is required.

The radio nuclide $^{11}C$ decays according to
$^{11}_{6}C\, \rightarrow\, ^{11}_5B\, +\, e^+\, +\, \nu$ : $T_{1/2}$ $= 20.3\ min$
The maximum energy of the emitted positron is $0.960\ MeV$ .
$m\, (^{11}_{6}C)\, =\, 11.011434\, u$ and $m\, (^{11}_{6}B)\, =\, 11.009305\, u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.

$^{11}_{6}C\, \rightarrow\, ^{11}_{6}B\, +\, e^+\, +\, \nu\, +\, Q$

$Q\, =\, \left [ m_N\left ( ^{11}_6C \right )\, -\, m_N\left ( ^{11}_6B \right )\, -\, m_e \right ]\, c^2$
Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add

$6m_e$ in case of

$^{11}C$ and

$5m_e$ in case of

$^{11}B$ . Hence,

$Q\, =\, \left [ m\left ( ^{11}_6C \right )\, -\, m\left ( ^{11}_6B \right )\, -\, 2m_e \right ]\, c^2$ (Note

$m_e$ has been doubled)
Using given masses, Q = 0.961 MeV.
Q =

$E_d\, +\, E_e\, +\, E_\nu$
The daughter nucleus is too heavy compared to

$e^+$ and

$\nu$ , so it carries negligible energy (

$E_d$

$\approx$ 0). If the kinetic energy (

$E_\nu$ ) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum

$E^e$

$\approx$ Q.

Find the Q-value and the kinetic energy of the emitted $\alpha$ - Particle in the $\alpha$ -decay of (a) $^{220}_{86}Rn$ and (b) $^{226}_{88}Ra$ .
Given:
$m\, (^{226}_{88}Ra)\, =\, 226.02540\, u$ ,
$m\, (^{222}_{86}Rn)\, =\, 222.01750\, u$ ,
$m\, (^{216}_{84}Po)\, =\, 216.00189\, u$

(a)

$^{226}_{88}{Ra} \rightarrow ^{222}_{86}{Rn} + ^{4}_{2}{\alpha}$

$^{222}_{86}{Rn} \rightarrow ^{216}_{84}{Po} + ^{4}_{2}{\alpha}$

$Q= \Delta m \times 931 MeV$

$\Delta m= 226.02540-222.01750-4.002603= 0.005297$

Q = 4.93 MeV,

$E_\alpha = 4.85 MeV$

Kinetic energy

$\displaystyle = \frac{Mass \:no.\: after\: decay}{Mass\: no.\: before\: decay} \times Q$

$K.E= \displaystyle \frac{222}{226} \times 4.93 = 4.85 MeV$

(b)

$^{222}_{86}{Rn} \rightarrow ^{216}_{84}{Po} + ^{4}_{2}{\alpha}$

$Q= \Delta m \times 931 MeV$

$\Delta m= 220.01137-216.00189-4.002603= 0.006877$

Q = 6.4 MeV,

$E_\alpha = 6.29 MeV$

Kinetic energy

$\displaystyle = \frac{Mass\: no.\: after\: decay}{Mass\: no.\: before\: decay} \times Q$

$K.E= \displaystyle \frac{216}{222} \times 6.4 = 6.29 MeV$

Consider the fission of $^{238}_{92}U$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are $^{140}_{58}Ce$ and $^{99}_{44}Ru$ . Calculate Q for this fission process. The relevant atomic and particle masses are:
$m(^{238}_{92}U) = 238.05079 u$
$m(^{140}_{58}Ce )\, =\,139.90543\, u$
$m(^{99}_{44}U) =98.90594 u$ .

The reaction for the fission of

$_{93}^{238}U$ by fast neutrons is as shown

$_{93}^{238}U +_{0}^{1}n \rightarrow _{58}^{140}Ce + _{44}^{99}U$

$Q\,=\, [m(^{238}_{92}U)\, +\, m_N\, -\, m\, (^{140}_{58}Ce)\, -\, m(^{99}_{44}Ru)] \times$ 931 .5 MeV

$Q\,=\, [ 238.05079\, +\, 1.00893\, - 139.90543 \, - 98.90594] \times$ 931 .5 MeV

$Q\,=\, 0.24835 \times$ 931 .5 MeV

$Q=\, 231.1\,$ MeV

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an $\alpha$ -particle. Consider the following $^{223}_{88}Ra\, \rightarrow\, ^{219}_{86}Rn\, +\, ^{4}_2He$
Calculate the Q-values for these decays and determine that both are energetically allowed.

(i) In case of first process,

$^{223}_{88} Ra \rightarrow \ ^{209}{82} Pb + \, ^{14}_6 C + Q$

The mass defect occured in reaction is,

$\Delta m = mass\ of\ Ra^{223} - (mass \ of \ Pb^{209} + mass \ of \ C^{14})$

$= 223.01850 - (208.98107 + 14.00324)$

$= 0.03419 u$

So, the amount of energy released is given by:

$Q = \Delta m\times931\ MeV$

$Q = 0.03419 \times 931 MeV = 31.83 MeV$

(ii) Now, in case of second reaction,

$^{223}_{88} Ra \rightarrow \ ^{219}_{86} Rn + \, ^4_2 He + Q$

Mass defect is given as:

$\Delta m = mass\ of\ Ra^{223} - (mass\ of\ Rn^{219} + mass\ of\ He^4)$

$=223.01850 - (219.00948 + 4.00260)$

$= 0.00642 u$

$\therefore$ the Energy released in the reaction is:

$Q = 0.00642 \times 931 MeV = 5.98 MeV$

Since both reactions have positive Q-value so both reactions are possible.

For the $\beta^+$ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K shell, is captured by the nucleus and a neutrino is emitted).
$e^+\, +\, ^A_ZX\, \rightarrow\, ^A_{Z - 1}Y\, +\, v$
Show that if $\beta^+$ emission is energetically allowed, electron capture is necessarily allowed but not vice versa.

Consider the competing processes:

$^A_ZW\, \rightarrow\, ^A_{Z - 1}\, Y\, +\, e^+\, +\, v_e\, +\, Q_1$ (positron cature)

$e^-\, +\, ^A_ZW\, \rightarrow\, ^A_{Z - 1}Y\, +\, v_e\, +\, Q_2$ (electron capture)

$Q_1\, =\, [m_N(^A_ZX)\, -\, m_N\, (^A_{Z - 1}Y)\, -\, m_e]\, c^2$

$=\, [m_N\,(^A_ZX)\, -\, Zm_e\, -\, m(^A_{Z - 1}Y)\,-\, (Z\,-\,1)\,m_e\,-\, m_e ]\, c^2$

$=\, [m_N(^A_ZX)\, -\, m_N\, (A_{Z - 1}Y)\, -\, 2m_e]\,c^2$

$Q_2\, =\, [m_N(^A_ZX)\, +\, m_e\, -\, m_N\, (^A_{Z - 1}Y)]\,c^2\, =\, [m(^A_ZX)\, -\, m\, (^A_{Z - 1}Y)]\, c^2$
This means

$Q_1\, >\, 0$ implies

$Q_2\, >\, 0$ , but

$Q_2\, >\, 0$ does not necessarily mean

$Q_1\, >\, 0$ .

Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstance. calculate the ground state energy of a $He-$ atom

For H atom

$Z = 1$ and

$n = 1$

$\therefore E_n = \dfrac{- m_ge^4}{8\varepsilon _0^{2} 1^2h^2} = 13.6 eV$
For He- atom,

$Z = 4$ and

$n = 1$

$\therefore E_n = \dfrac{- m_ge^4}{8\varepsilon _0^{2} 1^2h^2} = -4 \times 13.6 = 54.4 eV$

A neutron with kinetic energy $K=10\ MeV$ activates a nuclear reaction
$n+^{ 12 }{ C }\quad \rightarrow {^{ 9 } Be+\alpha }$
whose threshold ${ E }_{ th }=6.17\ MeV$ . Find the kinetic energy of the alpha particles outgoing at right angles to the direction of the incoming neutron.
[Take $u=931.1\ MeV$ ]

Conservation of energy gives,

${ m }_{ 1 }{ c }^{ 2 }+{ m }_{ 2 }{ c }^{ 2 }+{ K }_{ 1 }={ m }_{ 3 }{ c }^{ 2 }+{ m }_{ 4 }{ c }^{ 2 }+{ K }_{ 3 }+{ K }_{ 4 }$
or

$Q+{ K }_{ 1 }={ K }_{ 3 }+{ K }_{ 4 }$ ... (

$1$ )
Momentum conservation along

$x-$ axis.

$\sqrt { 2{ m }_{ 1 }{ K }_{ 1 } } =\sqrt { 2{ m }_{ 4 }{ K }_{ 4 } } \cos { \theta }$
or,

${ m }_{ 1 }{ K }_{ 1 }={ m }_{ 4 }{ K }_{ 4 }\cos ^{ 2 }{ \theta }$ ...(

$2$ )
Along

$y-$ axis

$\sqrt { 2{ m }_{ 3 }{ K }_{ 3 } } =\sqrt { 2{ m }_{ 4 }{ K }_{ 4 } } \sin { \theta }$
or,

${ m }_{ 3 }{ K }_{ 3 }={ m }_{ 4 }{ K }_{ 4 }\sin ^{ 2 }{ \theta }$ (

$3$ )
Adding (

$2$ ) and (

$3$ )

${ m }_{ 1 }{ K }_{ 1 }+{ m }_{ 3 }{ K }_{ 3 }={ m }_{ 4 }{ K }_{ 4 }$
or,

${ K }_{ 4 }=\dfrac { { m }_{ 1 } }{ { m }_{ 4 } } { K }_{ 1 }+\dfrac { { m }_{ 3 } }{ { m }_{ 4 } } { K }_{ 3 }$
Substituting the value of

${ K }_{ 4 }$ in equation(

$1$ ), we get

$Q+\left( 1-\dfrac { { m }_{ 1 } }{ { m }_{ 4 } } \right) { K }_{ 1 }=\left( 1+\dfrac { { m }_{ 3 } }{ { m }_{ 4 } } \right) { K }_{ 3 }$
Here,

$Q=\dfrac { { -E }_{ th } }{ \left( 1+\dfrac { { m }_{ 1 } }{ { m }_{ 2 } } \right) } =\dfrac { -6.17 }{ \left( 1+\dfrac { 1 }{ 12 } \right) } =-6.17\left( \dfrac { 12 }{ 13 } \right) =-5.69\ MeV$
Thus

$\left( 1+\dfrac { 4 }{ 9 } \right) { K }_{ 3 }=-5.69+\left( 1-\dfrac { 1 }{ 9 } \right) \left( 10 \right)$

${ K }_{ 3 }=\dfrac { 9 }{ 13 } \left[ -5.69+8.89 \right] =2.21\ MeV$
1038461_1013736_ans_107c985cbe2a4249a96d76ef0a5bc30c.png

A radio nuclide $A_1$ with decay constant $\lambda_1$ transforms into a radio nuclide $A_2$ with decay constant $\lambda_2$ . Assuming that at the initial moment, the preparation contained only the radio nuclide $A_1$
(a) Find the equation describing accumulation of radio nuclide $A_2$ with time.
(b) Find the time interval after which the activity of radio nuclide $A_2$ reaches its maximum value.

(a) Suppose

$N_1$ and

$N_2$ are the number of two radionuclides

$A_1 , A_2$ at time t.

Then

$\dfrac{d N_1}{dt} = -\lambda_1 N_1$

$(1)$

$\dfrac{d N_2}{dt} = \lambda_1 N_1 - \lambda_2 N_2$

$(2)$

From

$(1)$

$N_1 = N_{10} e^{-\lambda_1 t}$

where

$N_{10}$ is the initial number of nuclides

$A_1$ at time

$t = 0$

From (2)

$\left(\dfrac{d N_2}{dt} + \lambda_2 N_2\right) e^{\lambda_2 t} = \lambda_1 N_{10} e^{-(\lambda_1 - \lambda_2) t}$

$(N_2 e^{\lambda_2 t}) =$ const

$\dfrac{\lambda_1 N_{10}}{\lambda_1 - \lambda_2} e^{-(\lambda_1 - \lambda_2)t}$

since

$N_2 = 0$ at

$t = 0$

constant

$N_2 = \dfrac{\lambda_1 N_{10}}{\lambda_1 - \lambda_2}$

Thus =

$\dfrac{\lambda_1 N_{10}}{\lambda_1 - \lambda_2} (e^{-\lambda_2 t - e^{-\lambda_1 t}})$

(b) The activity of nuclide

$A_2$ is

$\lambda_2 N_2$ . This is maximum when

$N_2$ is maximum.

