# NCERT Solutions for Class 12 Physics Chapter 13

## NCERT Solutions For Class 12 Physics Chapter 13 PDF Free Download

NCERT solutions for class 12 chapter 13 Nuclei is provided here with an aim to help students in their CBSE class 12 board exam preparations. All the solutions present in these NCERT solutions are prepared by our subject experts under the guidelines on NCERT and as per the latest CBSE syllabus 2018-19.

Ncert solutions for class 12 physics chapter nuclei provided here has answers to the textbook questions along with nuclei class 12 important questions, exemplary problems, worksheets and exercises which will help you gain complete knowledge on the topics Nuclei.

## Class 12 Physics NCERT solutions for Nuclei

One of the best methods to prepare physics is by taking notes and writing down all the important formulas and points in a copy. The notes will also help students during the time of revision. Once you get an idea about the type of question being asked in examination it is easy for you to collect important points and prepare class 12 physics chapter 13 notes.

### Subtopics of class 12 chapter 13 Nuclei

1. Introduction
2. Atomic Masses And Composition Of Nucleus
3. Size Of The Nucleus
4. Mass-energy And Nuclear Binding Energy
1. Mass – Energy
2. Nuclear binding energy
5. Nuclear Force
2. Alpha decay
3. Beta decay
4. Gamma decay
7. Nuclear Energy
1. Fission
2. Nuclear reactor
3. Nuclear fusion – energy generation in stars
4. Controlled thermonuclear fusion.

### Class 11 Physics NCERT Solutions Nuclei Important Questions

Q 13.1 (a) Lithium has two stable isotopes $_{3}^{6}\textrm{Li}$ and $_{3}^{7}\textrm{Li}$ have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

(b) Two stable isotopes of Boron , $_{5}^{10}\textrm{B}$ and $_{5}^{11}\textrm{B}$ . Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.811 u. Find the abundances of $_{5}^{10}\textrm{B}$ and $_{5}^{11}\textrm{B}$.

A1 :

(a) Mass of $_{3}^{6}\textrm{Li}$ lithium isotope , m1 = 6.01512 u

Mass of $_{3}^{7}\textrm{Li}$ lithium isotope , m = 7.01600 u

Abundance of $_{3}^{6}\textrm{Li}$ , η1= 7.5%

Abundance of $_{3}^{7}\textrm{Li}$ , η2= 92.5%

The atomic mass of lithium atom is given as:

$m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }$ $m =\frac{ 6.01512 \times 7.5 + 7.1600 \times 92.5 }{ 92.5 + 7.5 }$

= 6.940934 u

(b) Mass of boron isotope $_{5}^{10}\textrm{B}$ m = 10.01294 u

Mass of boron isotope $_{5}^{11}\textrm{B}$ m = 11.00931 u

Abundance of $_{5}^{10}\textrm{B}$, η1 = x%

Abundance of $_{5}^{11}\textrm{B}$, η2= (100 − x)%

Atomic mass of boron, m = 10.811 u The atomic mass of boron atom is given as:

$m =\frac{ m_1 n_1 + m_2 n_2 }{ n_1 + n_2 }$ $10.811 =\frac{ 10.01294 \times x + 11.00931 \times ( 100 – x ) }{ x + 100 -x }$

1081.11 = 10.01294x + 1100.931 – 11.00931 x

x = 19.821/0.99637 = 19.89 %

And 100 − x = 80.11%

Hence, the abundance of $_{5}^{10}\textrm{B}$ is 19.89% and that of $_{5}^{11}\textrm{B}$ is 80.11%.

Q 13.2 :The three stable isotopes of neon: $_{10}^{20}\textrm{Ne}$, $_{10}^{21}\textrm{Ne}$ and $_{10}^{22}\textrm{Ne}$ have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Atomic mass of $_{10}^{20}\textrm{Ne}$, m1= 19.99 u

Abundance of $_{10}^{20}\textrm{Ne}$, η1 = 90.51%

Atomic mass of $_{10}^{21}\textrm{Ne}$, m2 = 20.99 u

Abundance of $_{10}^{21}\textrm{Ne}$, η2 = 0.27%

Atomic mass of $_{10}^{22}\textrm{Ne}$, m3 = 21.99 u

Abundance of $_{10}^{22}\textrm{Ne}$, η3 = 9.22%

The average atomic mass of neon is given as:

$m=\frac{ m_1 n_1 + m_2 n_2 + m_3 n_3 }{ n_1 + n_2 + n_3 }$

= $\frac{ 19.99 \times 90.51 + 20.99 \times 0.27 + 21.99 \times 9.22 }{ 90.51 + 0.27 + 9.22 }$

= 20.1771 u

Q 13.3: Obtain the binding energy in MeV of a nitrogen nucleus $_{7}^{14}\textrm{N}$, given m($_{7}^{14}\textrm{N}$)=14.00307 u

Ans:

Atomic mass of nitrogen $_{7}^{14}\textrm{N}$, m = 14.00307 u

A nucleus of $_{7}^{14}\textrm{N}$nitrogen contains 7 neutrons and 7 protons.

