In $$Q-26$$, find the initial activity of sample.

$$6.79\ \times 10^{16}\ disintegration/sec$$

$$5.79\ \times 10^{15}\ disintegration/sec$$

$$10\ \times 10^{15}\ disintegration/sec$$

MCQ Questions for Class 12 Medical Physics Nuclei Quiz 8

What is the

$Q-value$ of the reaction?

$P + \,^{7}Li \rightarrow \, ^{4}He +\, ^{4}He$
The atomic masses of

$^{1}H, ^{4}He$ and

$^{7}Li$ are

$1.007825u, 4.0026034u$ and

$7.016004u$ respectively.

0%
$17.35\ MeV$
0%
$18.06\ MeV$
0%
$177.35\ MeV$
0%
$170.35\ MeV$

Explanation

The total mass of the initial particles

$m_{i} = 1.007825 + 7.016004$

$= 8.023829u$
and the total mass of final particles

$m_{f} = 2\times 4.002603$

$= 8.005206u$
Difference between initial and final mass of particles

$\triangle m = m_{i} - m_{f}$

$= 8.023829 - 8.005206$

$= 0.018623u$
The

$Q-value$ is given by

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

$= 0.018623\times 931.5$

$= 17.35\ MeV$ .

Two radioactive sources

$X$ and

$Y$ of half lives

$1h$ and

$2h$ respectively initially contain the same number of radioactive atoms. At the end of

$2h$ , their rates of disintegration are in the ratio of :

0%
$4:3$
0%
$3:4$
0%
$1:2$
0%
$2:1$

Explanation

Rate of disintegration

$\alpha$ number of atoms left

For

$X \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

For

$Y \dfrac{N}{N_0} = \left(\dfrac{1}{2}\right)^1 = \dfrac{1}{2}$

$\therefore \dfrac{R_X}{R_Y} = \dfrac{\dfrac{N_0}{4}}{\dfrac{N_0}{2}}=\dfrac{1}{2}$

The binding energy per nucleon of

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

$7.97 MeV$ and that of

$_{ 8 }{ { O }^{ 17 } }$ is

$7.75 MeV$ . The energy required to remove one neutron from

$_{ 8 }{ { O }^{ 17 } }$ is ___________

$MeV$ .

0%
$3.62$
0%
$3.52$
0%
$4.23$
0%
$7.86$

Explanation

$_{ 8 }{ { O }^{ 16 } }\longrightarrow$

$_{ 8 }{ { O }^{ 17 } }+_{ 0 }{ { n }^{ 1 } }$
∴ energy to remove neutron

= binding energy of

$_{ 8 }{ { O }^{ 17 }}$ –binding energy of

$_{ 8 }{ { O }^{ 16 }}$ = (17 × 7.75) – (16 × 7.79) = 4.23 MeV.

Consider the nuclear reaction

$X^{200}\rightarrow\,A^{110}+B^{90}+ energy$
If the binding energy per nucleon for

$X , \,A$ and

$B$ is

$7.4 MeV , 8.2 MeV$ and

$8.2 MeV$ respectively . What is the energy released?

0%
$200 MeV$
0%
$160 MeV$
0%
$110 MeV$
0%
$90 MeV$

Explanation

Energy released

$=(E_A+E_B)-E_X$

$=(110\times 8.2+90 \times 8.2)-200 \times 7.4$

$=1640-1480$

$=160 MeV$

$^{234}U$ has

$92$ protons and

$234$ nucleons total in its nucleus. It decays by emitting an alpha particle. After the decay it becomes.

0%
$^{232}U$
0%
$^{232}Pa$
0%
$^{230}Th$
0%
$^{230}Ra$

Explanation

$\displaystyle ^{234}_{92}U\overset{a}{\rightarrow} {^{230}_{90}Th}+\alpha$
The mass number of thorium is

$230$ and its atomic number, Z is

$90$ .
The mass number of radium is

$226$ and it atomic number is

$88$ . As it is given

$^{230}Ra$ , students can make a mistake. Also the value of Z is not given. As at one instant, only one

$\alpha$ particle can be emitted, unless it is given two successive emission of

$\alpha$ particle, Ra cannot be obtained.