That happens when

$\dfrac{d N_2}{dt} = 0$

This require

$\lambda_2 e^{-\lambda_2t_m = \lambda_1 e^{-\lambda_2 t_m}}$

$t_m = \dfrac{ln (\lambda_1/\lambda_2)}{\lambda_1 - \lambda_2}$

A stationary $Pb^{200}$ nucleus emits an $\alpha -$ particle with a $K.E.$ , $K=5.77\ MeV$ . Find the recoil velocity of daughter nucleus. What fraction of the total energy liberated in this decay is accounted for by the daughter nucleus?

The momentum of the

$\alpha -$ particle is given by,

$p_{\alpha}=\sqrt {2m_{\alpha k}}$ . . .(

$1$ )
Let the recoiled momentum of the daughter nucleus be

$p_{d}=m_{d}\ v_{d}$ where

$m_{d}$ and

$v_{d}$ are the mass and velocity of daughter nucleus respectively. Using the principle of conservation of momentum we get,

$P_{d}=P_{\alpha}=\sqrt {2m_{\alpha k}}$

$\Rightarrow \quad v_{ d }=\dfrac { \sqrt { { 2m }_{ \alpha }k } }{ { m }_{ d } }$ . . .(

$2$ )

$\Rightarrow \quad v_{ d }=\dfrac { 1 }{ 196 } \sqrt { \dfrac { 2\times\ 4\times\ k }{ m_{ p } } }$

$=\dfrac { 2 }{ 196 } \sqrt { \dfrac { 2k }{ m_{ p } } } \quad \quad \quad$ where

$m_{p}$ is the mass of the proton.

$\Rightarrow \quad v_{d}=3.39 \times 10^{5}\ m/s$
Let the

$K.E.$ of the daughter nucleus be

$K$ , then,

$\dfrac {k}{k}=\dfrac {m_{\alpha}}{m_{d}}$ as the momenta are same.

$\therefore \quad \dfrac {k}{k+k}=\dfrac {m_{\alpha}}{m_{\alpha}+_{d}}$
But

$K+K$ is the total energy,

$K_{t}$ (say).

$\therefore \dfrac { k }{ k_{t} }=\dfrac { m_{ \alpha } }{ m_{ \alpha }+m_{ d } }$

$\Rightarrow \quad k=\dfrac { m_{ \alpha } }{ m_{ \alpha }+m_{ d } } k_{1}=\dfrac {4}{196+4}k_{t}$

$\Rightarrow \quad k=0.02\ k_{t}$

$\Rightarrow \dfrac {k}{k_{t}}=0.02$ .

Show that the minimum energy needed to separate a proton from a nucleus with $Z$ protons and $N$ neutrons is
$\triangle E = (M_{Z - 1, N} + M_{H} - M_{Z, N})C^{2}$
where $M_{Z, N} =$ mass of an atom with $Z$ protons and $N$ neutrons in the nucleus and $M_{H} =$ mass of a hydrogen atom. This energy is known as proton - separation energy.

1978101_1173797_ans_89d4bca8d2a04b8eb4e295243807af88.jpeg

Suppose the potential energy between electron and proton at a distance $r$ is given by $U=-ke$ in (r), where $k$ is a positive constant. Use Bohr's theory to obtain the energy of $n^{th}$ energy level for such an atom.

Since H atom is a bounded system U cannot be positive

$U=\dfrac { { -ke }^{ 2 }2 }{ { 3r }^{ 3 } }$

Electrostatic force of attraction at a distance r will be

$F=\dfrac { -dU }{ dr } =\dfrac{ { -ke }^{ 2 } }{ { r }^{ 4 } }$

$\dfrac { { mv }^{ 2 } }{ r } =\dfrac { { -ke }^{ 2 } }{ { r }^{ 4 } }$

${ mv }^{ 2 }=\dfrac { { ke }^{ 2 } }{ { r }^{ 3 } }$

$\dfrac { 1 }{ 2 } { mv }^{ 2 }=\dfrac { { ke }^{ 2 } }{ { 2r }^{ 3 } }$

$KE=\dfrac { { ke }^{ 2 } }{ { 2r }^{ 3 } }$

$E=KE+U$

$=ke\dfrac { 2 }{ { 2r }^{ 3 } } +\left( -ke\dfrac { 2 }{ { 3r }^{ 3 } } \right)$

$=\dfrac { { ke }^{ 2 } }{ { 6r }^{ 3 } }$

In the fusion reaction,
$_{1}^{2}H + _{1}^{2}H \rightarrow _{2}^{3}He + _{0}^{1}n$
The masses of deuteron, helium and neutron are $2.015u, 0.017u$ and $1.009u$ respectively. If $1\ kg$ of deuteron completely undergoes fusion, then calculate free energy.
$[1u = 9.31\ MeV/c^{2}]$ .

1966712_1348840_ans_7d67f67e34aa48d3ac48a53b1c911e3f.jpeg

The mass defect for the nucleus of lithium is $0.058135$ $amu$ . What is the binding energy(to the nearest integer) per nucleon for lithium in $MeV$ will be_______.

2089775_1390732_ans_0c3bcc7e7429493d8314402fa63d39b3.jpg

In a nuclear reactor, fission is produced in $1g$ of $_{ }^{ 235 }{ U }(235.0349a.m.u)$ . In assuming that $_{ 53 }^{ 92 }{ Kr }(91.8673a.m.u)$ and $_{ 36 }^{ 141 }{ Ba }(140.9139a.m.u)$ are produced in all reactions and no energy is lost, write the complete reaction and calculate the total energy produced in kilowatt hour. Given: $(1 a.m.u=931MeV)$

The nuclear fission reaction in the above process is

$_{ 92 }^{ 235 }{ U }+_{ 0 }^{ 1 }{ n }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+3_{ 0 }^{ 1 }{ n }$

The sum of the masses before reaction is

$235.0439+1.0087=236.0526a.m.u$

The sum of masses after reaction is

$140.9139+91.8973+(1.0087)=235.8375a.m.u$

Mass defect

$\Delta m=236.0526-235.8373=0.2153a.m.u$

Energy released in the fission of

$_{ }^{ 235 }{ U }$

$\Delta E=\Delta m\times 931.2MeV=0.2153\times 931.2=200MeV$

Number of atoms in

$1g$ of

$_{ }^{ 235 }{ U }$

$\quad N=\cfrac { 6.02\times { 10 }^{ 23 } }{ 235 } =2.56\times { 10 }^{ 21 }$

Energy released in fission of

$1g$ of

$_{ }^{ 235 }{ U }$

$E=200\times 2.56\times { 10 }^{ 21 }MeV$

$\Rightarrow E=5.12\times { 10 }^{ 23 }MeV$

$\Rightarrow E=8.2\times { 10 }^{ 10 }J=\cfrac { 8.2\times { 10 }^{ 10 } }{ 3.6\times { 10 }^{ 6 } } kWh=2.28\times { 10 }^{ 4 }kWh$

The nuclear reaction
$n+_{ 5 }^{ 10 }{ B }\rightarrow _{ 3 }^{ 7 }{ Li }+_{ 2 }^{ 4 }{ He }$
is observed to occur even when very slow-moving neutrons $({M}_{n}=1.0087a.m.u)$ strike a boron atom at rest. For a particular reaction in which ${K}_{n}=0$ , the helium $({M}_{He}=4.0026a.m.u)$ is observed to have a speed of $9.30\times {10}^{6}m{s}^{-1}$ . Determine (a) the kinetic energy of the lithium $({M}_{Li}=7.0160a.m.u)$ and (b) the $Q$ value of the reaction.

(a) Since the neutron and boron are both initially at rest. The total momentum before the reaction is zero and afterward also it is zero. Therefore

${ M }_{ Li }{ n }_{ Li }={ M }_{ He }{ v }_{ He }$

We solve this for

${v}_{Li}$ and substitute it into the equation for kinetic energy. We can use classical kinetic energy with little error, rather than relativistic formulas, because

${ n }_{ He }=9.30\times { 10 }^{ 6 }m{ s }^{ -1 }$ is not close to the speed of light

$c$ and

${n}_{Li}$ will be even less since

${ n }_{ He }=9.30\times { 10 }^{ 6 }m{ s }^{ -1 }$ . Thus we can write

${ K }_{ Li }=\cfrac { 1 }{ 2 } { M }_{ Li }{ v }_{ Li }^{ 2 }=\cfrac { 1 }{ 2 } { M }_{ Li }{ \left( \cfrac { { M }_{ He }{ v }_{ He } }{ { M }_{ Li } } \right) }^{ 2 }=\cfrac { { M }_{ He }^{ 2 }{ v }_{ He }^{ 2 } }{ 2{ M }_{ Li } }$

We put in numbers, changing the mass in

$u$ to

$kg$ and recalling that

$1.60\times {10}^{-13}J=1MeV$

${ K }_{ Li }=\cfrac { { \left( 4.0026 \right) }^{ 2 }\left( 1.66\times { 10 }^{ -27 } \right) { \left( 9.30\times { 10 }^{ 6 } \right) }^{ 2 } }{ 2\left( 7.0160 \right) \left( 1.66\times { 10 }^{ -27 } \right) } =1.02MeV$

(b) We are given the data

${ K }_{ \alpha }={ K }_{ X }=0;Q={ K }_{ Li }+{ K }_{ He }$ where

${ K }_{ He }=\cfrac { 1 }{ 2 } { M }_{ He }{ v }_{ He }^{ 2 }$

$K_{He}=\cfrac { 1 }{ 2 } { \left( 4.0026 \right) }^{ 2 }\left( 1.66\times { 10 }^{ -27 } \right) { \left( 9.30\times { 10 }^{ 6 } \right) }^{ 2 }=2.87\times { 10 }^{ -13 }J=1.80MeV$
hence

$Q=1.02MeV+1.80MeV=2.82MeV$

Calculate the ground state $Q$ value of the induced fission reaction in the equation
$n+_{ 92 }^{ 235 }{ U }\rightarrow _{ 92 }^{ 235 }{ U }^{\ast}\rightarrow _{ 52 }^{ 134 }{ Te }+3n$
if the neutron is thermal. A thermal neutron is in thermal equilibrium with its environment; it has an average kinetic energy given by $(3/2)kT$ .

Given $m(n)=1.0087a.m.u$ , $M(_{ }^{ 235 }{ U })=235.0439a.m.u$ ; $M(^{ 99 }{ Zr })=98.916a.m.u$ ;
$M(_{ }^{ 134 }{ Te })=133.9115a.m.u$

Kinetic energy of a thermal neutron can be neglected; even for a temperature of

${10}^{6}K$ , the thermal energy is only

$130eV$ .

The

$Q$ value of the above reaction is given by the equation

$Q=\Delta m\times 931.2MeV$

Here

$\Delta m$ is the mass defect of the reaction, given by

$\Delta m=\left[ M(_{ }^{ 235 }{ U })+m(n) \right] -\left[ M(_{ }^{ 99 }{ Zr })+M(_{ }^{ 134 }{ Te })+3m(n) \right]$

$\Delta m =\left[ 235.0439-98.9165-133.9115-2(1.0087) \right] a.m.u=0.1985a.m.u$

Thus, energy released is

$Q=0.1985\times 931.2=184.84MeV$
Even with a thermal neutron of negligible kinetic energy, a tremendous amount of energy is released.