Hence, the mass defect of this nucleus, ∆m = 7mH + 7mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m = 7 × 1.007825 + 7 × 1.008665 − 14.00307

= 7.054775 + 7.06055 − 14.00307

= 0.11236 u

But 1 u = 931.5 MeV/c2

∆m = 0.11236 × 931.5 MeV/c2

Hence, the binding energy of the nucleus is given as:

Eb = ∆mc2

Where, c =Speed of light

$E_b = 0.11236 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}$

= 104.66334 Mev

Hence, the binding energy of a nitrogen nucleus is 104.66334 MeV.

Q 13.4: Find the binding energy of the nuclei $_{23}^{56}\textrm{Fe}$ and $_{83}^{209}\textrm{Bi}$ in units of MeV from the following

data:

m ($_{23}^{56}\textrm{Fe}$) = 55.934939 u m($_{83}^{209}\textrm{Bi}$) = 208.980388 u

Ans :Atomic mass of $_{23}^{56}\textrm{Fe}$, m1 = 55.934939 u

$_{23}^{56}\textrm{Fe}$ nucleus has 26 protons and (56 − 26) = 30 neutrons

Hence, the mass defect of the nucleus, ∆m = 26 × mH + 30 × mn − m1

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m = 26 × 1.007825 + 30 × 1.008665 − 55.934939

= 26.20345 + 30.25995 − 55.934939

= 0.528461 u

But 1 u = 931.5 MeV/c2

∆m = 0.528461 × 931.5 MeV/c2

The binding energy of this nucleus is given as:

Eb1 = ∆mc2

Where, c = Speed of light

$E_b = 0.528461 \times 931.5 \left ( \frac{MeV}{c^{2}} \right ) \times c^{2}$

= 492.26 MeV

Average binding energy per nucleon = $\frac{492.26}{56} = 8.79 MeV$

Atomic mass of $_{83}^{209}\textrm{Bi}$, m2 = 208.980388 u

$_{83}^{209}\textrm{Bi}$ nucleus has 83 protons and (209 − 83) 126 neutrons.

Hence, the mass defect of this nucleus is given as:

∆m’ = 83 × mH + 126 × mn − m2

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m’ = 83 × 1.007825 + 126 × 1.008665 − 208.980388

= 83.649475 + 127.091790 − 208.980388

= 1.760877 u

But 1 u = 931.5 MeV/c2

∆m’ = 1.760877 × 931.5 MeV/c2

Hence, the binding energy of this nucleus is given as:

Eb2 = ∆m’c2

= 1.760877 × 931.5 $\left ( \frac{MeV}{c^{2}} \right ) \times c^{2}$

= 1640.26 MeV

Average binding energy per nucleon = 1640.26/209 = 7.848 MeV

Q 13.5: A coin of mass 3.0 g. Find the amount of nuclear energy to separate the neutrons and protons from each other, assuming that the coin has atoms of $_{29}^{63}\textrm{Cu}$ with mass = 62.92960 u.

Ans :

Mass of a copper coin, m’ = 3 g

Atomic mass of $_{29}^{63}\textrm{Cu}$ atom, m = 62.92960 u

The total number of $_{29}^{63}\textrm{Cu}$ atoms in the coin, $N = \frac{N_A \times m’}{Mass \; number}$

Where,

NA = Avogadro’s number = 6.023 × 1023 atoms /g

Mass number = 63 g

$N = \frac{6.023 \times 10^{23}\times 3}{63}$ = 2.868 x 1022 atoms

$_{29}^{63}\textrm{Cu}$ nucleus has 29 protons and (63 − 29) 34 neutrons

Mass defect of this nucleus, ∆m’ = 29 × mH + 34 × mn − m

Where,

Mass of a proton, mH = 1.007825 u

Mass of a neutron, mn = 1.008665 u

∆m’ = 29 × 1.007825 + 34 × 1.008665 − 62.9296

= 0.591935 u

Mass defect of all the atoms present in the coin, ∆m = 0.591935 × 2.868 × 1022

= 1.69766958 × 1022 u

But 1 u = 931.5 MeV/c2

∆m = 1.69766958 × 1022 × 931.5 MeV/c2

Hence, the binding energy of the nuclei of the coin is given as:

Eb = ∆mc2

= 1.69766958 × 1022 × 931.5

= 1.581 × 1025 MeV

But 1 MeV = 1.6 × 10−13 J

Eb = 1.581 × 1025 × 1.6 × 10−13

= 2.5296 × 1012 J

This much energy is required to separate all the neutrons and protons from the given

coin.