The binding energy per nucleon of

$C^{12}$ is

$E_{1}$ and that of

$C^{13}$ is

$E_{2}$ . The energy required to remove one neutron from

$C^{13}$ is

0%
$12E_{1} + 12E_{2}$
0%
$13E_{2} - 12E_{1}$
0%
$12E_{2} - 13E_{1}$
0%
$13E_{2} + 12E_{1}$

Explanation

Binding energy of

$C^{12}$ is

$BE_{1} = 12\times E_{1}$
Binding energy of

$C^{13}$ is

$BE_{2} = 13\times E_{2}$
Thus, energy required to remove one neutron from

$C^{13}$ is

$\triangle E = BE_{2} - BE_{1} = 13E_{2} - 12E_{1}$ .

A nuclear reaction along with the masses of the particles taking part in it is as follows:

$A+B\longrightarrow C+D+QMeV$

$\underset { amu }{ 1.002 } +\underset { amu }{ 1.004 } \longrightarrow \underset { amu }{ 1.001 } +\underset { amu }{ 1.003 } +Q$
The energy

$Q$ liberated in the reaction is

0%
$1.234MeV$
0%
$1.862MeV$
0%
$0.931MeV$
0%
$0.800MeV$

Explanation

by relation

$Q=\sum { { m }_{ r } } -\sum { { m }_{ p } } =2.006-2.004$

$\therefore Q=0.002\times 931=1.86MeV\quad$

Li nucleus has three protons and four neutrons. Mass of lithium nucleus is 7.03.6005 amu. Mass of proton is 1.007277 amu and mass of neutron is 1.008665 amu. Mass defect for lithium nucleus in amu is

0%
0.04048
0%
0.04050
0%
0.04052
0%
0.04055

Explanation

Mass defect = mass of nucleons - mass of nucleus

$= (3 \times 1.007277 + 4 t008665)- 7.016005$

$= 0.040486\, amu$

$\approx 0.04050$

The mass of

$_{7}N^{15}$ is

$15.0011\ amu$ , mass of

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

$15.99492\ amu$ and

$m_{P} = 1.00783\ amu$ . Determine binding energy of last proton of

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

0%
$2.13\ MeV$
0%
$0.13\ MeV$
0%
$10\ MeV$
0%
$12.13\ MeV$

Explanation

$M(_{8}O_{16}) = M(_{7}N^{15}) + 1m_{P}$
binding energy of last proton

$= M(N^{15}) + m_{P} - M(_{1}O^{16})$

$= 15.00011 + 1.00783 - 15.99492$

$= 0.01302\ amu = 12.13\ MeV$ .

Which of the following statements is correct?

0%
The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons
0%
The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons
0%
In nuclear fission, energy is released by fusion two nuclei of medium mass(approximately $100$ amu)
0%
None of the above

If M(A, Z),

$M_p$ and

$M_n$ denote the masses of the nucleus, proton and neutron respectively in units of

$u(1u=931.5 MeV/c^2)$ and BE represents its binding energy in MeV, then.

0%
$M(A, Z)=ZM_p+(A-Z)M_n-BE$
0%
$M(A, Z)=ZM_p+(A-Z)M_n+BE/c^2$
0%
$M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2$
0%
$M(A, Z)=AM_p+(A-Z)M_n+BE$

Explanation

$ZM_p +(A-Z)M_n= M(A,Z)$

Mass effect=

$\dfrac{B.E.}{c^2}$

$M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2$

Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third of the corresponding Balmer series is equal to 108.5 nm

0%
54.4 eV
0%
13.6 eV
0%
112.4 eV
0%
None of these

Explanation

Binding energy in the ground state

$E_b=\frac{E_0 Z^2}{n^2}$

$=\frac{E_0Z^2}{1^2}$

$=E_0Z^2$

Wave number of third Balmer line is

$v=\frac{E_0 Z^2}{hc} (\frac{1}{2^2}-\frac{1}{5^2})$

$=\frac{E_0Z^2}{hc}. \frac{21}{100}$

Wavelength of third Balmer line

$\lambda=\frac{1}{v}$

$=\frac{hc}{E_0Z^2}. \frac{100}{21}$

$=\frac{19.86*10^{-26}}{E_b} . \frac{100}{21}$

$E_b=\frac{19.86*10^{-26}} {\lambda} . \frac{100}{21}$

$=\frac{19.86*10^{-26}}{108.5*10^{-9}} . \frac{100}{21}$

$=8.7$ X

$10^-18 J$

$=\frac{8.7*10^{-18}}{1.6*10^-19} eV$

$=54.4 eV$

An energy of

$24.6eV$ is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is

0%
$38.2$
0%
$49.2$
0%
$51.8$
0%
$79.0$

Explanation

When one electron from neutral helium atom is removed, it becomes hydrogen like. The ionisation energy of hydrogen like atom is given by