In the fission of $_{ 94 }^{ 239 }{ Pu }$ by a thermal neutron, two fission fragments of equal masses and sizes are produced and four neutrons are emitted. Find the force between the two fission fragments at the moment they are produced. Given ${R}_{0}=1.1fermi$

Total mass number of

$_{ 94 }^{ 239 }{ Pu }+$ neutron (thermal)

$=239+1=240$ . Since

$4$ neutrons are produced, the mass number of each fragment

$(A)=\cfrac { 240-4 }{ 2 } =118$ . The atomic number of each fragment

$=94/2=47$ . Therefore, charge of each fragment is

$q=47\times 1.6\times { 10 }^{ -19 }=7.52\times { 10 }^{ -18 }C\quad$
The radius of each nucleus of the fragment is

$R={ R }_{ 0 }{ (A) }^{ 1/3 }=1.1\times { 10 }^{ -15 }\times { \left( 118 \right) }^{ 1/3 }=5.395\times { 10 }^{ -15 }m$
distance between the centres of the two fragments at the moment they are produced is

$r=2\times 5.395\times { 10 }^{ -15 }=10.79\times { 10 }^{ -15 }m$
The electrostatic force between them is

$F=\cfrac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \cfrac { { q }^{ 2 } }{ { r }^{ 2 } } =9\times { 10 }^{ 9 }\cfrac { { \left( 7.52\times { 10 }^{ -18 } \right) }^{ 2 } }{ { \left( 10.79\times { 10 }^{ -15 } \right) }^{ 2 } } =4.37\times { 10 }^{ 3 }N$

A nuclear reaction is given as
$p+_{ }^{ 15 }{ N }\rightarrow _{ Z }^{ A }{ X }+n$

(a) Find $A,Z$ and identify the nucleus $X$
(b) Find the $Q$ value of the reaction
(c) If the proton were to collide with the $_{ }^{ 15 }{ N }$ at rest, find the minimum KE needed by the proton to initiate the above reaction.
(d) If the proton has twice the energy in (c) and the outgoing neutron emerges at an angle of ${90}^{o}$ with the direction of the incident proton, find the momentum of the nucleus $X$

[Given, $m(p)=1.007825u,m(_{ }^{ 15 }{ C })=15.0106u$
$m(_{ }^{ 16 }{ N })=16.0061u,m(_{ }^{ 15 }{ N })=15.000u$
$m(_{ }^{ 16 }{ O })=15.9949u,m(n)=1.008665u$ ,
$m(_{ }^{ 15 }{ O })=15.0031u\quad and\quad 1u=931.5MeV$ ]

(a) The nucleus is identified by:

$Z=8,A=15\Rightarrow X={ _{ 8 }^{ }{ O } }^{ 15 }\quad$

(b)

$Q=\left[ m(p)+m({ N }^{ 15 })-m({ O }^{ 15 })-m(n) \right] { c }^{ 2 }=-3.67MeV$

(c)

${ K }_{ th }=-Q\left( 1+\cfrac { { m }_{ p } }{ { m }_{ N } } \right) =3.9MeV$

(d) Now

${ E }_{ k }=2\times { K }_{ th }=2\times 3.9MeV=7.8MeV$

$Q=-3.63MeV$
Conservation of momentum:

${ p }_{ 0 }\cos { \theta } =\sqrt { 2{ m }_{ p }{ E }_{ k } } ;{ p }_{ 0 }\sin { \theta } ={ p }_{ n }$

Conservation of energy:

$\cfrac { { p }_{ n }^{ 2 } }{ 2{ m }_{ n } } +\cfrac { { p }_{ 0 }^{ 2 } }{ 2{ m }_{ 0 } } ={ E }_{ k }+W$

$\Rightarrow { p }_{ n }^{ 2 }=\cfrac { { E }_{ k }\left( 1-\cfrac { { m }_{ p } }{ { m }_{ 0 } } \right) }{ \cfrac { 1 }{ 2{ m }_{ 0 } } +\cfrac { 1 }{ 2{ m }_{ n } } }$

$\therefore { p }_{ n }=79.4MeV/c$

${ p }_{ 0 }=145MeV/c$

$\theta ={ 33 }^{ o }\quad$

(a) How many atoms are contained in $1.0 \ kg$ of pure $^{235}U$ ?
(b) How much energy,in joules,is released by the complete fissioning of $1.0 \ kg$ of $^{235}U$ ? Assume $Q = 200 \ MeV$ .
(c) For how long would this energy light a $100 \ W$ lamp?

1974674_1780837_ans_e8b35fa1ac004913b3114320373d07b1.jpg

The radioactive nuclide $^{99}Tc$ can be injected into a patients bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a cow containing $^{99}Mo$ , a radioactive nuclide that decays to $^{99} Tc$ with a half-life of 67 h. Once a day, the cow is milked for its $^{99}Tc$ , which is produced in an excited state by the $^{99}Mo$ ; the $^{99} Tc$ de-excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half-life of 6.0 h. (a) By what process does $^{99} Mo$ decay to $^{99}Tc$ ? (b) If a patient is injected with an $8.2 \times 10^7 Bq$ sample of $^{99}Tc$ , how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected $^{99}Tc$ is 38 per second at a certain time, how many excitedstate $^{99}Tc$ are located in the tumor at that time?

1806365_1781541_ans_3cbf08732dd74ee5bf9e936140aa58c1.jpg

Calculate the disintegration energy $Q$ for the fission of $^{52}Cr$ into two equal fragments. The masses you will need are

$^{52}Cr$ $51.94051 \ u$ ^{26}Mg $25.98259 \ u$

1875963_1781577_ans_c7260d9c10cb4e0abde11db2071b0a92.jpeg

A typical chest x-ray radiation dose is 250 $\mu Sv$ , delivered by x-rays with an RBE factor of 0.Assuming that the mass of the exposed tissue is one-half the patients mass of 88 kg, calculate the energy absorbed in joules.

1792495_1781781_ans_989c16d50fef406c80e6e078ae33fe59.jpg

Some uranium samples from the natural reactor site were found to be slightly enriched in $^{235}U$ , rather than depleted.Account for this in terms of neutron absorption by the abundant isotope $^{238}U$ and the subsequent beta and alpha decay of its products.

1785360_1781729_ans_daad116e4b8f4eb7a6e4053003d8f127.jpg

A 75 kg person receives a whole-body radiation dose of $2.4 \times 10^{-4} Gy$ , delivered by alpha particles for which the RBE fac-tor isCalculate (a) the absorbed energy in joules and the dosee in (b) sieverts and (c) rem.

(a)Absorbed energy is given by:

$E= D \times M$ where D is radiation dose and M is mass of person.

$E=2.4\times 10^{-4}\times 75=18\,mJ$

(b) dose is given by:

$E= D \times f$ where D is radiation dose and f is RBE fac-tor

$E=2.4\times 10^{-4}\times 12=2.9\times 10^{-3}\,Sv$

$1 Sv= 100 \,rem$

$2.9\times 10^{-3} Sv= 0.29 rem$

so dose =

$0.29 \,rem$

A free neutron decays according to Eq. $(\triangle = M-A)$ . If the neutronhydrogen atom mass difference is $840 \mu u$ , what is the maximum kinetic energy $K_{max}$ possible for the electron producedin a neutron decay?

1785819_1781691_ans_d7d43e14e0174b51a36b37dcf5ed5ab3.jpg

A certain stable nuclide, after absorbing a neutron, emits an electron and the new nuclide splits spontaneously into two alpha particles. Identify the nuclide.

1792457_1781776_ans_9bd7e8b0fc89446c93590d5c73e56a49.jpg

A particular rock is thought to be 260 million years old. If it contain 3.70 mg of $^{238}U$ , how much $^{206}Pb$ should it contain? SeeProblem 61.

1788488_1781734_ans_fe8abf4b31c94972a294a4b88fc250b3.jpg

What is the $Q$ of the following fusion process?
$^2 H_1 + ^1 H_1 \rightarrow ^3 He_2 + photon$
Here are some atomic masses.

$^2 H_1$ $2.014 102 \ u$
$^1 H_1$ $1.007 825 \ u$
$^3 He_2$ $3.016 029 \ u$

1791021_1781760_ans_b2f7af56345342c8b34fb5b05b3ec175.jpg

What mass of phosphorus is needed to dope $1.0 g$ of silicon so that the number density of conduction electrons in the silicon is increased by a multiply factor of $10^6$ from the $10^{16} m^3$ in pure silicon.

Sample Problem — “Doping silicon with phosphorus” gives the fraction of silicon
atoms that must be replaced by phosphorus atoms. We find the number the silicon atoms
in 1.0 g, then the number that must be replaced, and finally the mass of the replacement
phosphorus atoms. The molar mass of silicon is

$M_{si}=28.086 g/mol$ , so the mass of one
silicon atom is

$m_{0,Si}=M_{Si}/N_A=(28.086g/mol)/(6.022\times 10^{23}mol^{-1})=4.66\times 10^{-23}g$
and the number of atoms in 1.0 g is

$N_{Si}=m_{Si}/m_{0,Si}=(1.0g)/(4.66\times 10^{-23}g)=2.14\times 10^{22}$
According to the Sample Problem, one of every

$5\times 10^6$ silicon atoms is replaced with a
phosphorus atom. This means there will be

$N_P=(2.14\times 10^{22})/(5\times10^6)=4.29\times 10^{15}$
phosphorus atoms in 1.0 g of silicon. The molar mass of phosphorus is

$M_p = 30.9758 g/mol$ ,so the mass of a phosphorus atom is

$m_{0,P}=M_p/N_A=(30.9758 g/mol)/(6.022\times 10^{-23}mol^{-1})=5.14\times 10^{-23}g.$
The mass of phosphorus that must be added to

$1.0$ g of silicon is

$m_P=N_Pm_{0,P}=(4.29\times 10^{15})(5.14\times 10^{-23}mol^{-1})=2.2\times 10^{-7}g$

The thermal energy generated when radiation from radionuclides is absorbed in matter can serve as the basis for a small power source for use in satellites, remote weather stations, and other isolated locations. Among the many fission products that may be extracted chemically from the spent fuel of a nuclear reactor is $^{90}Sr \ (T_{1/2} = 29 \ y)$ . This isotope is produced intypical large reactors at the rate of about $18 \ kg/y$ . By its radioactivity,the isotope generates thermal energy at the rate of $0.93 \ W/g$ . (a) Calculate the effective disintegration energy $Q_{eff}$ associated with the decay of a $^{90}Sr$ nucleus. (This energy $Q_{eff}$ includes contributions from the decay of the $^{90}Sr$ daughter products in its decay chain but not from neutrinos, which escape totally from the sample.) (b) It is desired to construct a power source generating $150 \ W$ (electric power) to use in operating electronic equipment in an underwater acoustic beacon. If the power source is based on the thermal energy generated by 90Sr and if the efficiency of the thermalelectric conversion process is $5.0 \%$ , how much $^{90}Sr$ is needed?

1785817_1781682_ans_a61880861d2d45d098337c0f80dccfc0.jpg

Calculate the Q value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is 30.0 MeV. What is the barrier height for $14C$ emission? (Be careful about the nuclear radii.) The needed atomic masses are
$223{Ra} \rightarrow 209{pb} + 14C and 223{Ra} \rightarrow 219{Rn} + 4{He}$
Calculate the Q value for the (a) first and (b) second decay and determine that both are energetically possible. (c) The Coulomb barrier height for alpha-particle emission is 30.0 MeV. What is the barrier height for $^{14} C$ emission? (Be careful about the nuclear radii.) The needed atomic masses are
$^{223}Ra$ 223.018 50 u $^{14}C$ 14.003 24 u
$^{209}Pb$ 208.981 07 u $^4He$ 4.002 60 u
$^{219}Rn$ 219.009 48 u

1779589_1781598_ans_2aa91068b520417c949ddf5741e329b5.jpg

If the unit for atomic mass were defined so that the mass of $^ 1H$ were exactly 1.000 000 u, what would be the mass of (a) $^{12}C$ (actual mass 12.000 000 u) and (b) $^{238}U$ (actual mass 238.050 785 u)?

1794505_1781801_ans_8fd128258acd443fb187f932c2392fc2.jpg

Find the disintegration energy Q for the decay of $^{49}V$ by K-electron capture . The needed data are $$ m_V = 48.948 52 u , $m_Ti = 48.947 87 u$ , and $E_K = 5.47 keV$ .

