MCQ Questions for Class 12 Medical Physics Nuclei Quiz 11

The temperature required for the process of nuclear fusion is nearly:

0%
$1000 K$
0%
$10^{4} K$
0%
$10^{5} K$
0%
$10^{7} K$

Explanation

To make fusion possible, a high temperature of approximately

$10^7$ K and high pressure is required.

Nuclear binding energy is equivalent to

0%
Mass of proton
0%
Mass of neutron
0%
Mass of nucleus
0%
Mass detect of nucleus

The accepted unit of atomic and molecular mass is:

0%
kilogram
0%
gram
0%
pound
0%
atomic mass unit

Nuclear reaction are given as

$(i) \Box \ (n,p)_{15}p^{32} (ii)\Box (p,a)_8O^{16} \ (iii) _7\Box^4 (\ \ p) _6C^{14}$
missing particles (in box

$\Box$ ) in these reactions are respectively

0%
$S,\ F,\ _0n^1$
0%
$F,\ S,\ _0n^1$
0%
$Be,\ F,\ _0n^1$
0%
None of these

Explanation

$(i)$

$_{16}S^{32}+_0n^1 \rightarrow _{15}p^{32}+_1H^1$

$(ii)$

$_9F^{19}+_1H^1 \rightarrow _2He^4 + _8O^{16}$

$(iii)$

$_7N^{14}+_0n^1 \rightarrow _6C^{14}+_1H^1$

Which of the following particles are constituents of the nucleus

0%
Protons and electrons
0%
Protons and electrons
0%
Neutrons and electrons
0%
Neutrons and positrons

Assertion : Separation of isotope is possible because of the difference in electron numbers of isotope
Reason : Isotope of an element can be separated by using a mass spectrometer.

0%
If both assertion and reason are true and the reason is the correct explanation of the assertion.
0%
If both assertion and reason are true but reason is not the correct explanation of the assertion.
0%
If assertion is true but reason is false.
0%
If the assertion and reason both are false
0%
If assertion is false but reason is true.

Which substance is discovered by the Prafulla Chandra Ray the inventor of Indian Chemical Industry ?

0%
Sodium Chloride
0%
Mercurus Nitrate
0%
Nitre
0%
Sal Ammonia

What type of energy is used to run submarine?

0%
Wind energy
0%
Ocean energy
0%
Atomic energy
0%
None of these

The nucleus with highest binding energy per nucleon is:

0%
$^{238}_{92}U$
0%
$^{4}_{2}He$
0%
$^{16}_{8}O$
0%
$^{56}_{26}Fe$

Explanation

$_{26}Fe^{56}$ as maximum binding energy per nucleon.

A nuclear reaction of

$40\%$ efficiency has

$1014$ decays / second. If the energy obtained per fission is

$250 \,MeV,$ then power of reactor is:

0%
$2 \,kW$
0%
$4 \,kW$
0%
$1.6 \,kW$
0%
$3.2\,kW$

Explanation

Per fission received energy

$-250\ MeV$

$=250 \times 10^6 \times 1.6 \times 10^{-19}\ J$

Number of fissions per second

$=10^{14}$

$\therefore$ Per second received energy

$=250 \times 10^6 \times 1.6 \times 10^{-19}\ \times 10^{14}$

As efficiency is

$40\%$ , so output power

$P=250 \times 10^6 \times 1.6 \times 10^{-5} \times \dfrac{40}{100}$

$=1.6 \times 10^3 \ watt=1.6\ kW$

In the decay of

$_{92}^{238}$ into

$_{82}^{206}Pb$ the number of emitted

$\alpha$ and

$\beta$ particles are respectively :

0%
$8,8$
0%
$6,6$
0%
$6,8$
0%
$8,6$

Explanation

Number of

$\alpha-$ particles

$=\dfrac{Difference\ in\ mass\ number}{4}$
Number of

$\alpha-$ particles

$=\dfrac{238-206}{4}$

$=\dfrac{32}{4}$

$=8$
Number of

$\beta$ particles

$=2$ (Number of

$\alpha$ -particles)

$-$ difference in mass number

$2(8)-(92-82)$

$=16-10$

$=6$

On absorbing energy

$^{22}Ne$ nucleus decays into a-particle and an unknown nucleus. The unknown nucleus is :

0%
Oxygen
0%
Boron
0%
Silicon
0%
Carbon

Explanation

$_{22}Ne \rightarrow 2(_2He^4)+_6C^{14}$
So, unknown nucleus is carbon.