${ E }_{ ion }={ Z }^{ 2 }Rhc$
for helium,

$Z=2$

$Rhc=13.6eV$

$\therefore$ Energy required to liberate second electron from helium

$\quad { E }_{ 2 }={ E }_{ ion }={ 2 }^{ 2 }\times 13.6eV=-54eV$

$\therefore$ Energy required to remove both the electrons from helium is

$={ E }_{ 1 }+{ E }_{ 2 }=(24.6+54.4)eV=79.0eV$

Consider the following statements (

$X$ and

$Y$ stand for two different elements)
(I)

$_{32}X^{65}$ and

$_{33}Y^{65}$ are isotopes.
(II)

$_{42}X^{86}$ and

$_{42}Y^{85}$ are isotopes.
(III)

$_{85}X^{174}$ and

$_{88}Y^{177}$ have the same number of neutrons.
(IV)

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

$_{94}Y^{235}$ are isobars.
The correct statements are

0%
II and IV only
0%
I, II and IV only
0%
II, III and IV only
0%
I, II, III and IV

Explanation

Two atoms are said to be isotope if they have same atomic number but different mass number.
So,

$_{42}X^{86}$ and

$_{42}Y^{85}$ are isotopes.
Number of neutrons in

$_{85}X^{174}$

$= 174 - 85 = 89$
Number of neutrons in

$_{88}Y^{177}$

$= 174 - 85 = 89$
So both have same number of neutrons.
Two atoms are said to be isobars if they have different atomic number but same mass number.
So,

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

$_{94}Y^{235}$ are isobars.

Mass defect for the helium nucleus is

$0.0303\ amu$ . Binding energy per nucleon for this (in

$MeV$ ) will be_______

0%
$28$
0%
$7$
0%
$4$
0%
$1$

Explanation

Mass defect of helium nucleus

$\Delta M = 0.0303 \ amu$
Binding energy

$E =\Delta Mc^2 = \Delta M\times 931.5 \ MeV$

$\implies \ E = 0.0303\times 931.5 = 28 \ MeV$
Number of nucleon in helium nucleus

$=4$
Thus binding energy per nucleon

$= \dfrac{28}{4} =7 \ MeV$

Atomic mass of

$_{ 6 }^{ 13 }{ C }$ is

$13.00335$ amu and its mass number is

$13.0$ . If amu

$=931$ MeV, binding energy of the neutrons present in the nucleus is

0%
3.10 $\mathrm { MeV }$
0%
974 $\mathrm { MeV }$
0%
97.4 $\mathrm { MeV }$
0%
9.974 $\mathrm { MeV }$

Explanation

Binding energy is the product of mass deflect and

$c$ square

$.$

Now

$,$

$13.00335 - 13 = 0.00335\,u.$

$\Rightarrow B.E = 931 \times 0.00335 = 3.10\,MeV$

Hence,

option

$(A)$ is correct answer.

When the radioactive isotope

$_{88}Ra^{226}$ decays in a series by emission of three alpha

$(\alpha)$ and a beta

$(\beta)$ particle, the isotope X which remains undecay is?

0%
$_{83}X^{214}$
0%
$_{84}X^{218}$
0%
$_{84}X^{220}$
0%
$_{87}X^{223}$

Explanation

Due to emission of

$3\alpha$ particles, the mass number is reduced by

$12$ while atomic number is decreased by

$6$ . Due to emission of

$\beta$ particle, the atomic number is increased by one while there is no change in mass number. Now, resulting mass will be decreased by

$12$ and atomic number is decreased by

$5$ . So, isotope X is

$_{83}X^{214}$ .

Which of the following nucleus is fissionable by slow neutrons:

0%
$-{92}U^{238}$
0%
$-{93}Np^{239}$
0%
$-{92}U^{235}$
0%
$_{2}I \, le^4$

Explanation

$_{-92}U^{238}$ is highly fissionable by slow neutrons.

An unknown stable nuclide after absorbing a neutron emits an electron, and the new nuclide splits spontaneously into two alpha particles. The unknown nuclide can be

0%
$_4Be^9$
0%
$_3Li^7$
0%
$_2He^2$
0%
$_5B^{10}$

Explanation

A neutron is a result of beta decay and as nucleus perform beta decay, it increases one proton. And the consequent two alpha decay also suggest that there are atleast 4 protons required.