Q 13.6: Write nuclear reaction equations for

(i) α-decay of$_{88}^{226}\textrm{Ra}$  (ii) α-decay of$_{94}^{242}\textrm{Pu}$

(iii) β−-decay of $_{15}^{32}\textrm{P}$ (iv) β -decay of$_{83}^{210}\textrm{Bi}$

(v) β+-decay of$_{6}^{11}\textrm{C}$ (vi) β -decay of$_{43}^{97}\textrm{Tc}$

(vii) Electron capture of$_{54}^{120}\textrm{Xe}$

Ans:

α is a nucleus of helium $_{2}^{4}\textrm{He}$ and β is an electron (e− for β− and e+ for β+). In every α decay, there is a loss of 2 protons and 4 neutrons. In every β + decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every β – decay, there is a gain of 1 proton and an anti-neutrino is emitted from the nucleus.

For the given cases, the various nuclear reactions can be written as:

(i) $_{88}^{226}\textrm{Ra} \rightarrow\; _{86}^{222}\textrm{Rn} \; + \; _{2}^{4}\textrm{He}$

(ii) $_{ 94 }^{ 242 }\textrm{ Pu } \rightarrow\; _{ 92 }^{ 238 }\textrm{ U } \; + \; _{ 2 }^{ 4 }\textrm{ He }$

(iii) $_{15}^{ 32 }\textrm{P} \rightarrow\; _{ 16 }^{ 32 }\textrm{ S } \; + \; e^- +\bar{v}$

(iv) $_{83}^{ 210 }\textrm{B} \rightarrow\; _{ 84 }^{ 210 }\textrm{ PO } \; + \; e^- +\bar{v}$

(v) $_{6}^{ 11 }\textrm{C} \rightarrow\; _{ 5 }^{ 11 }\textrm{ B } \; + \; e^+ + \; v$

(vi) $_{43}^{ 97 }\textrm{Tc} \rightarrow\; _{ 42 }^{ 97 }\textrm{ MO } \; + \; e^+ + \; v$

(vii) $_{54}^{ 120 }\textrm{Xe}+ \; e^+ \rightarrow\; _{ 53 }^{ 120 }\textrm{ I } \; + \; v$

Q 13.7: How much time will it take for the activity of a radioactive isotope of half life T years to come down to:

(a) 3.125% of its original value

(b) 1% of its original value?

Ans:

Half-life of the radioactive isotope = T years

Original amount of the radioactive isotope = N0

(a) After decay, the amount of the radioactive isotope = N

It is given that only 3.125% of N0 remains after decay. Hence, we can write:

$\frac{N}{N_0} = 3.125 \% = \frac{3.125}{100}=\frac{1}{32}$

But,  $\frac{N}{N_0} = e^{-\lambda t}$

Where, λ = Decay

Constant t = Time

$e^{-\lambda t} = \frac{1}{32}$ $-\lambda t = ln 1 – ln 32$ $-\lambda t = 0 – 3.4657$

Since, $-\lambda = \frac{0.693}{T}$

t = $\frac{3.466}{\frac{0.693}{T}} \approx \; 5 T years$

Hence, the isotope will take about 5T years to reduce to 3.125% of its original value.

(b) After decay, the amount of the radioactive isotope = N

It is given that only 1% of N0 remains after decay. Hence, we can write:

$\frac{N}{N_0} = 1 \% = \frac{1}{100}$ $\frac{N}{N_0} = e^{-\lambda t}$ $e^{-\lambda t} = \frac{1}{100}$ $-\lambda t = ln1 – ln100$ $e^{-\lambda t} =$ 0 – 4.6052

t = 4.6052/$\lambda$

Since, λ = 0.693/T

$t = \frac{4.6052}{\frac{0.693}{T}}$ = 6.645 T years

Hence, the isotope will take about 6.645T years to reduce to 1% of its original value.

Q 13.8: The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive$_{6}^{14}\textrm{C}$  present with the stable carbon isotope $_{6}^{12}\textrm{C}$ . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of$_{6}^{14}\textrm{C}$ , and the measured activity, the age of the specimen can be approximately estimated. This is the principle of $_{6}^{14}\textrm{C}$ dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation.

Ans :

Decay rate of living carbon-containing matter, R = 15 decay/min

Let N be the number of radioactive atoms present in a normal carbon- containing matter.

Half life of $_{6}^{14}\textrm{C},T_{\frac{1}{2}}$ = 5730 years

The decay rate of the specimen obtained from the Mohenjo-Daro site:

R’ = 9 decays/min

Let N’ be the number of radioactive atoms present in the specimen during the Mohenjo-Daro period.