The binding energy per nucleon for a deuterium nucleus is

$1.115\ MeV$ . Mass defect for this nucleus is about :

0%
$2.23\ u$
0%
$0.0024\ u$
0%
$0.027\ u$
0%
Data is insufficient

Explanation

For

$_1H^2$ mass number

$A=2$
Per nucleon binding energy of deuterium

$\bar E_b=1.115\ MeV$
Binding energy

$E_b=\bar E_b \times A$

$E_b=1.115\ MeV \times 2$

$E_b=2.230\ MeV$
Mass number

$\Delta m=\dfrac{2.230}{931}$

$\Delta m=0.0024\ u$

What is atomic mass of

$He$ in amu

$(u)$ units?

0%
$8$
0%
$9$
0%
$2$
0%
$4$

Explanation

The atomic mass of an element is the average mass of an element measured in an atomic mass unit (amu). The atomic mass of

$He$ is

$4\ u$ .

Option

$D$ is correct.

Which nuclear reaction takes place in an atom bomb?

0%
Nuclear fission
0%
Nuclear fusion
0%
Radioactivity
0%
Combustion

A stationary body explodes into two fragments each of rest mass 1kg that move apart at speed of 0.6c relative to the original body. The rest mass of the original body is:-

0%
2 kg
0%
2.5 kg
0%
1.6 kg
0%
2.25 kg

Explanation

Relativistic mass of a body moving with velocity

$v$ can be given by:

$m=\dfrac{m_0c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}$

Where

$m_0$ = rest mass of particle

Applying energy conservation,

Initial mass energy = Final mass energy

$m_0c^2=\dfrac{m_{01}c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}+\dfrac{m_{02}c^2}{\sqrt{1-\left(\dfrac{v}{c}\right)^2}}\Rightarrow m_0=\dfrac{1}{0.8}+\dfrac{1}{0.8}=2.5kg$

Where

$m_{01}$ = rest mass of particle one fragment = 1

$kg$

$m_{02}$ = rest mass of particle two fragment = 1

$kg$

$v$ = velocity of each fragment

The correct statement is

0%
The nucleus $^{6}_{3}Li$ can emit an alpha particle
0%
The nucleus $^{210}_{84}Po$ can emit a proton.
0%
Deuteron and alpha particle can undergo complete fusion.
0%
The nuclei $^{70}_{30} Zn$ and $^{82}_{34}Se$ can undergo complete fusion.

Explanation

For fusion reaction to take place, sum of masses of lighter nuclei must be greater than mass of heavy nucleus formed.

For fusion reaction :

$^2_1H + ^4_2He\rightarrow ^6_3Li$

Sum of mass of lighter nuclei

$M_l = 2.014102 + 4.002603 = 6.016705$ u

Mass of

$^6_3Li$

$M_h = 6.015123$ u

$M_l>M_h$ , thus above reaction is possible.

The energy required to separate the typical middle mass nucleus

$^{120}_{50}Sn$ into its constituent nucleons is :
(Mass of

$^{120}_{50}Sn=$

$119.902199\ amu$ ; mass
of proton

$=1.007825\ amu$ and mass of neutron

$=1.008665\ amu$ )

0%
$921\ MeV$
0%
$821\ MeV$
0%
$1021 \ MeV$
0%
$1121\ MeV$

Explanation

$Z = 50, A-Z = 120-50 = 70$

$\Delta m = Z.m_p + (A-Z)m_n - M$

$\Delta m = [50\times 1.007825 + 70\times 1.008665 - 119.902199]$

$= 1.095601\ MeV$

$E = 1.095601\times 931.478\ MeV$

$=1020.53\ MeV$

$\approx$

$1021\ MeV$

In the reaction

$_7N^{14} +$

$_2He^4 \rightarrow _8O^{17} +$

$_1H^1$ . The minimum energy of

$\alpha$ -particle is

$M_N=$

$14.00307$ amu

$M_{He}=$

$4.00260$ amu

$M_o =$

$16.99914$ amu

$M_H=$

$1.00783$ amu

0%
$1.21MeV$
0%
$1.62MeV$
0%
$1.89MeV$
0%
$1.96MeV$

Explanation

Mass on reactant side

$= (14.00307 + 4.00260) amu$

Mass on product side

$= (16.99914 + 1.00783) amu$

So, mass defect

$= 1.3 \times10^{-3} amu.$

Energy of the

$\alpha$ particle

$= 931.5\times1.3\times10^{-3} MeV$

$= 1.21 MeV$

The kinetic energy (in keV) of the alpha particle, when the nucleus

$^{210}_{84}Po$ at rest undergoes alpha decay, is

0%
$5319$
0%
$5422$
0%
$5707$
0%
$5818$

Explanation

$^{210}_{84}Po\rightarrow_{2}^{4}\mathrm{H}\mathrm{e}+_{82}^{206}$ Pb

$\mathrm{Q}=(209.982876-4.002603-205.97455)\mathrm{C}^{2}$

$=5.422\mathrm{M}\mathrm{e}\mathrm{V}$

From conservation of momentum:

$\sqrt{2\mathrm{K}_{1}(4)}=\sqrt{2\mathrm{K}_{2}(206)}$

$4\mathrm{K}_{1}=206\mathrm{K}_{2}$

$\displaystyle \mathrm{K}_{1}=\frac{103}{2}\mathrm{K}_{2}$

$\mathrm{K}_{1}+\mathrm{K}_{2}=5.422$

$\displaystyle \mathrm{K}_{1}+\frac{2}{103}\mathrm{K}_{1}=5.422$

$\Rightarrow \displaystyle \frac{105}{103}\mathrm{K}_{1}=5.422$

$\mathrm{K}_{1}=5.319\mathrm{M}\mathrm{e}\mathrm{V}=5319\mathrm{K}\mathrm{e}\mathrm{V}$

Atomic masses of

$^{14}_{7}N$ and

$^{16}_{8}O$ are

$14.008$ amu and

$16.000$ amu respectively. The mass of

$^{1}_{1}H$ atom is 1.007825 amu and the mass of neutron is

$1.008665$ amu. Pick the correct option

0%
$^{14}_{7}N$ is more stable than $^{16}_{8}O$
0%
$^{16}_{8}O$ is more stable than $^{14}_{7}N$
0%
Both are not stable
0%
Both are equally stable

Explanation

Mass of 7 protons and 7 neutrons=

$7(1.007825+1.008665)=14.11543amu$

Mass of

$_7^{14}N=14.008amu$

Hence mass lost as energy in formation of nitrogen atom=

$0.10743amu$

Mass of 8 protons and 8 neutrons=

$8(1.007825+1.008665)amu=16.13192amu$

Mass of

$_8^{16}O=16.000amu$

Hence mass lost as energy in formation of oxygen atom=

$0.13192amu$

Since more mass has been lost in formation of oxygen atom, oxygen atom is more stable.

The mass of chlorine (

$_{17}Cl^{35}$ ) atom is 34.98 amu, mass of proton = 1.007825 amu, mass of neutron= 1.008665 amu. Then binding energy is :

0%
$287.83$ Mev
0%
$287.83$ joules
0%
$8.22$ Mev
0%
$8.22$ joule

Explanation

$\Delta m = (17 \times 1.007825) + (18 \times 1.008665) - 34.98$

$= 0.308995 amu$

$= 0.308995 \times 931.478 \dfrac{MeV}{c^{2}} ....... [$ using

$1 amu = 931.478 \dfrac{MeV}{c^{2}}]$

$= 287.82 \dfrac{MeV}{c^{2}}$

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

$= 287.82 MeV$

Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy

$1.5 kT$ , when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature

$T$ required for them to reach a separation of

$4 \times 10^{-15}$ m is in the range

0%
$1.0\times 10^{9}\mathrm{K}<\mathrm{T}<2.0\times 10^{9}\mathrm{K}$
0%
$2.0\times 10^{9}\mathrm{K}<\mathrm{T}<3.0\times 10^{9}\mathrm{K}$
0%
$3.0\times 10^{9}\mathrm{K}<\mathrm{T}<4.0\times 10^{9}\mathrm{K}$
0%
$4.0\times 10^{9}\mathrm{K}<\mathrm{T}<5.0\times 10^{9}\mathrm{K}$

Explanation

Conservation of energy:

$KE = PE$

$2 (1.5kT) = \cfrac{e^2}{4 \pi \epsilon _o d}$

$3kT = \cfrac{(1.6 \times 10^{-19})^2}{4 \pi (8.84 \times 10^{-12} (4 \times 10^{-15}}$

$T =1.4 \times 10^9 K$

A radioactive nucleus undergoes a series of decays according to the sequence

$A\overset{\beta}{\rightarrow}A_1\overset{\alpha}{\rightarrow}A_2\overset{\alpha}{\rightarrow}A_3$
If the mass number and atomic number of

$A_3$ are

$172$ and

$69$ respectively, then the mass number and atomic number of A is

0%
$56, 23$
0%
$180, 72$
0%
$120, 52$
0%
$84, 38$

Explanation

$^A_Z{A} \rightarrow A_1 \rightarrow A_2 \rightarrow _69 {A}_3^{172}$

Mass balance,

$A = 0 + 4 + 4 + 172$

$= 180$
Charge balance,

$Z= -1 + 2 + 2 + 69$

$= 72$

Parto of the uranium decay series is shown

$_{92}U^{238} \rightarrow _{90}Th^{234} \rightarrow _{91}Pa^{234}\rightarrow$

$_{92}U^{234} \rightarrow _{90}Th^{230}\rightarrow _{88}Ra^{226}$
How many pairs of isotopes are there in the above series :

0%
1
0%
2
0%
3
0%
0

Explanation

Atoms of the element having same atomic number and different mass number are known as isotopes.