${ _{ 3 }{ { Li }^{ 7 } } }$ best suit the condition because 1 beta decay increases the proton to 4.

Ans:

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

The first controlled nuclear chain reaction took place in which year?

0%
1895
0%
1910
0%
1942
0%
1945

What form of radiation most closest resembles X-rays?

0%
alpha
0%
beta
0%
gamma
0%
neutron

A radioactive nucleus of mass

$M$ emits a photon of frequency

$v$ and the nucleus recoils. The recoil energy will be:

0%
$hv$
0%
$M{ c }^{ 2 }-hv$
0%
$\dfrac { { h }^{ 2 }{ v }^{ 2 } }{ 2M{ c }^{ 2 } }$
0%
$Zero$

Explanation

Given,

Mass of nucleus $=M$

Frequency of photon $=\upsilon$

Velocity of nucleus $={{V}_{2}}$

Momentum of photon $P=\dfrac{h}{\lambda }=\dfrac{h\nu }{c}$

From conservation of momentum

Initial momentum = final momentum

$0+0=M{{V}_{2}}+P$

${{V}_{2}}=\dfrac{P}{M}=\dfrac{h\nu }{Mc}$

Recoil energy $E=\dfrac{1}{2}M{{V}^{2}}=\dfrac{1}{2}M{{\left( \dfrac{h\nu }{Mc} \right)}^{2}}=\dfrac{{{h}^{2}}{{\nu }^{2}}}{2M{{c}^{2}}}$

Recoil energy

$E=\dfrac{{{h}^{2}}{{\nu }^{2}}}{2M{{c}^{2}}}$

$\gamma$ ray photon produces an electron positron pair. If the rest mass energy of electron is

$0.51\ Mev$ and the total kinetic energy of electron - positron pair is

$0.78\ MeV$ then the energy of

$\gamma$ - ray photon is

$Mev$ is

0%
$0.78$
0%
$1.8$
0%
$1.28$
0%
$0.28$

Explanation

Here total energy is that of the photon.

Positron has mass equal to the electron.

Hence its mass-energy also be equal.

Hence Total mass-energy is

$1.02Me_V$

KE + this energy

$=$ photon energy.

$1.02+0.78=1.8Me_V$

A nuclear transformation is denoted by

$X(n, \alpha)_{3}^{7}Li$ . Which of the following is the nucleus of element

$X$ ?

0%
$_{6}^{12}C$
0%
$_{5}^{10}B$
0%
$_{5}^{9}B$
0%
$_{4}^{11}Be$

Explanation

$_{y}^{x}\textrm{X}$ +

$_{0}^{1}\textrm{n}$

$\rightarrow$

$_{2}^{4}\textrm{He}$ +

$_{3}^{7}\textrm{Li}$

On balancing atomic number and mass number on both sides, we get

x=10 and y=5

so correct option is

$_{5}^{10}\textrm{B}$

Rn decays into Po by emitting an

$\alpha$ particle with half life of 4 days. A sample contains

$6.4\ \times \ { 10 }^{ 10 }$ atoms of Rn. After 12 days, the number of atoms of Rn left in the sample will be

0%
$3.2\ \times \ { 10 }^{ 10 }$
0%
$0.53\ \times \ { 10 }^{ 10 }$
0%
$2.1\ \times \ { 10 }^{ 10 }$
0%
$0.8\ \times \ { 10 }^{ 10 }$

Explanation

$\begin{array}{l}\text { Half Life }=t_{1 / 2}=4 \text { days } \\\text { Total time }=12 \text { days } \\T=n \times t_{1 / 2}, \quad n= \text {no. of half lives}\\\qquad \begin{array}{c}12=n \times 4 \\\text { }\end{array}\\n=3\end{array}$

$\begin{aligned}N &=N_{0} \times\left(\frac{1}{2}\right)^{n}..... \text { radioactivity equation }\\N_{0}&=6+4 \times 10^{10} ...\text { Given }\\N&=6.4\times10^{10}\left(\frac{1}{2}\right)^{3}\\&=\frac{6.4\times10^{10}}{8}\\&=0.8\times10^{10}\end{aligned}$

Which symbol is used to represent the unit of atomic mass, amu?

0%
u
0%
A
0%
M
0%
n

$^{12}_6C$ and

$^{14}_6C$ are examples of isobars.