Therefore, we can relate the decay constant, λ and time, t as:

$\frac{N}{N’}=\frac{R}{R’}= e^{-\lambda t}$ $e^{-\lambda t} = \frac{9}{15}=\frac{3}{5}$ $-\lambda t = log_e\frac{3}{5} = -0.5108$ $t = \frac{0.5108}{\lambda}$

But ${\lambda} = \frac{0.693}{T_\frac{1}{2}}=\frac{0.693}{5730}$ $t = \frac{0.5108}{\frac{0.693}{5730}}= 4223.5 years$

Hence, the approximate age of the Indus-Valley civilization is 4223.5 years.

Q 13.9: Calculate the amount of $_{27}^{60}\textrm{Co}$ required to create a radioactive source of strength 8.0 mCi considering that the half-life of $_{27}^{60}\textrm{Co}$  is 5.3 years.

Ans:

The strength of the radioactive source is given as:

$\frac{dN}{dt} = 8.0 mCi$

= 8 x 10-3 x 3.7 x 1010

= 29.6 x 107 decay/s

Where,

N = Required number of atoms

Half-life of $_{ 27 }^{ 60 }\textrm{ Co }$ = 5.3 years

= 5.3 × 365 × 24 × 60 × 60

= 1.67 × 108 s

For decay constant λ, we have the rate of decay as:

$\frac{dN}{dt} = \lambda N$

Where, $\lambda = \frac{0.693}{T_\frac{1}{2}} = \frac{0.693}{1.67 \times 10^{8}}s^{-1}$ $N = \frac{1}{\lambda} \frac{dN}{dt}$ $= \frac{29.6 \times 10^{ 7}}{\frac{ 0.693 }{ 1.67 \times 10^{ 8 }}}$

= 7.133 x 1016 atoms

For $_{27}^{60}\textrm{Co}$ :

Mass of 6.023 × 1023 (Avogadro’s number) atoms = 60 g

Mass of atoms 7.133 x 1016 atoms = $= \frac{60\times7.133\times10^{16}}{6.023\times10^{23}}$

= 7.106 x 10-6 g

Hence, the amount of $_{27}^{60}\textrm{Co}$ necessary for the purpose is 7.106 × 10−6 g.

Q 13.10 :If 28 years is the half life of $_{ 38 }^{ 90 }\textrm{ Sr }$. What will be the disintegration rate of a quantity of 15 mg of said $_{ 38 }^{ 90 }\textrm{ Sr }$isotope ?

Ans:r

Half life of $_{ 38 }^{ 90 }\textrm{ Sr }$, $t_{\frac{1}{2}}$ = 28 years

= 28 × 365 × 24 × 60 × 60

= 8.83 × 108 s

Mass of the isotope, m = 15 mg

90 g of $_{ 38 }^{ 90 }\textrm{ Sr }$ atom contains 6.023 × 1023 (Avogadro’s number) atoms.

Therefore, 15 mg of $_{ 38 }^{ 90 }\textrm{ Sr }$ contains :

$\frac{6.023 \times 10^{23} \times 15 \times 10 \times 10^{-3}}{90}$

i.e 1.038 x 1020 number of atoms

Rate of disintegration, $\frac{dN}{dt}= \lambda N$

Where,

$\lambda$ = decay constant = $\frac{0.693}{8.83 \times 10^{8}}s^{-1}$ $\frac{dN}{dt}=\frac{0.693\times 1.0038 \times 10^{20}}{8.83 \times 10^{8}}$

= 7.878 x 1010 atoms/s

Hence, the disintegration rate of 15 mg of $_{ 38 }^{ 90 }\textrm{ Sr }$ is

7.878 × 1010 atoms/s.

Q 13.11: Obtain approximately the ratio of the nuclear radii of the gold isotope $_{ 79 }^{ 197 }\textrm{Au}$ and the $_{ 47 }^{ 107 }\textrm{Ag}$  silver isotope .

Ans:

Nuclear radius of the gold isotope $_{ 79 }^{ 197 }\textrm{Au}$ = RAu

Nuclear radius of the silver isotope $_{ 47 }^{ 107 }\textrm{Ag}$ = RAg

Mass number of gold, AAu = 197

Mass number of silver, AAg = 107

The ratio of the radii of the two nuclei is related with their mass numbers as :

$\frac{ R_{ Au }}{ R_{ Ag }}=\left (\frac{ R_{ Au }}{ R_{ Ag }} \right )^{\frac{ 1 }{ 3 } }$ $=\left (\frac{ 197 }{ 107 } \right )^{\frac{ 1 }{ 3 } }$

= 1.2256

Hence, the ratio of the nuclear radii of the gold and silver isotopes is about 1.23.