Thus

$_{92}U^{238}$ is an isotope of

$_{92}U^{234}$ whereas

$_{90}Th^{234}$ is an isotope of

$_{90}Th^{230}$ .

Hence there are two pairs of isotopes.

The above is a plot of binding energy per nucleon

$\mathrm{E}_{\mathrm{b}}$ , against the nuclear mass

$\mathrm{M};\mathrm{A},\ \mathrm{B},\ \mathrm{C},\ \mathrm{D},\ \mathrm{E},\ \mathrm{F}$ correspond to different nuclei. Consider four reactions :
(i)

$\mathrm{A}+\mathrm{B}\rightarrow \mathrm{C}+\epsilon$
(ii)

$\mathrm{C}\rightarrow \mathrm{A}+\mathrm{B}+\epsilon$
(iii)

$\mathrm{D}+\mathrm{E}\rightarrow \mathrm{F}+\epsilon$
(iv)

$\mathrm{F}\rightarrow \mathrm{D}+\mathrm{E}+\epsilon$
where

$\epsilon$ is the energy released ? In which reaction is

$\epsilon$ positive ?

0%
(i) and (iv)
0%
(i) and (iii)
0%
(ii) and (iv)
0%
(ii) and (iii).

Explanation

The reactions for which product is/are more stable than reactant, energy & will be positive. (Energy will be released)

As B.E/nuclear decides stability

$\therefore$ Stability

$:C>B$

$C>A$

$D>F$

and

$E>F$

$\left( i \right) A+B\longrightarrow C+\varepsilon$

$\left( iv \right) F\longrightarrow D+E+\varepsilon$

are the reactions for which

$\varepsilon$ will be positive

Match list I and list II.
List I List II
A.

$_{1}^{2}H+^{3}_{1}H\rightarrow ^{4}_{2}He +^{1}_{0}n+17.6 MeV$ 1. Artificial radioactivity

$^{235}_{92}U+^{1}_{0}n\rightarrow^{143}_{56}Ba+^{90}_{36}Kr+3^{1}_{0}n+200Mev$ 2. Isodiaphers

$^{23}_{13}Al+^{4}_{2}He\rightarrow ^{30}_{14}Si+^{1}_{1}H$ 3. Atom Bomb

$^{m}_{Z}A\overset{-\alpha}{\rightarrow} ^{m-4}_{Z-2}B$ 4. Nuclear Fissio Nuclear Fusion

0%
A - 5; B - 3,4; C - 1; D - 2
0%
A - 2; B - 3,4; C - 1; D - 5
0%
A - 1,5; B - 3; C - 4; D - 2
0%
A - 3,4; B - 1; C - 5; D - 4,2

During a

$\beta$ positive decay experiment it is observed that kinetic energy of

$\beta$ positive particle is 50% of Q value of the reaction, then select the correct alternative

0%
Almost remaining 50% energy must be in form of kinetic energy of daughter nucleus.
0%
Almost remaining 50% energy must be lost in form of heat.
0%
Almost remaining 50% energy must be taken away by neutrino.
0%
Almost remaining 50% energy must be taken away by antineutrino.

Explanation

$_{Z}X^{A}\rightarrow _{Z-1}Y^{A}+_{+1}e^{0}+v+Q$
Almost all energy came from Q value is distributed between

$_{+1}e^{0}$ &

$v.$

A helium atom, a hydrogen atom and a neutron have masses of

$4.003 u$ ,

$1.008 u$ and

$1.009 u$ (unified atomic mass units), respectively. Assuming that hydrogen atoms and neutrons can fuse to form helium, what is the binding energy of a helium nucleus?

0%
$2.01 u$
0%
$3.031 u$
0%
$1.017 u$
0%
$0.031 u$

Explanation

$2 \:_1H^1 + 2n \rightarrow _2 He ^4$
Binding energy

$=$ Mass defect

$=-m_{He} + 2m_n + 2m_H$

$=2(1.008) + 2(1.009) -4.003$

$=0.031 u$

When the number of nucleons in nuclei increase, the binding energy per nucleon

0%
increases continuously with mass number
0%
decreases continuously with mass number
0%
remains constant with mass number
0%
first increases and then decreases with increase of mass number

A beam of

$16 MeV$ deutrons from a cyclotron falls on a copper block. The beam is equivalent to a current of 15

$\mu A$ . At what rate do the deutrons strike the block?