0%
True
0%
False

A nuclear transformation is denoted by

$X(n, \alpha) \rightarrow _{3}^{7}Li$ . Which of the following is the nucleus of element

$X$ ?

0%
$_{6}^{12}C$
0%
$_{5}^{10}B$
0%
$_{5}^{9}B$
0%
$_{4}^{11}Be$

Explanation

$\begin{aligned}& X(n, \alpha) \longrightarrow{ }_{3}^{7} L_{i} \\X &+n \quad \longrightarrow \quad{ }_{3}^7 L_{i}+\alpha \\\alpha &={ }_2^{4} H_{e}\\n&=^0n\end{aligned}$

$\begin{array}{l}X+{ }^{1} n \longrightarrow _{3}^{7} Li+{ }^{4}_2 H_{e} \\X \text { must have Atomic mass }=10\end{array}$

$\text { and Atomic no. }=5$

${ }^{10}_5 X={ }^{10}_5 B$

The binding energies per nucleon of deuteron

$(_{1}H^{2})$ and helium atom

$(_{2}He^{4})$ 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

ATQ, following reaction takes place,

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

{-Q: energy reloaded}

$\rightarrow 2({ H }_{ 1 }^{ 2 })\rightarrow { H }_{ 2 }^{ 4 }\quad -Q\\$

So, ATQ binding energy per nucleon of deuteron=1.1MeV.

Therefore, There are 2 nucleon in deuteron atom, hence total binding energy

$=1.1\times 2\\ =2.2MeV$

Therefore, for left hand part, BE of

${ H }_{ 1 }^{ 2 }=2.2MeV$

Similarly, BE per nucleon of helium atom= 7eV.

Number of nucleons in helium

$=4$

Total BE of nucleus

$==7\times 4=28MeV$

Now, substituting the values of BE in the equation to get the value of Q,

$\Rightarrow 2(2.2)\rightarrow 28-Q\\ \rightarrow Q=28-4.4=23.6MeV$

In the nuclear fusion reaction

$_{ 1 }^{ 2 }{ H }+_{ 1 }^{ 3 }{ H }\rightarrow _{ 2 }^{ 4 }{ He }+n\quad$ given that the repulsive potential energy between the two nuclei is

$-7.7\times { 10 }^{ -14 }J$ , the temperature at which the gases must be heated to initiate the reaction is nearly (Boltzmann's constant

$k=1.38\times { 10 }^{ -23 }J/K$ )

0%
${ 10 }^{ 7 }K\quad$
0%
${ 10 }^{ 5 }K$
0%
${ 10 }^{ 3 }K$
0%
${ 10 }^{ 9 }K$

Explanation

From conservation of energy

$\quad 2\cfrac { 1 }{ 2 } m{ v }_{ }^{ 2 }=7.7\times { 10 }^{ -14 }$

$\therefore v=\sqrt { \cfrac { 7.7\times { 10 }^{ -14 } }{ m } } =\sqrt { \cfrac { 3kT }{ m } } \quad \quad \therefore T\approx { 10 }^{ 9 }K$

A radioactive nuclide X decays into nuclei T and Z by simultaneous disintegration as shown. Effective decay constant for the disintegration is

0%
${ \lambda }_{ 1 }+{ \lambda }_{ 2 }$
0%
$\dfrac { { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } }$
0%
$\dfrac { { \lambda }_{ 1 }+{ \lambda }_{ 2 } }{ 2 }$
0%
$\dfrac { 2{ \lambda }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } }$

Explanation

$\begin{array}{l}\frac{d y}{d t}=\lambda_{1} x---\text { for } x \rightarrow y \\\frac{d z}{d t}=\lambda_{2} x--- \text { for } x \rightarrow z \\\text { Now adding }\\\left(\lambda_{1}+\lambda_{2}\right) x=\frac{d y}{d t}+\frac{d z}{d t}\longrightarrow(i) \\\frac{d y}{d t}+\frac{d z}{d t}=-\frac{d x}{d t}\longrightarrow\text { (iii) }\end{array}$

$\begin{array}{l}\text { Rate of decrease of } x \\\text { Would be equal to } \\\text { net increase rale } \\\text { of } y \text { and } z \text { . } \\-\frac{d x}{d t}=\left(\lambda_1+\lambda_{2}\right)x\\\Rightarrow \lambda_{\text {effective }}=\lambda_{1}+\lambda_{2}\end{array}$

2006219_850779_ans_97396bc5f02540b3992bdc969ca57576.PNG

Decreasing order of atomic weight is correct of the elements given below?