1794511_1781806_ans_86544e7ee881439591e727f121736ed5.jpg

Verify the $Q$ values reported in equations,
$^2H + \ ^2H \rightarrow \ ^3He + n$ $(Q=+3.27 \ MeV)$ ,
$^2H + \ ^2H \rightarrow \ ^3H + \ ^1H$ $(Q=+4.03 \ MeV)$ , and
$^2H + \ ^3H \rightarrow \ ^4He + n$ $(Q=+17.59 \ MeV)$ ,
The needed masses are
$^1H$ $1.007825 \ u$ $^4He$ $4.002603 \ u$
$^2H$ $2.014102 \ u$ $n$ $1.008665 \ u$
$^3H$ $3.016049 \ u$

1796056_1781838_ans_b19498b6abc54fb5948211f6a924a149.jpg

The above figure shows an early proposal for a hydrogen bomb. The fusion fuel is deuterium, $^2H$ . The high temperature and particle density needed for fusion are provided by an atomic bomb trigger" that involves a $^{235}U$ or $^{239}Pu$ fission fuel arranged to impress an imploding, compressive shock wave on the deuterium. The fusion reaction is
$5 \ ^2H \rightarrow ^3He + ^4H + ^1H + 2n$
$^1H$ $1.007 825 \ u$ $^4He$ $4.002 603 \ u$
$^2H$ $2.014 102 \ u$ $e^{\pm}$ $0.000 548 6 \ u$
$^3He$ $3.016 029 \ u$
(b) Calculate the rating (The bombs rating is the magnitude of the released energy, specified in terms of the mass of TNT required to produce the same energy release. One megaton of TNT releases $2.6 \times 10^{28} \ MeV$ of energy) of the fusion part of the bomb if it contains $500 \ kg$ of deuterium, $30.0 \%$ of which undergoes fusion.

1797938_1781821_ans_8512f13c61a2464fafd6cf288f0b0127.jpg

Consider a $^{238}U$ nucleus to be made up of an alpha particle $(^4He)$ and a residual nucleus $(^{234}Th)$ . Plot the electrostatic potential energy U(r), where r is the distance between these particles. Coverthe approximate range $10 fm < r <100 fm$ and compare your plotwith that of Fig. 42-10.

1794500_1781798_ans_d75902d2303c410fb1eb16c639d0cfc6.jpg

In certain stars the carbon cycle is more effective than the proton-proton cycle in generating energy. This carbon cycle is

$^{12}C + ^1H \rightarrow ^{13}N + \gamma$ $Q_1 = 1.95 \ MeV$ ,
$^{1}N \rightarrow ^{13}C + e^+ + v$ $Q_2 = 1.19$ ,
$^{13}C + ^1H \rightarrow ^{14}N + \gamma$ $Q_3 = 7.55$ ,
$^{14}N + ^1H \rightarrow ^{15}O + \gamma$ $Q_4 = 7.30$ ,
$^{15}O \rightarrow ^{15}N + e^+ + v$ $Q5 = 1.73$ , $^{15}N + ^1H \rightarrow ^{12}C + ^4He$ $Q_6 = 4.97$ .
(a) Show that this cycle is exactly equivalent in its overall effects to the proton-proton cycle of the above figure. (b) Verify that the two cycles,as expected,have the same $Q$ value.

1797940_1781831_ans_eca0cf3e1c724892881ddd38827a55d0.jpg

Show that the energy released when three alpha particles fuse to form $^{12}C$ is $7.27 \ MeV$ . The atomic mass of $^4He$ is $4.0026 \ u$ , and that of $^{12}C$ is $12.0000 \ u$ .

1797521_1781819_ans_ba4d141d347a422d9e9ebd608b875038.jpg

Nuclei with magic no. of proton $Z=2,8,20,28,50,52$ and magic no. of neutrons $N=2,8,20,28,50,82$ and $126$ are found to be very stable
(i) Verify this by calculting the proton separation energy ${S}_{p}$ for ${ _{ 50 }^{ }{ Sn } }^{ 120 }\left( Z=50 \right)$ and ${ _{ 51 }^{ }{ Sb } }^{ 121 }\left( Z=51 \right)$ . The protn separation energy for a nuclide is the minimum energy required to separate the least tightly bound proton from a nucleus of that nuclide. It is given by
${S}_{p}=[M{Z}_{-1,n}+{M}_{{H}^{-}}-{M}_{ZN}]{c}^{2}$
Given $_{ }^{ 119 }{ In }=118.9058u,_{ }^{ 120 }{ Sn }=119.902199u,_{ }^{ 121 }{ Sb }=120.903842u,_{ }^{ 1 }{ H }=1.0078252u$
(ii) What does the existence of magic number indicate?

(i)

${s}_{p}=[{M}_{Z-1,N}+{M}_{H}-{M}_{z,N}]{c}^{2}$

Here in this formula

${M}_{Z-1}$ is the mass of atom of

$Z-1$ atomic number

${M}_{Z}$ is the mass of atom of mass number

$Z$

$\therefore$

${M}_{Z-1}=$ Mass of atom whose atomic number is

$50-1=49$

It is

${ _{ 49 }^{ }{ In } }^{ 119 }$ in this case

${ M }_{ Z-1 }={ _{ 49 }^{ }{ In } }^{ 119 }=118.9058;N=119-49=70$

${s}_{p}$ for

${ _{ 50 }^{ }{ Sn } }^{ 120 }={ c }^{ 2 }\left[ 118.9058+1.0078252-199.902199 \right] =0.0114362{ c }^{ 2 }$

Now for

${s}_{p}$ of

$\quad { _{ 51 }^{ }{ Sb } }^{ 121 }$

${ s }_{ p }=\left[ { M }_{ Z-1,N }+{ M }_{ H }-{ M }_{ z,N } \right] { c }^{ 2 }\Rightarrow Z=51,Z-1=50$

${M}_{Z-1}=$ Mass of

$_{ 50 }^{ }{ Sn }=199.902199u\quad$

$\therefore$

${s}_{p}$ for

${ _{ 51 }^{ }{ Sb } }^{ 121 }=\left[ 199.902199+1.0078252-120.903824 \right] { c }^{ 2 }=0.0059912{ c }^{ 2 }$

$\quad \therefore { s }_{ p }\left( \quad { _{ 50 }^{ }{ Sn } }^{ 120 } \right) >{ s }_{ p }\left( { _{ 51 }^{ }{ Sb } }^{ 121 } \right)$

(ii) the existence of magic numbers indicate that the shell structure of nucleus is similar ot the shell structure of atom. this also expains the peaks in binding energy per nucleon curve.

Consider the decay $\Lambda^{o} \rightarrow p | \pi^{-}$ with the $\Lambda^{o}$ at rest.
(a) Calculate the disintegration energy. What is the kinetic energy of (b) the proton and (c) the pion?

1765774_1785493_ans_86779d5b76344e4b891cf64923f6bd36.jpg

Calculate the disintegration energy of the reactions
(a) $\pi^{+} + p \rightarrow \sum^{+} + K^{+}$ and (b) $K^{-} + p \rightarrow \Lambda^{o} + \pi^{o}$ .

1765770_1785425_ans_e841911ab75d4c8f9be67d19fa610c15.jpg

What Is the longest wavelength of a photon that can ionize a hydrogen atom in its ground state? Specify the type of radiation.

Since the energy of the incident photon

$=hv=\dfrac{hc}{\lambda}=13.6\,eV$

$\lambda =\dfrac{6.6\times10^{-34}\times3\times10^8}{13.6\times 1.6\times10^{-19}}$

$\lambda=0.910 \times 10^{-10}\,m$

This radiation is in the ultraviolet region.

A radioactive source emits three type of radiations. From which part of the atom do these radiations come?

Gamma radiation and x rays are examples of electromagnetic radiation. Gamma radiation originates in the nucleus while x rays come from the electronic part of the atom. Beta and alpha radiation are examples of particulate radiation.

Draw the graph showing the variation of binding energy per nucleon with the mass number for a large number of nuclei $2 < A < 240.$ What are the main inferences from the graph ? How do you explain the constancy of binding energy in the range $30 < A < 170$ using the property that the nuclear force is short-ranged ? Explain with the help of this plot the release of energy in the processes of nuclear fission and fusion.

The variation of binding energy per nucleon versus mass number is shown in figure. Inferences from graph
(1) The nuclei having mass number below

$20$ and above

$180$ have relatively small binding energy and hence they are unstable.
(2) The nuclei having mass number

$56$ and about

$56$ have maximum binding energy -

$5.8 \,MeV$ and so they are most stable.
(3) Some nuclei have peaks, e.g.,

$^{4}_{2}He, ^{12}_{6}C, ^{12}_{6}O;$ this indicates that these nuclei are relatively more stable than their neightbours.
(i) Explanation of constancy of binding energy: Nuclear force is short ranged, so every nucleon interacts with its neighbours only, therefore binding energy per nucleon remains constant.
(ii) Explanation of nuclear fission: When a heavy nucleus

$(A \geq 235 \,say)$ breaks into two lighter nuclei (nuclear fission), the binding energy per nucleon increases i.e, nucleons get more tightly bound. This implies that energy would be released in nuclear fission.
(iii) Explanation of nuclear fusion: When two very nuclei

$(A \leq 10)$ join to form a heavy nucleus, the binding is energy per nucleon of fused heavier nucleus more than the binding energy per nucleon of lighter nuclei, so again energy would be released in nuclear fusion.
1655426_1784923_ans_3922200bfdd643b787c5630eb099bfde.png

1655426_1784923_ans_3922200bfdd643b787c5630eb099bfde.png

'Radioactivity is a nuclear phenomenon'. Comment on this statement.

Any physical change (such as change in pressure and temperature) or chemical change (such as excessive heating, freezing, action of strong electric and magnetic fields, chemical treatment, oxidation etc.) does not change the rate of decay and the nature of radiation emitted by the substance. This shows clearly that the phenomenon of radioactivity cannot be due to the orbital electrons which could easily be affected by such changes. Therefore the radioactivity should be the property of the nucleus. Thus the radioactivity is a nuclear phenomenon.

Explain mass defect and binding energy. Explain the main results obtained from the graph between binding energy per nucleon and mass number.