0%
$9.4 \times 10^9$
0%
$9.4 \times 10^7$
0%
$9.4 \times 10^{11}$
0%
$9.4 \times 10^{13}$

Explanation

Given:

$I = 15\mu A$

$=1.5\times 10^{-5}A$ and

$|e|=1.6\times 10^{-19}C$

Using,

$I = \dfrac{q}{t}$ where

$q=n|e|$

$\therefore$ Rate of deutrons striking the block,

$\dfrac{dn}{dt} = \dfrac{I}{|e|} =\dfrac{1.5\times 10^{-5}}{1.6\times 10^{-19}} = 9.4\times 10^{13}$

In an

$\alpha-decay$ , the kinetic energy of

$\alpha-particle$ is

$48\ MeV$ and Q value of the reaction is

$50\ MeV$ . The mass number of the mother nucleus is: - (Assume that daughter nucleus is in ground state)

0%
$96$
0%
$100$
0%
$104$
0%
none of these

Explanation

Let the reaction be:

$M_1 \rightarrow \alpha + M_2 +Q$

By conservation of energy:

$Q = KE_1+KE_2$

$KE_1 = 48\ MeV$

$KE_2 = 50 -48 = 2\ MeV$

Now conservation of momentum:

$M_ {\alpha}V_{\alpha} = M_2V_2$
Squaring both the sides and dividing by

$2$ .
Also,

$KE = \cfrac{1}{2}mv^2 = mp/2$

$m_{\alpha} =4$

Substituting:

$4 \times 48 = M_2 \times 2$

$M_2 = 96$

Now by conservation of mass:

$M_1 = M_{\alpha} + M_2$

$M_1 =100$

The binding energy of an electron in the ground state of

$He\$ atom is

$E_0=24.6 eV.$ The energy required to remove both the electrons from the atom is

0%
$24.6eV$
0%
$79.0eV$
0%
$54.4eV$
0%
None of these

Explanation

Given, Energy required to remove electron in ground state from a neutral atom is

$24.6eV$ , which makes it

${He}^{+}$ hydrogen like atom.
Energy required to remove electron from singly ionized helium is

$13.6\times{2}^{4}$ .
Therefore, Energy required to remove both the electrons is

$24.6+ 54.4= 79.0eV$

Find the binding energy of valence electron in the ground state of a

$Li$ atom if the wavelength of the sharp series is known to be

$\lambda_{1}$

$= 813 nm$ and the short wave cutoff wavelength

$\lambda_{2}$ ,

$= 350 nm$

0%
$3.54 eV$
0%
$5.32 eV$
0%
$1.5 eV$
0%
$4.32 eV$

Explanation

For the first line of sharp series

$(3s\rightarrow 2p)$ in a

$Li$ atom

$\displaystyle \frac{hc}{\lambda}=\frac{hR}{(3+\alpha _0)^2} + \frac{hR}{(2+\alpha_1)^2}$ ...........(1)

For the short wave cut off wavelength of the same series

$\displaystyle \frac{hc}{\lambda_{2}}= \frac{hR}{(2+\alpha_1)^2}$ ...............(2)

Subtracting (1) from (2), we get:

$(3+\alpha _0)^2$

$=\displaystyle \frac{hR\lambda_{1}\lambda_{2}}{hc(\lambda_{1}-\lambda_{2})}=\frac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}$

$\displaystyle 3+\alpha _{0}=\sqrt\frac{R\lambda _{1}\lambda _{2}}{2\pi c(\lambda _{1}-\lambda _{2})}$

In the ground state BE of electrons is:

$=\displaystyle \dfrac{hR}{(2+\alpha _{0})^{2}}$

$\displaystyle =\dfrac{hR}{\left (\sqrt{\dfrac{R\lambda_{1}\lambda_{2}}{2\pi c(\lambda_{1}-\lambda_{2})}-1} \right )^{2}}=5.32 eV$

The above is a plot of binding energy per nucleon

$E_{b},$ against the nuclear mass M; A, B,C, D, E, correspond to different nuclei. Consider four reactions :
(i)

$A+B\rightarrow C+\varepsilon$
(ii)

$C\rightarrow A+B+\varepsilon$
(iii)

$D+E\rightarrow F+\varepsilon$
(iv)

$F\rightarrow D+E+\varepsilon ,$
where

$\varepsilon$ is the energy released. In which reactions

$\varepsilon$ positive?