0%
$Fe > Co > Ni$
0%
$Ni > Co > Fe$
0%
$Co > Ni > Fe$
0%
$Co > Fe > Ni$

Explanation

Atomic weight of Iron

$\approx 56$ u

Atomic weight of Cobalt

$\approx 59$ u

Atomic weight of Nickel

$\approx 58.7$ u

So, the order of atomic weights is

$Co>Ni>Fe$ .
Option

$C$ is correct.

A collective name for nucleons and other elementary particles that decay into nucleons by the emission of mesons is?

0%
Baryon
0%
Boson
0%
Fermion
0%
Hadron

Explanation

$\begin{array}{l}\text { A collective name for } \\\text { nucleons and other elementry } \\\text { particles that decay into } \\\text { nucleons by the emission } \\\text { of mesons is Hadron. }\end{array}$

The mass of

$_{ 17 }{ { CI }^{ 35 } }$ is

$34.9800$ amu. Calculate its binding energy (mass of

$_{ 0 }{ { n }^{ 1 } }=1.008665$ amu

$_{ 1 }{ { H }^{ 1 } }=1.007825$ amu):

0%
$2.88 MeV$
0%
$28.8 MeV$
0%
$280 MeV$
0%
$2880 MeV$

Explanation

There are 17 protons and 18 neutrons in the nucleus of chlorine

$\Delta$ m=Mass of the nucleons combined- Actual mass of the nucleus

$=17\times 1.0007825+18\times 1.0086-34.9800$

$=35.288995-34.98$

$=0.308995$
B.E.

$=\Delta m(931.5)=279.45$

The above is a plot of binding energy per nucleon

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

$M$ ;

$A,B,C,D,E,F$ 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 is

$\varepsilon$ positive.

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

A nucleus of mass

$M+\Delta m$ is at rest and decays into two daughter nuclei of equal mass

$\cfrac{M}{2}$ each. Speed of light is

$c$ . The binding energy per nucleon for the parent nucleus is

${E}_{1}$ and that for the daughter nuclei is

${E}_{2}$ . Then :

0%
${ E }_{ 2 }>2{ E }_{ 1 }$
0%
${ E }_{ 1 }>{ E }_{ 2 }$
0%
${ E }_{ 2 }>{ E }_{ 1 }$
0%
${ E }_{ 1 }=2{ E }_{ 2 }$

Explanation

$\begin{array}{l} \text { let } x \rightarrow 2 y \\ \text { mass } \rightarrow M+\Delta m \quad 2 \times \frac{M}{2}=M \\ \Delta m \text { mass is lost as energy } \\ \Rightarrow \text { energy is released } \\ \Rightarrow \text { energy of } x>\text { energy of y } \\ \qquad E_{1}>E_{2} \\ \text { option } B \end{array}$

If the binding energy of the electron in a hydrogen atom is

$13.6\ eV$ , the energy required to remove the electron from the first excited state of

${Li}^{2+}$ is:

0%
$30.6\ eV$
0%
$13.6\ eV$
0%
$3.4\ eV$
0%
$122.4\ eV$

Explanation

$E$ of

${ Li }^{ ++ }$ at

$n=2=\cfrac { \left( 13.6 \right) { 3 }^{ 2 } }{ { 2 }^{ 2 } } =30.6eV$

When

${ 10 }^{ 20 }$ electrons are removed from a neutral metal plate through some process, the charge on it becomes ______

0%
$-1.6C$
0%
$+16C$
0%
$+1.6C$
0%
${10}^{-19}C$

Explanation

$Q={ 10 }^{ 20 }\times 1.6\times { 10 }^{ -19 }\\ \therefore Q=+16C$

A nuclide

$A$ undergoes

$\alpha$ decay and another nuclide

$B$ undergoes

$\beta$ decay. Then

0%
all the $\alpha-$ particles emitted by $A$ will have almost the same speed
0%
the $\alpha-$ particles emitted by $A$ may have widely different speeds
0%
all the $\beta-$ particles emitted b $B$ will have almost the same speed
0%
the $\beta-$ particles emitted by $B$ may have widely different speeds.

A small quantity of solution containing

$^{24}Na$ radio-nuclide (half life

$15$ hours) of activity

$1.0$ micro-curie is injected into the blood of a person. A sample of the blood of volume

$1\ cm^{3}$ taken after

$5\ hours$ shows an activity of

$296$ disintegrations per minute. Determine the total volume of blood in the body of the person. Assume that the radio active solution mixes uniformly in the blood of the person.
(

$1\ curie = 3.7\times 10^{10}$ disintegrations per second).