Q 13.12: Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of

(a) $_{88}^{226}\textrm{Ra}$

(b) $_{86}^{220}\textrm{Rn}$

$m ( _{ 88 }^{ 226 }\textrm{ Ra })$= 226.02540 u, $m ( _{ 86 }^{ 222 }\textrm{ Rn })$ = 222.01750 u,

$m ( _{ 86 }^{ 220 }\textrm{ Rn })$= 220.01137 u, $m ( _{ 84 }^{ 216 }\textrm{ Po })$ = 216.00189 u.

(a) Alpha particle decay of $_{88}^{226}\textrm{Ra}$ emits a helium nucleus. As a result, its mass number reduces to 222 (226 − 4) and its atomic number reduces to 86 (88 − 2). This is shown in the following nuclear reaction.

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

Q-value of emitted α-particle = (Sum of initial mass − Sum of final mass) c2

Where, c = Speed of light

It is given that :

$m ( _{ 88 }^{ 226 }\textrm{ Ra })$= 226.02540 u

$m ( _{ 86 }^{ 222 }\textrm{ Rn })$ = 222.01750 u

$m ( _{ 2 }^{ 4 }\textrm{ He })$ = 4.002603 u

Q-value = [226.02540 − (222.01750 + 4.002603)] u c2

= 0.005297 u c2

But 1 u = 931.5 MeV/c2

Q = 0.005297 × 931.5 ≈ 4.94 MeV

Kinetic energy of the α-particle = $\left ( \frac{Mass \; number \; after \; decay }{Mass \; number \; before \; decay } \right ) \times Q$ $= \frac{222}{226} \times 4.94$ = 4.85 MeV

(b) Alpha particle decay of $_{86}^{220}\textrm{Rn}$

It is given that:

Mass of $_{86}^{220}\textrm{Rn}$ = 220.01137 u

Mass of $_{84}^{216}\textrm{Po}$ = 216.00189 u

Q-value =[ 220.01137 – ( 216.00189 + 4.00260 ) ] x 931.5

≈ 641 MeV

Kinetic energy of the α-particle = $= \left( \frac{ 220 – 4 }{220} \right ) \times 6.41$

= 6.29 MeV

Q 13.13: The radionuclide 11C decays according to $_{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{B} + e^{+} + v \; : T_{\frac{1}{2}} \; = \; 20.3 \; min$. The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:

$m(_{6}^{11}\textrm{C})$ = 11.011434 u and $m(_{5}^{11}\textrm{B})$ = 11.009305 u.

Calculate Q and compare it with the maximum energy of the positron emitted

Ans:

The given nuclear reaction is:

$_{6}^{11}\textrm{C}\rightarrow _{5}^{11}\textrm{B} + e^{+} + v$

Half life of $_{6}^{11}\textrm{C}$ nuclei, T1/2 = 20.3 min

Atomic mass of $m(_{6}^{11}\textrm{C})$ = 11.011434 u

Atomic mass of $m(_{5}^{11}\textrm{B})$ = 11.009305 u

Maximum energy possessed by the emitted positron = 0.960 MeV

The change in the Q-value (∆Q) of the nuclear masses of the $_{6}^{11}\textrm{C}$ $\Delta Q = [m(_{6}^{11}\textrm{C}) – [m'(_{5}^{11}\textrm{B}+m_e)]]c^2$               —–(1)

Where,

me = Mass of an electron or positron = 0.000548 u

c = Speed of light

m’ = Respective nuclear masses

If atomic masses are used instead of nuclear masses, then we have to add 6 me in the

case of $_{}^{11}\textrm{C}$ and 5 me in the case of $_{}^{11}\textrm{B}$.

= 11.011434 u

Hence, equation (1) reduces to:

$\Delta Q = [m(_{6}^{11}\text rm{C}) – m(_{5}^{11}\text rm{B})- 2m_e]c^{2}$

Here, $m(_{6}^{11}\text rm{C})$ and $m(_{5}^{11}\text rm{B})$ are the atomic masses.

∆Q = [11.011434 − 11.009305 − 2 × 0.000548] c2

= (0.001033 c2 ) u

But 1 u = 931.5 Mev/c2

∆Q = 0.001033 × 931.5 ≈ 0.962 MeV

The value of Q is almost comparable to the maximum energy of the emitted positron.

Q 13.14: The nucleus $_{10}^{23}\textrm{Ne}$ decays $\beta ^-$ by emission. Write down the$\beta$  decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:

$m(_{10}^{23}\textrm{Ne})$ = 22.994466 u

$m(_{11}^{23}\textrm{Na})$ = 22.989770 u.

Ans:

In $\beta ^-$ emission, the number of protons increases by 1, and one electron and an antineutrino is emitted from the parent nucleus.

$\beta ^-$ emission from the nucleus $_{10}^{23}\textrm{Ne}$.