1967921_1836790_ans_eed40bec726e41a7a845ca9daa485ba3.jpg

Find the Q-value and the kinetic energy of the emitted $\alpha$ - particle in the $\alpha$ - decay of (a) $_{ 88 }^{ 226 }{ Ra }$ (b) $_{ 86 }^{ 222 }{ Rn }$
Given
$m\left( _{ 88 }^{ 226 }{ Ra } \right) =226.02540u;\quad m\left( _{ 86 }^{ 222 }{ Rn } \right) =222.01750u;\quad m\left( _{ 86 }^{ 222 }{ Rn } \right) =220.01137u;\quad m\left( _{ 84 }^{ 216 }{ Po } \right) =216.00189u$

(a) The decay takes place as shown:

$\quad _{ 88 }^{ 226 }{ Ra }\rightarrow _{ 86 }^{ 222 }{ Rn }+_{ 2 }^{ 4 }{ He }\quad$

$m\left( _{ 88 }^{ 226 }{ Ra } \right) =226.02540u;\quad m\left( _{ 86 }^{ 222 }{ Rn } \right) =222.01750u;\quad m\left( _{ 2 }^{ 4 }{ He } \right) =4.002603amu$
Q-value of the reaction

$Q=\left[ { m }_{ N }\left( _{ 88 }^{ 226 }{ Ra } \right) -{ m }_{ N }\left( _{ 86 }^{ 222 }{ Rn } \right) -{ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) \right] \times 931.5MeV\quad$
Now,

${ m }_{ N }\left( _{ 88 }^{ 226 }{ Ra } \right) ={ m }_{ }\left( _{ 88 }^{ 226 }{ Ra } \right) -88{ m }_{ e }$

${ m }_{ N }\left( _{ 86 }^{ 222 }{ Rn } \right) ={ m }_{ }\left( _{ 86 }^{ 222 }{ Rn } \right) -86{ m }_{ e }\quad$

${ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) ={ m }_{ }\left( _{ 2 }^{ 4 }{ He } \right) -2{ m }_{ e }$

$\therefore Q=\left[ \left\{ { m }_{ N }\left( _{ 88 }^{ 226 }{ Ra } \right) -88{ m }_{ e } \right\} -\left\{ { m }_{ }\left( _{ 86 }^{ 222 }{ Rn } \right) -86{ m }_{ e } \right\} -\left\{ { m }_{ }\left( _{ 2 }^{ 4 }{ He } \right) -2{ m }_{ e } \right\} \right] \times 931.5MeV$

$Q=\left[ 226.02540-221.01750-4.002603 \right] \times 931.5MeV=0.005297\times 931.5MeV=4.934MeV$
The energy of the emitted

$\alpha$ - particle

${ E }_{ \alpha }=\cfrac { A-4 }{ A } Q=\cfrac { 222-4 }{ 222 } \times 4.934=4.845MeV$
(b) The decay takes place as below:

$\quad _{ 86 }^{ 222 }{ Rn }\rightarrow _{ 84 }^{ 216 }{ Po }+_{ 2 }^{ 4 }{ He }$

$m\left( _{ 86 }^{ 222 }{ Rn } \right) =220.01137u;\quad m\left( _{ 84 }^{ 216 }{ Po } \right) =216.00189u;\quad m\left( _{ 2 }^{ 4 }{ He } \right) =4.002603amu\quad$
Q- value of the reaction

$Q=\left[ m\left( _{ 86 }^{ 222 }{ Rn } \right) -m\left( _{ 84 }^{ 216 }{ Po } \right) -{ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) \right] \times 931.5MeV$

${ m }_{ N }\left( _{ 86 }^{ 222 }{ Rn } \right) ={ m }_{ }\left( _{ 86 }^{ 222 }{ Rn } \right) -86{ m }_{ e }\quad$

$\quad { m }_{ N }\left( _{ 84 }^{ 216 }{ Po } \right) ={ m }_{ }\left( _{ 84 }^{ 216 }{ Po } \right) -84{ m }_{ e }$

${ m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) ={ m }_{ }\left( _{ 2 }^{ 4 }{ He } \right) -2{ m }_{ e }$

$\therefore Q=\left[ \left\{ { m }_{ N }\left( _{ 86 }^{ 222 }{ Rn } \right) -86{ m }_{ e } \right\} -\left\{ { m }_{ N }\left( _{ 84 }^{ 216 }{ Po } \right) -84{ m }_{ e } \right\} -\left\{ { m }_{ N }\left( _{ 2 }^{ 4 }{ He } \right) -2{ m }_{ e } \right\} \right] \times 931.5MeV$

$Q=\left[ 220.01137-216.00189-4.002603 \right] \times 931.5MeV=0.006877\times 931.5=6.406MeV\quad$
The energy of the emitted

$\alpha$ - particle

${ E }_{ \alpha }=\cfrac { A-4 }{ A } Q=\cfrac { 222-4 }{ 222 } \times 6.406=6.29MeV$

The Q value of a nuclear reaction $A+b\rightarrow C+d$ is defined by $Q=({m}_{A}+{m}_{B}-{m}_{c}-{m}_{d})$ where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic.
(i) $_{ 1 }^{ 1 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 1 }^{ 2 }{ H }+_{ 1 }^{ 2 }{ H }$
(ii) $_{ 6 }^{ 12 }{ C }+_{ 6 }^{ 12 }{ C }\rightarrow _{ 10 }^{ 20 }{ Ne }+_{ 2 }^{ 4 }{ He }$
Atomic masses are given to be
$m\left( _{ 1 }^{ 2 }{ H } \right) =2.014102u;\quad m\left( _{ 1 }^{ 3 }{ H } \right) =3.016049;\quad m\left( _{ 6 }^{ 12 }{ C } \right) =12.00000u;\quad m\left( _{ 10 }^{ 20 }{ Ne } \right) =19.992439u$

(i)

$m\left( _{ 1 }^{ 2 }{ H } \right) =2.014102u;\quad m\left( _{ 1 }^{ 3 }{ H } \right) =3.016049;\quad m\left( _{ 1 }^{ 1 }{ H } \right) =1.007825amu$

$Q=\left[ m\left( _{ 1 }^{ 1 }{ H } \right) +m\left( _{ 1 }^{ 3 }{ H } \right) -2\times m\left( _{ 1 }^{ 2 }{ H } \right) \right] \times 931.5MeV=\left[ 1.007825+3.016049-2\times 2.014102 \right] \times 931.5MeV=-4.033MeV$

Since

$Q$ value is negative, the reaction is endothermic

(ii)

$\quad _{ 6 }^{ 12 }{ C }+_{ 6 }^{ 12 }{ C }\rightarrow _{ 10 }^{ 20 }{ Ne }+_{ 2 }^{ 4 }{ He }$

$m\left( _{ 6 }^{ 12 }{ C } \right) =12.00000amu;\quad m\left( _{ 10 }^{ 20 }{ Ne } \right) =19.992439amu;\quad m\left( _{ 2 }^{ 4 }{ He } \right) =4.002603amu$

$\therefore Q=\left[ 2\times m\left( _{ 6 }^{ 12 }{ C } \right) -m\left( _{ 10 }^{ 20 }{ Ne } \right) -m\left( _{ 2 }^{ 4 }{ He } \right) \right] \times 931.5MeV=\left[ 2\times 12.00000-19.992439-4.002603 \right] \times 931.5MeV=4.618MeV\quad$

Since the Q-value is positive, the reaction is exothermic

What is $\alpha$ -decay ? What type of energy spectrum do $\alpha$ -particles have ?

1967909_1836775_ans_4d3add7051a5433f80b6adc01569391b.jpg

Calculate binding energy for $^{56}_{26}Fe$ nucleus [ Given: mass of $^{56}_{26}Fe = 55.9349 \,u,$ mass of neutron $= 1.00867 \,u,$ mass of proton $= 1.00783 \,u$ and $1 \,u = 931 \,MeV/c^2$ ].

1967916_1836808_ans_d3b297af78754fa693cc034fa3b7e105.jpg

Describe $\alpha$ -decay in a radioactive nucleus. Explain that the energy spectrum of $\alpha$ -particles obtained from the decay is wide spectrum of group of energies.

1967911_1836800_ans_dfc06c98c10c4f0a8b33a24b5cc20e32.jpg

Find the energy Q liberated in $\beta^{-}$ and $\beta^{+}$ -decays and in K- capture if the masses of the parent atom $M_{p}$ , the daughter atom $M_{d}$ and an electron $m$ are known.

$\beta^{-}$ decay

$Z^{X^{A}} \rightarrow _{z + 1}Y^{A} + e^{-} + Q$

$Q = (M_{x} - M_{y} - m_{e})c^{2}$

$= [(M_{x} + Zm_{e}) - (M_{y} + Zm_{e} + m_{e})] c^{2}$
Since

$M_{p} , M_{d}$ are the masses of the atoms. The binding energy of the electrons in ignored.
In K capture

$e^{-}_{K} + _{z}X^{A} \rightarrow _{z - 1}Y^{A} + Q$

$Q = (M_{X} - M_{Y})c^{2} + m_{e} c^{2}$

$= (M_{x}^{c^{2}} + Zm_{e} c^{2}) - (M_{Y} c^{2} + (Z - 1)m_{e} c^{2})$

$= c^{2} (M_{p} - M_{d})$

In

$\beta^{+}$ decay

$_{Z}X^{A} \rightarrow _{z - 1} Y^{A} + e^{+} + Q$
Then

$Q = (M_{x} - M_{y} - m_{e}) c^{2}$

$= [M_{x} + Zm_{e}]c^{2} - [M_{y} + (Z - 1)m_{e}]c^{2} - 2 m_{e} c^{2}$

$= (M_{p} - M_{d} - 2m_{e})c^{2}$

Under certain circumstances, a nucleus can decay by emitting $a$ - particle more massive than an $a$ - particle. Consider the following decay processes:
$\quad _{ 88 }^{ 223 }{ Ra }\rightarrow _{ 82 }^{ 209 }{ Pb }+_{ 6 }^{ 14 }{ C }$
$\quad _{ 88 }^{ 223 }{ Ra }\rightarrow _{ 86 }^{ 219 }{ Rn }+_{ 2 }^{ 4 }{ He }$
Calculate the Q-values for these decays and determine that both are energetically allowed.

$_{ 88 }^{ 223 }{ Ra }\rightarrow _{ 82 }^{ 209 }{ Pb }+_{ 6 }^{ 14 }{ C }$

$Q=\left[ m\left( _{ 88 }^{ 223 }{ Ra } \right) -m\left( _{ 82 }^{ 209 }{ Pb } \right) -m\left( _{ 6 }^{ 14 }{ C } \right) \right] \times 931.5MeV=\left[ 223.01850-208.98107-14.00324 \right] \times 931.5MeV$

$=0.03419\times 931.5MeV=31.85MeV$

Since the Q-value is positive, the reaction is energetically allowed

For the decay process

$\quad _{ 88 }^{ 223 }{ Ra }\rightarrow _{ 86 }^{ 219 }{ Rn }+_{ 2 }^{ 4 }{ He }$

$Q=\left[ m\left( _{ 88 }^{ 223 }{ Ra } \right) -m\left( _{ 86 }^{ 219 }{ Rn } \right) -m\left( _{ 2 }^{ 4 }{ He } \right) \right] \times 931.5MeV=0.006417\times 931.5=5.98MeV$

Since Q-value s positive the reaction is energetically allowed

Determine the spectral symbol of an atomic singlet term if the total splitting of that term in a weak magnetic field of induction $B = 3.0\,kG$ amounts to $\Delta E = 104 \mu eV$ .

1971216_1855581_ans_0171db3fe32f47e5bd91a2d8925a0eec.jpg

Taking into account the motion of the nucleus of a hydrogen atom, find the expressions for the electron's binding energy in the ground state and for the Rydberg constant. How much (in per cent) do the binding energy and the Rydberg constant, obtained without taking into account the motion of the nucleus, differ from the more accurate corresponding values of these quantities ?

1881060_1854503_ans_ae8d4b9c6e1346a0a07a41c6e621eb3d.jpg

Calculate the binding energy of a K electron in vanadium whose L absorption edge has the wavelength $\lambda_{L} = 2.4\,nm$

1971330_1855562_ans_75424c96ee2546f7af559206c3ee001c.jpg

Find the binding energy of an L electron in titanium if the wavelength difference between the first line of the K series and its short-wave cut-off is $\Delta \lambda = 26\,pm$ .

1971196_1855563_ans_4676a24c58a944b9a9baafd2681327c2.jpg

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $_{ 20 }^{ 41 }{ Ca }$ $_{ 13 }^{ 27 }{ Al }$ from the following data
$m\left( _{ 20 }^{ 40 }{ Ca } \right) =39.962591u;\quad m\left( _{ 20 }^{ 41 }{ Ca } \right) =40.962278u;\quad m\left( _{ 13 }^{ 26 }{ Al } \right) =25.986895u;\quad m\left( _{ 13 }^{ 27 }{ Al } \right) =26.981541u$

For

$_{ 20 }^{ 41 }{ Ca }$ nucleus:

$=$ Neutron separation energy

$=\left[ \left\{ m\left( _{ 20 }^{ 40 }{ Ca } \right) +{ m }_{ n } \right\} -m\left( _{ 20 }^{ 41 }{ Ca } \right) \right] \times 931.5MeV=\left[ \left( 39.962591+1.008665 \right) -40.962278 \right] \times 931.5MeV=0.008978\times 931.5=8.36MeV\quad$

For

$_{ 13 }^{ 27 }{ Al }$

$=$ Neutron separation energy

$=\left[ \left\{ m\left( _{ 13 }^{ 26 }{ Al } \right) +{ m }_{ n } \right\} -m\left( _{ 13 }^{ 27 }{ Al } \right) \right] \times 931.5MeV=\left[ \left( 25.986895+1.008665 \right) -26.981541 \right] \times 931.5MeV=13.06MeV$

A stationary $Pb^{200}$ nucleus emits an alpha-particle with kinetic energy $T_{\alpha} = 5.77 \,MeV$ . Find the recoil velocity of a daughter nucleus. What fraction of the total energy liberated in this decay is accounted for by the recoil energy of the daughter nucleus ?