0%
$5.95\ Litre$ .
0%
$5\ Litre$ .
0%
$55\ Litre$ .
0%
$7\ Litre$ .

Statement 1: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement II: For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decrease with increasing Z

0%
Statement 1 is false, statement 2 is true
0%
Statement 1 is false, statement 2 is true, statement 2 is correct explanation for statement 1
0%
Statement 1 is false, statement 2 is true, statement 2 is not the correct explanation for statement 1
0%
Statement 1 is true, statement 2 is false

Explanation

$\begin{array}{l} \text { In nuclear fission or } \\ \text { fusion, energy is released. } \\ \text { Binding energy per nucleon } \\ \text { increase with increase } z \text { for } \\ \text { heavy nuclei and decrease with } \\ \text { decrease } z \text { for light nuclei. } \\ \text { Both are true } \\ \text { and correct explanation. } \end{array}$

A nuclear transformation is denoted by

$X\left( n,\alpha \right) \ _{ 3 }^{ 7 }{ Li }$ . Which of the following is the nucleus of element

$X$ ?

0%
${ _{ }^{ 12 }{ C } }_{ 6 }$
0%
$_{ 5 }^{ 10 }{ B }$
0%
$_{ 5 }^{ 9 }{ B }$
0%
$_{ 4 }^{ 11 }{ Be }$

Explanation

$X+n=_{ 2 }^{ 4 }{ He }+_{ 3 }^{ 7 }{ Li }\quad$

So. from conservation of mass no.

$X$ must be

$_{ 5 }^{ 10 }{ B }$

The energy of thermal neutron is about :

0%
$25\ MeV$
0%
$0.51\ MeV$
0%
$25\ meV$
0%
$1.02\ MeV$

Explanation

Quantitatively, the thermal energy per particle is about

$0.025$ electron volt—an amount of energy that corresponds to a neutron speed of about

$2,000$ metres per second and a neutron wavelength of about

$2 × 10^{-10}$ metre (or about two angstroms).

Binding Energy per nucleon of a fixed nucleus

$X^4$ is

$6\ MeV.$ It absorbs a neutron moving with

$KE = 2\ MeV,$ and converts into

$Y$ at ground state, emitting a photon of energy

$1\ MeV.$ The Binding Energy per nucleon of

$Y$ (

$in Mev$ ) is:

0%
$\dfrac{(6A + 1)}{(A + 1)}$
0%
$\dfrac{(6A - 1)}{(A + 1)}$
0%
$7$
0%
$\dfrac{7}{6}$

Explanation

$\begin{array}{l}\text { Given-* Binding Energy per nucleon of } X^{A}=6 \text { Mev } \\\Rightarrow \text { Binding Energy } B E \text { of } X^{A}=6 A \text { MeV } \\\text { we have - }\\\qquad\begin{array}{c}X^{A}+_1^1{} n \longrightarrow Y^{A+1}+\text { Photon (l Mev) } \\ \end{array}\end{array}$

$\begin{aligned}\text { Initial Energy } &=B E \text { of }X^{A}+\text { Energy of neutron } \\&=(6 A+2) \mathrm{MeV} \\\text { Final Energy }&=B E \text { of } Y^{A+1}+\text { Energy of photon } \\&=B E_{Y}+1\mathrm{MeV}\end{aligned}$

$\begin{array}{l}\text { By energy conservation- }\\\qquad\begin{aligned}(6 A+2) \mathrm{meV}=B E_{Y}+\text { 1 MeV }\\\Rightarrow B E \text { of } Y=(6 A+1) \text { MeV } \\\Rightarrow \text { Binding Energy per nucleon }=\\B E=\frac{6 A+1}{A+1} \text { Mev }\end{aligned} \\\text { Hence, opt (A) is correct. }\end{array}$

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 amu = 931 MeV)

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

Explanation

Given,

$m_p=1.007277 amu$ mass of proton

$m_n=1.00866amu$ mass of neutron

$m_c(ca-40)=39.97545 amu$

and

$1amu=931 MeV/c^2$

$Z=20$ No of proton for calcium

$N=20$ no. of neutron

The energy required to dissociate completely is called binding energy.