$_{ 10 }^{ 23 }\textrm{ Ne }\rightarrow _{ 11 }^{ 23 }\textrm{ Na } + e^{ – } + \bar{ v } + Q$

It is given that:

Atomic mass of $m(_{10}^{23}\textrm{Ne})$= 22.994466 u

Atomic mass of $m(_{11}^{23}\textrm{Na})$ = 22.989770 u

Mass of an electron, me = 0.000548 u

Q-value of the given reaction is given as:

$Q = [m(_{ 10 }^{ 23 }\textrm{Ne}) – [m(_{ 11 }^{ 23 }\textrm{Na}) + m_e]]c^{ 2 }$

There are 10 electrons and 11 electrons in $_{ 10 }^{ 23 }\textrm{Ne}$ and $_{ 11 }^{ 23 }\textrm{Na}$ respectively. Hence, the mass of the electron is cancelled in the Q-value equation.

Q  = [ 22.994466 – 22.989770 ] c2

= (0.004696 c2) u

But 1 u = 931.5 Mev/c2

Q = 0.004696 x 931.5 = 4.374 MeV

The daughter nucleus is too heavy as compared to e and $\bar{v}$. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.

Q 13.15: The Q value of a nuclear reaction A + b → C + d is defined by

Q = [ mA+ mb− mC− md]c2 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) $_{ 1 }^{ 1 }\textrm{H} + _{ 1 }^{ 3 }\textrm{H}\rightarrow _{ 1 }^{ 2 }\textrm{ H } + _{ 1 }^{ 2 }\textrm{H}$

(ii) $_{ 6 }^{ 12 }\textrm{C} + _{ 6 }^{ 12 }\textrm{C}\rightarrow _{ 10 }^{ 20 }\textrm{ Ne } + _{ 2 }^{ 4 }\textrm{H}$

Atomic masses are given to be

$m( _{ 1 }^{ 2 }\textrm{H} ) = \; 2.014102 \; u$

$m( _{ 1 }^{ 3 }\textrm{H} ) = \; 3.016049 \; u$

$m( _{ 6 }^{ 12 }\textrm{C} ) = \; 12.00000 \; u$

$m( _{ 10 }^{ 20 }\textrm{Ne} ) = \; 19.992439 \; u$

Ans

(i) The given nuclear reaction is:

$_{ 1 }^{ 1 }\textrm{H} + _{ 1 }^{ 3 }\textrm{H}\rightarrow _{ 1 }^{ 2 }\textrm{ H } + _{ 1 }^{ 2 }\textrm{H}$

It is given that:

Atomic mass $m( _{ 1 }^{ 1 }\textrm{H} ) = \; 1.007825 \; u$

Atomic mass $m( _{ 1 }^{ 3 }\textrm{H} ) = \; 3.016049 \; u$

Atomic mass $m( _{ 1 }^{ 2 }\textrm{H} ) = \; 1.007825 \; u$

According to the question, the Q-value of the reaction can be written as :

Q = [$m( _{ 1 }^{ 1 }\textrm{H} )$ + $m( _{ 1 }^{ 3 }\textrm{H} )$ – 2m$m( _{ 1 }^{ 2 }\textrm{H} )$ ] c2

= [ 1.007825 + 3.016049 – 2 x 2.014102] c2

Q = ( – 0.00433 c2) u

But 1 u = 931.5 MeV/c2

Q = -0.00433 x 931.5 = – 4.0334 MeV

The negative Q-value of the reaction shows that the reaction is endothermic.

(ii) The given nuclear reaction is:

$_{ 6 }^{ 12 }\textrm{C} + _{ 6 }^{ 12 }\textrm{C}\rightarrow _{ 10 }^{ 20 }\textrm{ Ne } + _{ 2 }^{ 4 }\textrm{H}$

It is given that:

Atomic mass of $m( _{ 6 }^{ 12 }\textrm{C} ) = \; 12.00000 \; u$

Atomic mass of $m( _{ 10 }^{ 20 }\textrm{Ne} ) = \; 19.992439 \; u$

Atomic mass of $m( _{ 2 }^{ 4 }\textrm{He} ) = \; 4.002603 \; u$

The Q-value of this reaction is given as :

Q = [ $m( _{ 6 }^{ 12 }\textrm{C} )$ – $m( _{ 10 }^{ 20 }\textrm{Ne} )$ – $m( _{ 2 }^{ 4 }\textrm{He} ) = \; 4.002603 \; u$ ] c2

= [ 2 x 12.0 – 19.992439 – 4.002603 ] c2

= [ 0.004958 c2] u

= 0.004958 x 931.5 = 4.618377 Mev

The positive Q-value of the reaction shows that the reaction is exothermic.