The momentum of the

$\alpha$ - particle is

$\sqrt{2 M_{\alpha} T}$ .

This is also the recoil momentum of the daughter nuclear in opposite direction.
The recoil velocity of the daughter nucleus is

$\dfrac{\sqrt{2M_{\alpha} T}}{M_{d}} = \dfrac{2}{196} \sqrt{\dfrac{2T}{M_{p}}} = 3.39 \times 10^{5}m/s$ .

The energy of the daughter nucleus is

$\dfrac{M_{\alpha}}{M_d} T$ and this represents a fraction

$\dfrac{M_{\alpha} / M_{d}}{1 + \dfrac{M_{\alpha}}{M_d}} = \dfrac{M_{\alpha}}{M_{\alpha} + M_{d}} = \dfrac{4}{200} = \dfrac{1}{50} = 0.02$

of total energy. Here

$M_{d}$ is the mass of the daughter nucleus.

For atoms of light and heavy hydrogen (H and D) find the difference
(a) between the binding energies of their electrons in the ground state;
(b) between the wavelengths of first lines of the Lyman series.

1881061_1854505_ans_587bf3cb64484980962e3338d213868b.jpg

Find the difference in binding energies of a neutron and a proton in a $B^{11}$ nucleus. Explain why there is the difference.

1971108_1855840_ans_12ae76a90e044094affd2c34cb6c27ee.jpeg

Making use of the tables of atomic masses, determine the energies of the following reactions:
(a) $Li^{7} (p , n) Be^{7}$ ;
(b) $Be^{9} (n , \gamma ) Be^{10}$ ;
(c) $Li^{7} (\alpha , n ) B^{10}$ ;
(d) $O^{16} (d , \alpha) N^{14}$

1971114_1855866_ans_f7b4fa7c3eaa4f6da36f7e930867e561.jpeg

Calculate in atomic mass units the mass of
(a) a $Li^{8}$ atom whose nucleus has the binding energy $41.3\,MeV$ ;
(b) a $C^{10}$ nucleus whose binding energy per nucleon is equal to $6.04\,MeV$ .

1971109_1855850_ans_f5774e8f410a479ab440c039eb8db6a5.jpeg

To activate the reaction $(n, \alpha)$ with stationary $B^{11}$ nuclei, neutrons must have the threshold kinetic energy $T_{th} = 4.0\,MeV$ . Find the energy of this reaction.

1971117_1855875_ans_20d49b9da53e458a99900703b9b8530f.jpg

What kinetic energy must a proton possess to split a deuteron $H^2$ whose binding energy is $E_{b} = 2.2\,MeV$ ?

1971116_1855873_ans_e3cd6a7bbf774befa642487621d4d7f2.jpeg

Find the energy required for separation of a $Ne^{20}$ nucleus into two alpha-particles and a $C^{12}$ nucleus if it known that the binding energies per one nucleon in $Ne^{20} , He^{4}$ , and $C^{12}$ nuclei are equal to $8.03 , 7.07$ and $7.68\,MeV$ respectively.

1971260_1855845_ans_4174a268e6be4cb5952cbe0732d4b19c.jpg

Making use of the tables of atomic masses, find:
(a) the mean binding energy per one nucleon in $O^{16}$ nucleus;
(b) the binding energy of a neutron and an alpha-particle in a $B^{11}$ nucleus.
(c) the energy required for separation of an $O^{16}$ nucleus into four identical particles.

1971259_1855839_ans_e80129f72b0a4e66b211c1ad63c0af2f.jpg

Find the binding energy of a nucleus consisting of equal numbers of protons and neutrons and having the radius one and a half times smaller than that of $Al^{27}$ nucleus.

1971258_1855838_ans_8cf8c90f57944edc8d3ffe0ee479a4eb.jpg

Assuming that the splitting of a $U^{235}$ nucleus liberates the energy of $200\,MeV$ , find:
(a) the energy liberated in the fission of one kilogram of $U^{235}$ isotope, and the mass of coal with calorific value of $30\,kJ/g$ which is equivalent to that for one kg of $U^{235}$ ;
(b) the mass of $U^{235}$ isotope split during the explosion of the atomic bomb with $30\,kt$ trotyl equivalent if the calorific value of trotyl is $4.1\,kJ/g$ .

1971110_1855858_ans_a94f51ec24f24b5091e4990375620031.jpeg

A negative pion with kinetic energy $T = 50\,MeV$ disintegrated during its flight into a muon and a neutrino. Find the energy of the neutrino outgoing at right angles to the pion's motion direction.

See the diagram. By conservation of energy

$\sqrt{m^{2}_{\pi} c^{4} + c^{2} p^{2}_{\pi}} = c p_{\nu} + \sqrt{m^{2}_{\mu} c^{4} + p^{2}_{\pi} c^{4} + p^{2}_{\mu} c^{4} + p^{2}_{\pi} c^{2} + c^{2} p^{2}_{\nu}}$

or

$\left (\sqrt{m^{2}_{\pi} c^{4} + c^{2} p^{2}_{\pi}} - cp_{\nu} \right )^{2} = m^{2}_{\mu} c^{4} + c^{2}p^{2}_{\pi} + c^{2} p^{2}_{\nu}$

or

$m^{2}_{\pi} c^{4} - 2 c\ p_{\nu} \sqrt{m^{2}_{\pi} c^{4} + c^{2}p^{2}_{\pi}} = m^{2}_{\mu} c^{4}$

Hence the energy of the neutrino is

$E_{\nu} = c\ p_{\nu} = \dfrac{m^{2}_{\pi}c^{4} - m^{2}_{\mu} c^{4}}{2 (m_{\pi} c^{2} + T)}$

on writing

$\sqrt{m^{2}_{\pi} c^{4} + c^{2} p^{2}_{\pi} + c^{2} p^{2}_{\pi}} = m_{\pi} c^{2} + T$

Substitution gives

$E_{\nu} = 21.93\,MeV$
1822361_1857826_ans_0d15a2e36fc94913bb4d968a0c4a38b3.png

A nuclear reaction is given as
$p+ \ ^{15}N\rightarrow \ _Z^AX+n$
a) Find $A, Z$ and identify the nucleus $X$ .
b) Find the $Q$ value of the reaction.
c) If the proton were to collide with the $^{15}N$ at rest, find the minimum $KE$ needed by the proton to initiate the above reaction

a) The nuclear reaction will be

$_1^1p+ \ _7^{15}N=\ _Z^AX+ \ _0^1n$

Now,

$1+15=A+1 \Rightarrow A=15$ and

$1+7=Z+0 \Rightarrow Z=8$

The nucleus would be isotope of oxygen i.e

$_8^{15}O$

b) The nuclear reaction will be

$_1^1p+\ _7^{15}N=\ _Z^AX+\ _0^1n$

Now,

$1+15=A+1 \Rightarrow A=15$ and

$1+7=Z+0 \Rightarrow Z=8$

The nucleus would be isotope of oxygen i.e

$_8^{15}O$

The atomic mass of proton

$=1.00727$ amu ; atomic mass of neutron

$=1.00866$ amu ; atomic mass of

$^{15}O=15.9994$ amu and atomic mass of

$^{15}N=15.000108$ amu

Sum of masses of reactants

$=1.00727+15.000108=16.0074$ amu

Sum of masses of products

$=15.9994+1.00866=17.0081$ amu

The Q-value

$=$ Mass difference

$\Delta M=17.0081-16.0074=1.0007$ amu

$=1.0007\times 931 MeV=931.65 MeV$

c) K.E. of proton to initiate the reaction is equal to Q value of reaction

$=\triangle m c^2$

$\triangle m={ m }_{ (p) }+{ m }_{ (N) }-{ m }_{ (X) }-{ m }_{ (n) }\\ =1.007825+15.000-\ ^{ 15 }{ O }-n\\ =1.007825+15-15.0031-1.008665\\ =-0.00394\\$

The K.E needed by the proton to initiate the reaction

$Q=-0.00394\times 931MeV\\ =-3.66MeV$

Negative sign signifies that this energy is supplied to initiate the reaction.

$^{238}U$	$238.05079 \ u$	$^{140}Ce$	$139.90543 \ u$
$n$	$1.00866 \ u$	$^{99}Ru$	$98.90594 \ u$

$^1H$	$1.007 825 \ u$	$^4He$	$4.002 603 \ u$
$^2H$	$2.014 102 \ u$	$e^{\pm}$	$0.000 548 6 \ u$
$^3He$	$3.016 029 \ u$

$^2 H_1$	$2.014 102 \ u$
$^1 H_1$	$1.007 825 \ u$
$^3 He_2$	$3.016 029 \ u$

$^1H$	$1.007825 \ u$	$^4He$	$4.002603 \ u$
$^2H$	$2.014102 \ u$	$n$	$1.008665 \ u$
$^3H$	$3.016049 \ u$

Nuclei - Class 12 Medical Physics - Extra Questions

Find the binding energy of an α\alpha-particle from the following data:Mass of helium nucleus=4.001265=4.001265 a.m.u; mass of proton =1.007277=1.007277 a.m.u. Mass of neutron =1.00866=1.00866 a.m.u (take 11 a.m.u=931.5=931.5 MeV)

The binding energies per nucleon for deutron (1H2)(_{1}H^{2}) and helium (2He4)(_{2}He^{4}) are 1.11.1 MeV and 7.07.0 MeV respectively. Calculate the energy released, when two deutrons fuse to form a helium nucleus (2He4)(_{2}He^{4}).

The binding energy per nucleon for 6C12_{6}C^{12} nucleons is 7.687.68MeV and that for 6C13_{6}C^{13} is 7.477.47 MeV. Calculate the energy required to remove a neutron from 6C13_{6}C^{13} nucleons.

What is (a) mass defect, and (b) binding energy in Oxygen ^{16}_{8}{O}^{16}_{8}{O}, whose nuclear mass is 15.99515.995 amu. (m_p=1.0078(m_p=1.0078amu;; m_n=1.0087m_n=1.0087amu)).

What are the different types of nuclear reactions?

Answer the following question:Identify the nature of the radioactive radiations emitted in each step of the decay process given below:^{A}_{Z}X \rightarrow ^{A-4}_{Z-1}W^{A}_{Z}X \rightarrow ^{A-4}_{Z-1}W

Answer the following question:A heavy nucleus splits into two lighter nuclei. Which one of the two-parent nucleus or the daughter nuclei has more binding energy per nucleon ?

Give reason for the release of 21.6\ MeV21.6\ MeV energy in the fusion of deuterium reaction.

Write one nuclear fusion reaction.

Why is a very high temperature required for the process of nuclear fusion? State the approximate temperature required.

Which of the isotope of Uranium is easily fissionable? Give reason.

The binding energy of _{ 17 }^{ }{ { Cl }^{ 35 } }_{ 17 }^{ }{ { Cl }^{ 35 } } nucleus is 298298 MeV. Find its atomic mass. The mass of hydrogen atom \left( _{ 1 }^{}{ { H }^{ 1 } } \right)\left( _{ 1 }^{}{ { H }^{ 1 } } \right) is 1.0081431.008143 a.m.u. Given 11 a.m.u. =931=931 MeV.

A ^{238}U^{238}U nucleus emits a 4.196 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual ^{234}Th^{234}Th nucleus into account.

We have seen that QQ for the overall proton-proton fusion cycle is 26.7 \ MeV26.7 \ MeV. How can you relate this number to the QQ values for the reactions that make up this cycle, as displayed in the above figure?

Calculate the Q values f the fusion reaction ^4He +^4 He = ^8Be ^4He +^4 He = ^8Be In such a fusion energetically favourable? the atomic mass of ^8 Be ^8 Be is 8.0053 u and that of ^4He ^4He is 4.0026 u.