$B.E=(Zm_p+Nm_n-m_c)c^2$ . . . .(1)

$B.E=(20\times 1.007277+20\times 1.00866-39.97545)c^2$

$B.E=0.34359\times 931MeV$

$B.E=319.6MeV$ This is for 1 atom of calcium-40

No. of atom in

$1g$ of Ca-40 is,

$N=\dfrac{1}{40}\times 6.6022\times 10^{23}$

$N=1.65055\times 10^{22}$ atom.

The total energy required is,

$E=N(B.E)$

$E=1.65055\times 10^ {22}\times319.6MeV$

$E=4.813\times 10^{24}MeV$

The correct option is A.

$Q-26$ , find the initial activity of sample.

0%
$6.79\ \times 10^{16}\ disintegration/sec$
0%
$5.79\ \times 10^{15}\ disintegration/sec$
0%
$10\ \times 10^{15}\ disintegration/sec$
0%
none of these.

The binding energy per nucleon of

$^7_3 Li$ and

$^4_2 He$ nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction

$^7_3 Li + ^1_1 He \longrightarrow ^4_2 H + ^4_2 He + Q$ the value of energy Q released is

0%
-2.4 MeV
0%
8.4 MeV
0%
17.3 MeV
0%
19.6 MeV

Explanation

The binding energy for

$_1H^1$ is around zero and also not give in the question so we can ignore it,

$Q=2(4\times 7.06)-7\times (5.60)$

$=(56.48-39.2)MeV$

$=17.28\ MeV$

$=17.3\ MeV$

$I$ excitation energy for the

$H-$ like (hypothetical) sample is

$24\ eV$ , then binding energy in

$III$ excired state is:

0%
$2\ eV$
0%
$3\ eV$
0%
$4\ eV$
0%
$5\ eV$

Explanation

$\text{First(I) excitation energy means energy difference between first(ground) state and second state. }$

If ground state energy is

$E_1$ then for

$n=2$ the energy becomes

$E_1/n^2$

Note- As the energy shown above is due to attractive force so it is

$negative$ in magnitude.

$E_2=E_1/2^2=E_1/4$

difference is

$E_1-E_2=3E_1/4$

Excitation energy will be

$negative$ of the energy difference.

Given that

$-3E_1/4=24eV$ so

$E_1=-32eV$

$\text{Third(III) excited state means fourth normal state i.e n=4 }$

Energy of

$4^{th} state$ is

$E_1/4^2=E_1/16=-32/16=-2eV$

Binding energy will be positive in nature so answer is

$2eV$

Binding energy per nucleon versus mass number curve for nuclei is shown in fig.W, X, Y, and Z are four nuclei indicated on the curve. The process that would release energy is

0%
${\rm{Y}} \to {\rm{2Z}}$
0%
${\rm{W}} \to {\rm{X + Z}}$
0%
${\rm{W}} \to {\rm{2Y}}$
0%
${\rm{X}} \to {\rm{Y + Z}}$

Explanation

If it is mass A

$Y\rightarrow 2Z$

Reactant

$R=60\times8.5=10meV$

Product

$P=2\times30\times5=300meV$

$\Delta E=-210meV$

Therefore Endothermic

If it was C

$W\rightarrow 2Y$

$R=120\times75=900meV$

$P=2\times60\times8.5=1020meV$

$\Delta E=120meV$

Therefore Exothermic

If it was D

$X\rightarrow Y+Z$

$R=90\times8=720meV\\P=60\times8.5+30\times5=66.0meV$

$\Delta E=-60meV$

Therefore Endothermic

A stationary

$U^{238}$ nucleus undergoes

$\alpha$ - decay. If the kinetic energy of product nucleus is

$E$ , the total energy released in the process is-

0%
$E$
0%
$58.5E$
0%
$59.5E$
0%
$60.5E$

Explanation

mass of alpha particle

$m_a=4m$ , where

$m$ is mass of a proton.

mass of Uranium nucleus is

$m_u=238m$ after emission the residue will have mass

$m_r=238-4=234m$

We know

$KE=momentum^2/2m$

so for residue nucleus, it should be like this

$E=p^2_r/2m_r$

or momentum of residue nucleus is

$p_r=\sqrt[2]{2m_rE}=\sqrt[2]{468mE}$

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

before and after emission so it should also be

$p_a=\sqrt[2]{468mE}$

If kinetic energy of alpha particle is

$E_a$ then

$E_a=p^2_a/2m_a=\dfrac{468mE}{2\times 4m}=58.5E$

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

$59.5E$

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 8 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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