0%
(i) and (iii)
0%
(ii) and (iv)
0%
(ii) and (iii)
0%
(i) and (iv)

Explanation

Lighter nuclei $A$ and $B$ can fuse to form a heavier nucleus C. In these fusion reactions, there is a mass defect giving rise to a release in energy. On the other hand, a heavy nucleus $F$ can be split into two nuclei $D$ and $E$ of moderate masses. In this fission reaction also, there is a mass defect which appears in the form of release of energy. Thus energy is released in the processes $(i)$ and $(iv)$ .

A fusion reaction consists of combining four protons into an

$\alpha$ - particle. The mass of

$\alpha$ - particle is

$4.002603 amu$ and that of proton is

$1.007825 amu$

0%
the equation $4p{_{1}}^{1} \rightarrow He{_{2}}^{4}$ does not satisfy conservation of charge
0%
the correct reaction equation may be $4p{_{1}}^{1} \rightarrow He{_{2}}^{4}+2\beta ^{+}+2\upsilon$ where $\beta ^{+}$ is positron and $\upsilon$ is the neutrino (zero rest mass and uncharged)
0%
loss of mass in the reaction is $0.028697 amu$
0%
the energy equivalent of the mass defect is $26.7 MeV$

Explanation

Mass defect,

$\Delta m=4m_{H}-m_{He}$

$= 4(1.007825)-4.002603\\ =0.028697 u$

$=0.028697 u \times 932 \dfrac{MeV}{u}\\ =26.7\ MeV$

Statement 1: Binding energy of

$_8O^{15}$ is less than

$_7N^{14}$
Statement 2 : Nuclear force is independent of the charge on the nucleons. For isobars, the difference in binding energies is mainly due to the difference in proton-proton coulombic repulsion

0%
Statement 1 is True, Statement 2 is True; Statement 2 is a correct explanation for Statement 1
0%
Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1
0%
Statement 1 is True, Statement 2 is False
0%
Statement 1 is False, Statement 2 is True.

The following deuterium reactions and corresponding reaction energies are found to occur

$^{14}N(d, p)^{15}N, Q=8.53 MeV$

$^{15}N(d, \alpha)^{13}C, Q=7.58 MeV$

$^{13}C(d, \alpha)^{11}B, Q=5.16 MeV$
The rotation

$^{14}N(d, p)^{15}N$ represents the reaction

$^{14}N+d\rightarrow ^{15}N+p$

$_2^4He=4.0026 amu, _1^2He=2.014 amu, _1^1H=1.0078 amu, n=1.0087 amu (1 amu=931 MeV)$
The Q values of the reaction

$^{11}B(\alpha, n)^{14}N$ is

0%
$0.5 eV$
0%
$0.5 MeV$
0%
$0.05 MeV$
0%
$0.05 eV$

Explanation

The following reactions are as follows-

$^{14}N (d,p)^{15}N : ^{14}N + d \rightarrow ^{15} N + p Q_1 = 8.53 MeV$ ..............(i)

$^{15}N (d,\alpha)^{13}C : ^{15}N + d \rightarrow ^{13} C + \alpha Q_2 = 7.58 MeV$ ..............(ii)

$^{13}C (d,\alpha)^{11}B : ^{13}C + d \rightarrow ^{11} B + \alpha Q_2 = 5.16 MeV$ ...........(iii)

Adding (i), (ii)

$(iii),$

$\implies$ $

$^{14}N + 3 d \rightarrow ^{11} B + 2\alpha +p$ ...........(a)

$Q$ value for this reaction

$Q_a = Q_1 + Q_2 + Q_3 = 21.27 MeV$

Now for reaction,

$3d \rightarrow p+ \alpha + n$ ...........(b)

$\therefore Q_b = [3\times 2.014 - 1.0078 + 4.0026 - 1.0087] \times 931 MeV = 21.32 MeV$

Now

$(b) - (a) \implies$

$^{11}B + \alpha \rightarrow ^{14} N + n$

$\therefore$ Q value for this reaction,

$Q = Q_b-Q_a = 21.32 - 21.27 = 0.05 MeV$

The rest mass of a deuteron is equivalent to an energy of

$1876$ MeV, that of a proton to

$939$ MeV, and that of a neutron to

$940$ MeV. A deuteron may disintegrate to a proton and a neutron if it

0%
emits an X-ray photon of energy $2$ MeV
0%
captures an X-ray photon of energy $2$ MeV
0%
emits an X-ray photon of energy $3$ MeV
0%
captures an X-ray photon of energy $3$ MeV