Q 13.16: Suppose, we think of fission of a$_{ 26 }^{ 56 }\textrm{Fe}$ nucleus into two equal fragments, $_{ 13 }^{ 28 }\textrm{Al}$ . Is the fission energetically possible? Argue by working out Q of the process. Given

$m( _{ 26 }^{ 56 }\textrm{Fe} ) = \; 55.93494 \; u$

$m( _{ 13 }^{ 28 }\textrm{Al} ) = \; 27.98191 \; u$

Ans:

The fission of can $_{ 26 }^{ 56 }\textrm{Fe}$ be given as :

$_{ 26 }^{ 56 }\textrm{Fe}$ -> 2 $_{ 13 }^{ 28 }\textrm{Al}$

It is given that:

Atomic mass of $m( _{ 26 }^{ 56 }\textrm{Fe} ) = \; 55.93494 \; u$

Atomic mass of $m( _{ 13 }^{ 28 }\textrm{Al} ) = \; 27.98191 \; u$

The Q-value of this nuclear reaction is given as :

Q = [$m( _{ 26 }^{ 56 }\text rm{Fe} ) – 2m( _{ 13 }^{ 28 }\text rm{Al} )$ ] c2

= [ 55.93494 – 2 x 27.98191 ]c2

= ( -0.02888 c2) u

But 1 u = 931.5 MeV/c2

Q = – 0.02888 x 931.5 = -26.902 MeV

The Q-value of the fission is negative. Therefore, the fission is not possible energetically.

For an energetically-possible fission reaction, the Q-value must be positive.

Q 13.17: The fission properties$_{ 94 }^{ 239 }\textrm{ Pu }$ of are very similar to those of$_{ 92 }^{ 235 }\textrm{ U }$ .The average energy released per fission is 180 MeV. How much energy is released if all the atoms in 1 kg of pure $_{ 94 }^{ 239 }\textrm{ Pu }$ undergo fission ?

Ans:

Average energy released per fission of $_{ 94 }^{ 239 }\textrm{ Pu }$ , Eav  = 180Mev

Amount of pure $_{ 94 }^{ 239 }\textrm{ Pu }$, m = 1 kg = 1000 g

NA= Avogadro number = 6.023 × 1023

Mass number of $_{ 94 }^{ 239 }\textrm{ Pu }$ = 239 g

1 mole of $_{ 94 }^{ 239 }\textrm{ Pu }$ contains NA atoms.

Therefore, mg of $_{ 94 }^{ 239 }\textrm{ Pu }$ contains $\left ( \frac{N_A}{Mass \; Number}\times m \right )$ atoms

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

Total energy released during the fission of 1 kg of $_{ 94 }^{ 239 }\textrm{ Pu }$ is calculated as:

E = Eav x 2.52 x 1024

= 180 x 2.52 x 1024

= 4.536 x 1026 MeV

Hence, 4.536 x 1026 MeV is released if all the atoms in 1 kg of pure $_{ 94 }^{ 239 }\textrm{ Pu }$ undergo fission.

Q 13.18: A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much $_{ 92 }^{ 235 }\textrm{ U }$  did it contain initially? Assume that the reactor operates 80% of the time, that all the energy generated arises from the fission of $_{ 92 }^{ 235 }\textrm{ U }$  and that this nuclide is consumed only by the fission process.

Half life of the fuel of the fission reactor, $t_{\frac{1}{2}} = 5$ years

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of 1 g of $_{ 92 }^{ 235 }\textrm{ U }$ nucleus, the energy released is equal to 200MeV.

1 mole, i.e., 235 g of $_{ 92 }^{ 235 }\textrm{ U }$ contains 6.023 × 1023 atoms.

1 g contains $_{ 92 }^{ 235 }\textrm{ U }$ => $\frac{6.023 \times 10^{23}}{235}$ atoms contains

The total energy generated per gram of $_{ 92 }^{ 235 }\textrm{ U }$ is calculated as:

$E = \frac{6.023 \times 10^{23}}{235} \times 200 MeV/g$ MeV/g

$= \frac{200 \times 6.023 \times 10^{ 23 }\times 1.6 \times 10^{-19}\times 10^{6}}{235}$

= 8.20 x 1016 J/g

The reactor operates only 80% of the time.

Hence, the amount of $_{ 92 }^{ 235 }\textrm{ U }$ consumed in 5 years by the 1000 MW fission reactor is

calculated as:

$= \frac{5 \times 80 \times 60 \times 60 \times 365\times 24 \times 1000 \times 10^6}{100\times 8.20 \times 10^{10}}$ $\approx 1538$ Kg

Initial amount of $_{ 92 }^{ 235 }\textrm{ U }$ = 2 × 1538 = 3076 kg

Q 13.19: How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium?