Lithium (Z = 3) has two stable isotopes ^6Li^6Li and ^7Li^7Li. When neutrons are bombarded on lithium sample, electrons and \alpha\alpha - particles are ejected. Write down the nuclear processes taking place.

How much quantity of radioactive substance will remain in decayed after 5 half-lives period ?

Hydrogen bomb is based on the principle of

The mass defect for the nucleus of _{2}He^{4}_{2}He^{4} is 0.0303 a.m.u0.0303 a.m.u. The approximate value of binding energy per nucleon in MeVMeV will be (1 \mathrm{a}.\mathrm{m}.\mathrm{u}=931\mathrm{M}\mathrm{e}\vee)(1 \mathrm{a}.\mathrm{m}.\mathrm{u}=931\mathrm{M}\mathrm{e}\vee)

In plants and animals, the ratio of C^{14}^{14} to C^{12}^{12} is _______.

The energy released during a nuclear reaction is called _______ energy

When an \alpha\alpha-particle is ejected, the atomic number of the atom decreases by ___________ .

If the binding energy per nucleon for _3_3Li^7^7 is 5.6 MeV, the total binding energy of a lithium nucleus is _______ MeV.

The mass of _{2}He^{4}_{2}He^{4} nnucleus is 0.030.03 amu less than the sum of the masses of 22 protons and 22 neutrons. What is the binding energy per nucleon?

Find out the binding energy per nucleon of an \alpha\alpha-particle in MeV. It is given that the masses of \alpha\alpha-particle, proton and neutron are respectively 4.001504.00150 a.m.u., 1.007281.00728 a.m.u. and 1.008671.00867 a.m.u.

Calculate the binding energy of an alpha-particle. Given mass of proton =1.0073=1.0073 a.m.u., mass of neutron =1.0087=1.0087 a.m.u., mass of \alpha\alpha-particle=4.0015=4.0015a.m.u.

The total energy of an electron in the first excited state of the hydrogen atom is -3.4-3.4eV. Find the out its (i) kinetic energy and (ii) potential energy in this state.

\alpha\alpha-particle is emitted from a _{88}^{226}Re_{88}^{226}Re nucleus. If the energy of \alpha\alpha-particle is 4.662 MeV, then what is the total free energy in this decay?

The masses pf ^{11}C and ^{11}B ^{11}C and ^{11}B are respectively 11.0114 u and 11.0093 u 11.0114 u and 11.0093 u . find the maximum energy a positron can have in the \beta^+ \beta^+ decay of ^{11}C to ^{11} B ^{11}C to ^{11} B

How much energy is released in the following reaction : ^7Li + p \rightarrow \alpha + \alpha ^7Li + p \rightarrow \alpha + \alpha tomic mass of ^Li = 7.0160 u ^Li = 7.0160 u and that of ^He = 4.0026 i=u ^He = 4.0026 i=u

Which one of the following cannot emit radiation and why ? Excited nucleus, excited electron.

The nuclide ^{14} C ^{14} C contains (a) how many protons and (b) how many neutrons?

What is the binding energy per nucleon of the europium isotope ^{152}_{63} Eu^{152}_{63} Eu? Here are some atomic masses and the neutron mass. ^{152} _{63}Eu^{152} _{63}Eu 151.921 742 u ^1H^1H 1.007 825 u n 1.008 665 u

In the reactions given below:(a) ^{11}_{6}C \rightarrow ^{z}_{y}B + x + v^{11}_{6}C \rightarrow ^{z}_{y}B + x + v(b) ^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{a}Ne + ^{c}_{b}He^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{a}Ne + ^{c}_{b}HeFind the values of x, y, z and a, b, c.

A nucleus of mass number A, has a mass defect \Delta m.\Delta m. Give the formula, for the binding energy per nucleon, of this nucleus.

The mass of H atom is less than sum of the masses of a proton and electron. Why is this?

Distinguish between isotopes and isobars. Give one example for each of the species.

What other name is given to nuclear fusion? Give reason.

Write the nuclear equations for:{\beta}^{+}{\beta}^{+}- decay of _{ 43 }^{ 97 }{ Tc }_{ 43 }^{ 97 }{ Tc }

Write the nuclear equations for:\alpha\alpha- decay of _{ 88 }^{ 226 }{ Ra }_{ 88 }^{ 226 }{ Ra }

Write the nuclear equations for:{\beta}^{-}{\beta}^{-}- decay of _{ 83 }^{ 210 }{ Bi }_{ 83 }^{ 210 }{ Bi }

What is radioactivity?

Fill in the blanks.The equation E=mc^2E=mc^2 is derived by the scientist _________.

Define binding energy of electron.

Four physical quantities are given in Column I and their order the values in Column II. Match the options appropriately :

When the number of nucleons in nucleus increase, the binding energy per nucleon first _________ and then _________ with increase of mass number.

The mass defect per nucleon is called ___________.

For greater stability, a nucleus should have greater value of binding energy per nucleon. Why?

State any two differences between chemical reactions and nuclear reactions.

Fill in the BlanksIn nuclear reactions we have the conservation of ___________.

Define specific binding energy.

In Column I consider each process just before and just after it occurs. Initial system is isolated from all other bodies. Consider all product particles (even those having rest mass zero) in the system. Match the system in Column I with the result they produce in Column II

Match column I with column II

In Column I some of the nuclear reactions are given. Match this with the energy involved in these reactions in Column II :

Draw a plot of BE/A versus mass number A for 2 \leq\leq A \leq\leq Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei.

Write the relation for binding energy (B.E) (in MeV) of a nucleus of mass \displaystyle _{Z}^{A}{M}\displaystyle _{Z}^{A}{M}, atomic number (Z) and mass number (A) in terms of the masses of its constituents - neutrons and protons.

The fission properties of ^{239}_{94}Pu^{239}_{94}Pu are very similar to those of ^{235}_{92}U^{235}_{92}U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure ^{239}_{94}Pu^{239}_{94}Pu undergo fission?

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :^2_1H\, +\, ^2_1\, \rightarrow\, ^3_2He\, +\, n\, +\, 3.27\, MeV^2_1H\, +\, ^2_1\, \rightarrow\, ^3_2He\, +\, n\, +\, 3.27\, MeV.

Write symbolically the nuclear \beta^+\beta^+ decay process of _{6}^{11}C_{6}^{11}C. Is the decayed product X an isotope or isobar of (_{6}^{11}C_{6}^{11}C)? Given the mass values m (_{6}^{11}C_{6}^{11}C) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

The mass of deutron (_1H^2)(_1H^2) nucleus is 2.0135532.013553 a.m.u. If the masses of proton and neutron are 1.0072751.007275 a.m.u. and 1.0086651.008665 a.m.u. respectively. Calculate the mass defect, the packing fraction, binding energy and binding energy per nucleon.

Find out the binding energy per nucleon of an \alpha\alpha-particle in MeV. It is given that the masses of \alpha\alpha-particles, proton and neutron are respectively 4.001504.00150 a.m.u., 1.007281.00728a.m.u. and 1.008671.00867 a.m.u.

The binding energy per nucleon of _{2}He^{4}_{2}He^{4} is 7.0757.075 MeV. What is its total binding energy?

_____________ is the energy released during nuclear reactions.

The mass of _{3}Li^{7}_{3}Li^{7} nucleus is 0.0420.042 a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

Calculate the (i) mass defect, (ii) binding energy and (iii) the binding energy per nucleon for a _{6}C^{12}_{6}C^{12} nucleus. Nuclear mass of _6C^{12}=12.000000_6C^{12}=12.000000 a.m.u., mass of hydrogen nucleus =1.007825=1.007825 a.m.u. and mass of neutron =1.008665=1.008665 a.m.u.

The binding energy per nucleon for C^{12}C^{12} is 7.687.68 MeV and that for C^{13}C^{13} is 7.477.47 MeV. What is the energy required to remove a neutron from C^{13}C^{13}?

With the help of one example, explain how the neutron to proton ratio changes during alpha decay of a nucleus.

Explain nuclear fusion with an equation.

The mass of deutron (_1H^2)(_1H^2) nucleus is 2.0135532.013553 a.m.u. If the masses of proton and neutron are 1.0072751.007275 a.m.u. and 1.0086651.008665 a.m.u. respectively, calculate the mass defect, the packing fraction , binding energy and binding energu per nucleon.

Obtain the binding energy of a nitrogen nucleus (_7N^{14})(_7N^{14}) from the following data: m_H=1.00783m_H=1.00783 a.m.u., m_n=1.0087m_n=1.0087 a.m.u.; m_N=14.00307m_N=14.00307 a.m.u.

What is the significance of binding energy per nucleon of a nucleus?

In our Nature, where is the nuclear fusion reaction taking place continuously?

What is the signeficance of binding energy per nucleon of a nucleus of a radioactive element?

Write one balanced reaction representing nuclear fusion.

Which reaction is responsible for solar energy? Name the major component of solar energy that reaches us.

Define mass defect of a nucleus. Binding energy of { _{ 8 }^{ }{ O } }^{ 16 }{ _{ 8 }^{ }{ O } }^{ 16 } is 127.5MeV127.5MeV. Write the value of its 'binding energy per nucleon'. Write the value of 1eV1eV energy in joule.

In a certain star, three alpha particles undergo fusion in a single reaction to form ^{12}_{6}C^{12}_{6}C nucleus. Calculate the energy released in this reaction in MeVMeV.Given: m(^{4}_{2}He) = 4.002604\ um(^{4}_{2}He) = 4.002604\ u and m(^{12}_{6}C) = 12.000000\ um(^{12}_{6}C) = 12.000000\ u

(a) What do you mean by QQ value of a nuclear reaction?(b) Write down the expression for QQ value in the class of \alpha\alpha decay.

What are isotopes and isobars?

Find the binding energy of an $\alpha$ -particle from the following data:
Mass of helium nucleus $=4.001265$ a.m.u; mass of proton $=1.007277$ a.m.u. Mass of neutron $=1.00866$ a.m.u (take $1$ a.m.u $=931.5$ MeV)

The binding energies per nucleon for deutron $(_{1}H^{2})$ and helium $(_{2}He^{4})$ are $1.1$ MeV and $7.0$ MeV respectively. Calculate the energy released, when two deutrons fuse to form a helium nucleus $(_{2}He^{4})$ .

The binding energy per nucleon for $_{6}C^{12}$ nucleons is $7.68$ MeV and that for $_{6}C^{13}$ is $7.47$ MeV. Calculate the energy required to remove a neutron from $_{6}C^{13}$ nucleons.

What is (a) mass defect, and (b) binding energy in Oxygen $^{16}_{8}{O}$ , whose nuclear mass is $15.995$ amu. $(m_p=1.0078$ amu $;$ $m_n=1.0087$ amu $)$ .

Answer the following question:
Identify the nature of the radioactive radiations emitted in each step of the decay process given below:
$^{A}_{Z}X \rightarrow ^{A-4}_{Z-1}W$

Answer the following question:
A heavy nucleus splits into two lighter nuclei. Which one of the two-parent nucleus or the daughter nuclei has more binding energy per nucleon ?

Give reason for the release of $21.6\ MeV$ energy in the fusion of deuterium reaction.

The binding energy of $_{ 17 }^{ }{ { Cl }^{ 35 } }$ nucleus is $298$ MeV. Find its atomic mass. The mass of hydrogen atom $\left( _{ 1 }^{}{ { H }^{ 1 } } \right)$ is $1.008143$ a.m.u. Given $1$ a.m.u. $=931$ MeV.

A $^{238}U$ nucleus emits a 4.196 MeV alpha particle. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual $^{234}Th$ nucleus into account.

We have seen that $Q$ for the overall proton-proton fusion cycle is $26.7 \ MeV$ . How can you relate this number to the $Q$ values for the reactions that make up this cycle, as displayed in the above figure?

Calculate the Q values f the fusion reaction $^4He +^4 He = ^8Be$
In such a fusion energetically favourable? the atomic mass of $^8 Be$ is 8.0053 u and that of $^4He$ is 4.0026 u.