Explanation

DEUTRON

$_1 H^2 = n+p$
Total energy of deutron

$= 1476\ MeV$
Energy of neutron

$+$ proton

$= 940 +939 = 1879\ MeV$
Energy required to disintegrate Deutron

$= 3\ MeV$

Assuming that about 20 MeV of energy is released per fusion reaction

$_1H^2+_1H^2\rightarrow _2He^4+E+$ other particles, then the mass of

$_1H^2$ consumed per day in a fusion reactor of power 1 megawatt will approximately be

0%
0.001 g
0%
1 g
0%
10.0 g
0%
1000 g

Explanation

$E = \Delta m c^2$
For one reaction
Mass Defect =

$\Delta m$

$=2(m_{H}) - m_{He} - m_n$

$=2(2.015)-3.017-1.009$ amu

$=0.004 amu$

$1 amu = 931.5 MeV/c^2$
Hence

$E = 0.004 \times 931.5 MeV = 3.724 MeV$

$E = 3.726 \times 1.6 \times 10^{-13} J =5.96 \times 10^{-13} J$
.
Total Requirement =

$1 MW = 10^6 J/s$
Total fusions required =

$\cfrac{10^6}{5.96 \times 10^{-13} } = 1.67 \times 10^{18}s^{-1}$
Mass rate =

$\cfrac{1.67 \times 10^{18}}{N_A} \times 2 \times 2 = 1.1 \times 10^{-5} g/s$
Total consumption in a day =

$M \times(24\times 3600) = 0.95 gms$

If mass of

$U^{235}=235.12142amu$ , mass of

$U^{236}=236.1205 amu$ , and mass of neutron

$1.008665 amu$ , then the energy required to remove one neutron from the nucleus of

$U^{236}$ is nearly about

0%
$75 MeV$
0%
$6.5 MeV$
0%
$1 eV$
0%
zero

Explanation

The total mass of

$U^{235}$ and a neutron is

$M_1=235.12142-1.008665=236.130\ amu$

Increases in mass when one neutron is removed from

$U^{236}=M_1-$ mass of

$U^{236}$

$=236.130-236.1205\\ =0.009\ amu \\ =0.009\times 931\ MeV \\ =8.9\ MeV\\ \sim 6.5\ MeV$

In the nuclear reaction given by

$_2He^4+_7N^{14}\rightarrow _1H^1+X$ the nucleus X is

0%
nitrogen of mass $16$
0%
nitrogen of mass $17$
0%
oxygen of mass $16$
0%
oxygen of mass $17$

Explanation

$_2 He^4 + _7N^{14} \rightarrow _1H^1 + _ZX^A$

Mass Balance:

$4 + 14 = 1+ A$

$A = 17$

Charge balance:

$2 + 7 = 1 +Z$

$Z =8$

Element of atomic no.

$8$ and mass

$17$ - Oxygen

The binding energies per nucleon of deutreron

$(_1H^2)$ and helium

$(_2He^4)$ atoms are

$1.1 MeV$ and

$7 MeV$ . If two deuteron atoms react to form a single helium atom, then the energy released is

0%
$13.9 MeV$
0%
$26.9 MeV$
0%
$23.6 MeV$
0%
$19.2 MeV$

Explanation

$2 _1 H^2 \rightarrow _2He ^4$
Energy released

$=$ Nuclear energy of Product

$-$ Nuclear energy of Reactant

$= 7 \times 4 - (1.1 \times 2)\times 2\ MeV$

$= 23.6 \ MeV$

What would be the energy required to dissociate completely

$1 g$ of

$Ca^{40}$ into its constituent particles?
Given: Mass of proton

$=1.007277 amu$ ,
Mass of neutron

$=1.00866 amu$ ,
Mass of

$Ca-40=39.97545 amu$
(Take

$1 amu=931 MeV)$

0%
$4.813\times 10^{24} MeV$
0%
$4.813\times 10^{24} eV$
0%
$4.813\times 10^{23} MeV$
0%
none of the above

Explanation

$Ca = 20p+20n$

For

$1$ atom of

$Ca$ :
Mass Defect

$= 20\times 1.007277 + 20\times 1.00866 -39.97545 =40.31-39.97545=0.34329\ amu$

Energy released for

$1$ atom of

$Ca = 319.6\ MeV$
Energy release for

$1\ gm$ of

$Ca = \cfrac{1}{39.975} \times 6.023 \times 10^{23} \times 319.6$

$= 4.81 \times 10^{24} MeV$

Calculate the binding energy of a deuteron atom, which consists of a proton and a neutron, given that the atomic mass of the deuteron is 2.014102 u

0%
0.002388 MeV
0%
2.014102 MeV
0%
2.16490 MeV
0%
2.224 MeV

Explanation

$\begin{array}{l} \text { atomic mass M(H) of hydorgen and nuclear } \\ \text { mass }\left(m_{n}\right) \text { are } \\ m(H)=1.007825 u \text { and } M_n=1.008665u \\ \text { mass defect } \\ \Delta m=[M(H)+M_n-M(D)] \end{array}$

$\begin{aligned} & 2.016490u-2.014102u \\ \Delta m &=0.002388u \end{aligned}$

$\begin{array}{l} \text { As 1u correspond to 931.49 Mev energy } \\ \text { therefore mass defect corresponds to } \end{array}$

$\begin{aligned} \text { energy } E_{b} &=0.002388 \times 931.49 \\ &=2.224 \mathrm{MeV} \\ \text { hence option } D & \text { is correct } \end{aligned}$

A stationary thorium nucleus

$(A=220, Z=90)$ emits an alpha particle with kinetic energy

$E_{\alpha}$ . What is the kinetic energy of the recoiling nucleus?

0%
$\dfrac {E_{\alpha}}{108}$
0%
$\dfrac {E_{\alpha}}{110}$
0%
$\dfrac {E_{\alpha}}{55}$
0%
$\dfrac {E_{\alpha}}{54}$

Explanation

$^{220} Z _{90} \rightarrow \alpha + X$

$m_X = 220 - m_{\alpha} = 216$

$E =\cfrac{p^2}{2m}$

$p = \sqrt{2mE}$

Momentum Conservation:

$\sqrt{2m_{\alpha}E_{\alpha}}= \sqrt{2m_{X}E_{X}}$

$m_{\alpha}E_{\alpha}= m_{X}E_{X}$

$4E_{\alpha}= 216E_{X}$

$E(X) = \cfrac{E_{\alpha}}{54}$

Rank the following nuclei in order from largest to smallest value of the binding energy per nucleon:
(i)

$_2^4He$ , (ii)

$_{24}^{52}Cr$ , (iii)

$_{62}^{152}Sm$ , (iv)

$_{80}^{100}Hg$ , (v)

$_{92}^{252}Cf$

0%
$E_{(i)} = E_{(ii)} = E_{(iii)} = E_{(iv)} = E_{(v)}$
0%
$E_{(v)} > E_{(iv)} > E_{(iii)} > E_{(ii)} > E_{(i)}$
0%
$E_{(i)} > E_{(ii)} > E_{(iii)} > E_{(iv)} > E_{(v)}$
0%
$E_{(ii)} > E_{(iii)} > E_{(iv)} > E_{(v)} > E_{(i)}$

Explanation

In this graph peak or the element with the maximum binding energy is This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.

so elements having less atomic number then Fe that is 26 in that range the with increase in the atomic number binding energy will increase

While we can see in this graph that element with the atomic number greater than that of Fe that is 26 with the increase in the atomic number binding energy will decrease .

so by using this concept we can compare the following result .

as Cr have atomic number 24 which is most close to 26 so it have maximum binding energy

now then Sm then Hg and then Cf and at last He

From the graph,

$^{52}_{24}Cr > ^{152}_{62}Sm > ^{100}_{80}Hg > ^{252}_{92}Cf > ^{4}_{2}He$

In the graph, it is seen that

$^{56}Fe$ has the maximum

$B\ E$ per nucleon

$\Rightarrow (c)$

Mark the correct statement(s)

0%
for an exothermic reaction, if Q value is $+12.56 MeV$ and the KE of incident particle is $2.44 MeV$ , then the total KE of products of reaction is $15.00$ MeV
0%
for an exothermic reaction, if Q value is $+12.56 MeV$ and the KE of incident particle is $2.44 MeV$ , then the total KE of products of reaction is $12.56$ MeV
0%
For an endothermic reaction, if we give the energy equal to $|Q|$ value of reaction, then the reaction will be carried out
0%
for an exothermic reaction, the BE per nucleon of products should be greater than the BE per nucleon of reactants

In the nuclear reaction presented above, the "other particles" might be

0%
an alpha particle, which consists of two protons and neutrons
0%
two protons
0%
one protons and one neutron
0%
two neutrons

Explanation

Difference in Mass number is

$236-140-94=2$ and atomic number is

$92-38-54=0$
which is most probably two neutrons.

Binding energy per nucleon is maximum

0%
for lighter order elements (low mass number)
0%
for heavier order elements (high mass number)
0%
for middle order elements
0%
equal for all order elements

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 11 - MCQExams.com

0:0:2

Practice Class 12 Medical Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.