Lithium (Z = 3) has two stable isotopes $^6Li$ and $^7Li$ . When neutrons are bombarded on lithium sample, electrons and $\alpha$ - particles are ejected. Write down the nuclear processes taking place.

The mass defect for the nucleus of $_{2}He^{4}$ is $0.0303 a.m.u$ . The approximate value of binding energy per nucleon in $MeV$ will be $(1 \mathrm{a}.\mathrm{m}.\mathrm{u}=931\mathrm{M}\mathrm{e}\vee)$

In plants and animals, the ratio of C $^{14}$ to C $^{12}$ is _______.

When an $\alpha$ -particle is ejected, the atomic number of the atom decreases by ___________ .

If the binding energy per nucleon for Li $^7$ is 5.6 MeV, the total binding energy of a lithium nucleus is _______ MeV.

The mass of $_{2}He^{4}$ nnucleus is $0.03$ amu less than the sum of the masses of $2$ protons and $2$ neutrons. What is the binding energy per nucleon?

Find out the binding energy per nucleon of an $\alpha$ -particle in MeV. It is given that the masses of $\alpha$ -particle, proton and neutron are respectively $4.00150$ a.m.u., $1.00728$ a.m.u. and $1.00867$ a.m.u.

Calculate the binding energy of an alpha-particle. Given mass of proton $=1.0073$ a.m.u., mass of neutron $=1.0087$ a.m.u., mass of $\alpha$ -particle $=4.0015$ a.m.u.

The total energy of an electron in the first excited state of the hydrogen atom is $-3.4$ eV. Find the out its (i) kinetic energy and (ii) potential energy in this state.

$\alpha$ -particle is emitted from a $_{88}^{226}Re$ nucleus. If the energy of $\alpha$ -particle is 4.662 MeV, then what is the total free energy in this decay?

The masses pf $^{11}C and ^{11}B$ are respectively $11.0114 u and 11.0093 u$ . find the maximum energy a positron can have in the $\beta^+$ decay of $^{11}C to ^{11} B$

How much energy is released in the following reaction :
$^7Li + p \rightarrow \alpha + \alpha$ tomic mass of $^Li = 7.0160 u$ and that of $^He = 4.0026 i=u$

The nuclide $^{14} C$ contains (a) how many protons and (b) how many neutrons?

What is the binding energy per nucleon of the europium isotope $^{152}_{63} Eu$ ? Here are some atomic masses and the neutron mass.
$^{152} _{63}Eu$ 151.921 742 u $^1H$ 1.007 825 u
n 1.008 665 u

In the reactions given below:
(a) $^{11}_{6}C \rightarrow ^{z}_{y}B + x + v$
(b) $^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{a}Ne + ^{c}_{b}He$
Find the values of x, y, z and a, b, c.

A nucleus of mass number A, has a mass defect $\Delta m.$ Give the formula, for the binding energy per nucleon, of this nucleus.

Write the nuclear equations for:
${\beta}^{+}$ - decay of $_{ 43 }^{ 97 }{ Tc }$

Write the nuclear equations for:
$\alpha$ - decay of $_{ 88 }^{ 226 }{ Ra }$

Write the nuclear equations for:
${\beta}^{-}$ - decay of $_{ 83 }^{ 210 }{ Bi }$

Fill in the blanks.
The equation $E=mc^2$ is derived by the scientist _________.

When the number of nucleons in nucleus increase, the binding energy per nucleon first _ and then _ with increase of mass number.

Fill in the Blanks
In nuclear reactions we have the conservation of ___________.

Draw a plot of BE/A versus mass number A for 2 $\leq$ A $\leq$ Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei.

Write the relation for binding energy (B.E) (in MeV) of a nucleus of mass $\displaystyle _{Z}^{A}{M}$ , atomic number (Z) and mass number (A) in terms of the masses of its constituents - neutrons and protons.

The fission properties of $^{239}_{94}Pu$ are very similar to those of $^{235}_{92}U$ . The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure $^{239}_{94}Pu$ undergo fission?

How long can an electric lamp of 100 W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as :
$^2_1H\, +\, ^2_1\, \rightarrow\, ^3_2He\, +\, n\, +\, 3.27\, MeV$ .

Write symbolically the nuclear $\beta^+$ decay process of $_{6}^{11}C$ . Is the decayed product X an isotope or isobar of ( $_{6}^{11}C$ )? Given the mass values m ( $_{6}^{11}C$ ) = 11.011434 u and m (X) = 11.009305 u. Estimate the Q-value in this process.

The mass of deutron $(_1H^2)$ nucleus is $2.013553$ a.m.u. If the masses of proton and neutron are $1.007275$ a.m.u. and $1.008665$ a.m.u. respectively. Calculate the mass defect, the packing fraction, binding energy and binding energy per nucleon.

Find out the binding energy per nucleon of an $\alpha$ -particle in MeV. It is given that the masses of $\alpha$ -particles, proton and neutron are respectively $4.00150$ a.m.u., $1.00728$ a.m.u. and $1.00867$ a.m.u.

The binding energy per nucleon of $_{2}He^{4}$ is $7.075$ MeV. What is its total binding energy?

The mass of $_{3}Li^{7}$ nucleus is $0.042$ a.m.u. less than the sum of masses of its nucleons. Find the binding energy per nucleon.

Calculate the (i) mass defect, (ii) binding energy and (iii) the binding energy per nucleon for a $_{6}C^{12}$ nucleus. Nuclear mass of $_6C^{12}=12.000000$ a.m.u., mass of hydrogen nucleus $=1.007825$ a.m.u. and mass of neutron $=1.008665$ a.m.u.

The binding energy per nucleon for $C^{12}$ is $7.68$ MeV and that for $C^{13}$ is $7.47$ MeV. What is the energy required to remove a neutron from $C^{13}$ ?

The mass of deutron $(_1H^2)$ nucleus is $2.013553$ a.m.u. If the masses of proton and neutron are $1.007275$ a.m.u. and $1.008665$ a.m.u. respectively, calculate the mass defect, the packing fraction , binding energy and binding energu per nucleon.

Obtain the binding energy of a nitrogen nucleus $(_7N^{14})$ from the following data: $m_H=1.00783$ a.m.u., $m_n=1.0087$ a.m.u.; $m_N=14.00307$ a.m.u.

Define mass defect of a nucleus. Binding energy of ${ _{ 8 }^{ }{ O } }^{ 16 }$ is $127.5MeV$ . Write the value of its 'binding energy per nucleon'. Write the value of $1eV$ energy in joule.

In a certain star, three alpha particles undergo fusion in a single reaction to form $^{12}_{6}C$ nucleus. Calculate the energy released in this reaction in $MeV$ .
Given: $m(^{4}_{2}He) = 4.002604\ u$ and $m(^{12}_{6}C) = 12.000000\ u$

(a) What do you mean by $Q$ value of a nuclear reaction?
(b) Write down the expression for $Q$ value in the class of $\alpha$ decay.

Calculate the binding energy and binding energy per nucleon (in MeV) of a nitrogen nucleus $(^{14}_7N)$ from the following data :
Mass of proton = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen nucleus = 14.00307 u

Calculate mass defect and binding energy per nucleon of $^{20}_{10}Ne,$ given
Mass of $^{20}_{10}Ne=19.992397$ u
Mass of $^1_1H=1.007825$ u
Mass of $^1_0n=1.008665$ u.

The binding energies per nucleon for deuterium and helium are $1.1$ MeV and $7.0$ MeV respectively. How much amount energy, in joules, will be liberated when $10^6$ deuterons take part in the reaction.

(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number $A$ .
(b) A radioactive isotope has a half-life of $10$ years. How long will it take for the activity to reduce to $3.125$ %?

The binding energy per nucleon for $_{6}C^{12}$ is $7.68\ MeV$ and that for $_{6}C_{13}$ is $7.47\ MeV$ . Calculate the energy required to remove a neutron from $_{6}C_{13}$ .

Write one balanced equation each to show.
(i) Nuclear fission
(ii) Nuclear fusion
(iii) Emission of $\beta^-$ (i.e., a negative beta particle).

Write a nuclear reaction when Uranium $238$ emits an alpha particle to form a Thorium(Th) nucleus.

Write the nuclear equations for:
$\alpha$ - decay of $_{ 92 }^{ 242 }{ Pu }$

The kinetic energy of an $\alpha -$ particle which flies out of the nucleus of a $Ra^{226}$ atom in radioactive disintegration is $4.78\ M\ ev$ . Find the total energy evolved during the escape of the $\alpha -$ particle.

How much energy will be released when $10\ kg$ of $U^{235}$ is completely converts into energy :

In a old rock, ratio of nuclei of uranium and lead is $1:1$ . Half life of uranium is $4.5 \times 10^n$ yrs. Let initially it contains only uranium nuclei. How old is the rock?

When a $U^{238}$ nucleus originally at rest, decay by emitting an alpha particle having a speed $1.5\times 10^{7}\ m^{-1}$ . Then calculate the kinetic energy of the residual nucleus.

Find the binding of the nucleus of lithium isotope $_3Li^7$ and hence find the binding energy per nucleon in it $(M_{_3Li^7} = 7.014353 \ \text{amu}, M_{_1H^1} = 1.007826$ , mass of neutron = $1.00867 \ u$ )

Find the binding energy of the nucleus of lithium isotope $_{3}{Li}^{7}$ and hence find the binding energy per nucleon in it.\left({ M }_{ _{ 3 }{ Li }^{ 7 } }=7.014353amu, { M }_{ _{ 1 }{ H }^{ 1 } }=1.007826, mass of netron=1.00867\ u\right)

Which process is responsible in Hydrogen bomb ?
If 20 N force is applied on a wall which doesn't move the wall than work done by wall is............

Obtain the binding energy (in $MeV$ ) of a nitrogen nucleus $(^{14}_{7}N)$ , given $m(^{14}_{7}N)=14.00307\ u$

The mass of the nucleus $_7N^{14}$ is $14.00000amu$ . Calculate the binding energy per nucleon of $_7N^{14}$ in MeV. Mass of proton $=1.00759amu$ , mass of neutron $=1.00898amu$ and $1 amu=931 MeV.$

A sample has $1000$ nuclear samples find after how much time will they remain $250$ .

Assume that the mass of a nucleus is approximately given by $M = Am_p$ , where $A$ is the mass number. Estimate the density of matter in $kgm^{-3}$ inside a nucleus. What is the specific gravity of nuclear matter?

Write the nuclear equations for:
${\beta}^{+}$ - decay of $_{ 6 }^{ 11 }{ C }$

Write nuclear reaction equations for
Electron capture of $_{ 54 }^{ 120 }{ Xe }$

${ _{ 92 }^{ }{ U } }^{ 238 }$ changes to ${ _{ 85 }^{ }{ At } }^{ 210 }$ by a series of $\alpha$ and $\beta$ decays. Find number of $\alpha$ - decays undergone (an integer)

Identify $X$ in the following nuclear reactions:
$\ ^{1}H+\ ^{0}Be\rightarrow X+n$

What is the minimum energy that is required to break a nucleus of $^{12} \mathrm{C}$ ( of mass $11.99671 \mathrm{u}$ ) into three nuclei of $^{4} \mathrm{He}$ (of mass $4.00151 \mathrm{u} \text { each }) ?$

What is the binding energy per nucleon of $^{262} Bh$ ? The mass of the atom is 262.1231 u.

What is the binding energy per nucleon of the rutherfordium isotope $^{259} _{104}Rf$ ? Here are some atomic masses and the neutron mass.
$^{259}_{104}Rf$ 259.105 63 u $^1H$ 1.007 825 u
n 1.008 665 u

The first of the following two reactions can occur, but
the second cannot.
Explain.
$K^0_S \rightarrow \pi^+ + \pi^-$ (can occur)

$\wedge^0 \rightarrow \pi^+ + \pi^-$ (cannot occur)

Compare and contrast the nature of $\alpha-, \beta-$ and $\gamma-$ radiations.

Answer the following question
In the following nuclear reaction
$n + ^{235}_{92}U \rightarrow ^{144}_{Z}Ba + ^{A}_{36}X + 3n,$
Assign the values of Z and A.