Take the fusion reaction as

$_{ 1 }^{ 2 }\textrm{ H } + _{ 1 }^{ 2 }\textrm{H} \rightarrow _{ 1 }^{ 3 }\textrm{ He } + n + 3.27 MeV$

Ans :

The given fusion reaction is:

$_{ 1 }^{ 2 }\textrm{ H } + _{ 1 }^{ 2 }\textrm{H} \rightarrow _{ 1 }^{ 3 }\textrm{ He } + n + 3.27 MeV$

Amount of deuterium, m = 2 kg

1 mole, i.e., 2 g of deuterium contains 6.023 × 1023 atoms.

2.0 kg of deuterium contains $= \frac{6.023 \times 10^{23}}{2}\times 2000 = 6.023 \times 10^{23}$

It can be inferred from the given reaction that when two atoms of deuterium fuse, 3.27 MeV energy is released.

Total energy per nucleus released in the fusion reaction:

$E = \frac{3.27}{2}\times 6.023 \times 10^{26} MeV$ $E = \frac{3.27}{2}\times 6.023 \times 10^{26} \times 1.6 \times 10^{-19} \times 10^{6}$

= 1.576 x 1014 J

Power of the electric lamp, P = 100 W = 100 J/s

Hence, the energy consumed by the lamp per second = 100 J

The total time for which the electric lamp will glow is calculated as:

= $\frac{1.576 \times 10^{14}}{100} s$

= $\frac{1.576 \times 10^{14}}{100} s$ years

Q 13.20: Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)

Ans :

When two deuterons collide head-on, the distance between their centers, d is given as:

Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m

d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C

Potential energy of the two-deuteron system:

$v = \frac{ e^{ 2 } }{ 4 \pi \epsilon_0 d }$

Where,

$\epsilon_0$= Permittivity of free space

$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} Nm^2C^{-2}$

Therefore,

$V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} } J$ $V = \frac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} \times (1.6 \times 10^{-19})}eV$

= 360 keV

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

Q 13.21: From the relation R = R0A1/3, where R0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A).

Ans :

We have the expression for nuclear radius as:

R = R0A1/3

Where,

R0 = Constant.

A = Mass number of the nucleus

Nuclear matter density, $\rho = \frac{Mass \; of \; the \; nucleus}{Volume \; of \; the \; nucleus}$

Let m be the average mass of the nucleus.

Hence, mass of the nucleus = mA

Therefore , $\rho = \frac{ mA }{\frac{ 4 }{ 3 }\pi R^{ 3 }}$

= $\frac{ 3mA }{4 \pi( R_0 A^{ \frac{1}{3} })^{3}}$

= $\frac{ 3mA }{4 \pi R_0^3 A}$

= $\frac{ 3m }{4 \pi R_0^3 }$

Hence, the nuclear matter density is independent of A. It is nearly constant.

Q 13.22: 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).

$e^+ + _{Z}^{A}\textrm{X} \rightarrow _{ Z-1 }^{A}\textrm{Y}+v$

Show that if$\beta^{+}$ emission is energetically allowed, electron capture is necessarily allowed but not vice−versa.

Ans :

Let the amount of energy released during the electron capture process be Q1. The

nuclear reaction can be written as:

$e^+ + _{Z}^{A}\textrm{X} \rightarrow _{ Z-1 }^{A}\textrm{Y} + v +Q_1$      —–(1)

Let the amount of energy released during the positron capture process be Q2. The nuclear reaction can be written as:

$e^+ + _{Z}^{A}\textrm{X} \rightarrow _{ Z-1 }^{A}\textrm{Y} + v +Q_2$      —–(2)

$m_n(_{Z}^{A}\textrm{X}) = Nuclear \; mass \; of \;_{Z}^{A}\textrm{X}$ $m_n(_{Z-1}^{A}\textrm{X}) = Nuclear \; mass \; of \;_{Z-1}^{A}\textrm{Y}$ $m(_{Z}^{A}\textrm{X}) = Atomic \; mass \; of \;_{Z}^{A}\textrm{X}$ $m(_{Z-1}^{A}\textrm{X}) = Atomic \; mass \; of \;_{Z-1}^{A}\textrm{Y}$

me = Mass of an electron

c = Speed of light

Q-value of the electron capture reaction is given as:

Q1 = [$m_n(_{Z}^{A}\textrm{X})$ + me – $m_n(_{Z-1}^{A}\textrm{Y})$ ] c2

= [ $m_n(_{Z}^{A}\textrm{X})$ – Zme + $m_n(_{Z-1}^{A}\textrm{Y})$ – ( Z – 1 )me ] c2

= [$m_n(_{Z}^{A}\textrm{X})$ – $m_n(_{Z-1}^{A}\textrm{Y})$ ] c2           —–(3)

Q – value of the positron capture reaction is given as :

Q2 = [ $m_n(_{Z}^{A}\textrm{X})$ – $m_n(_{Z-1}^{A}\textrm{Y})$ -me ] c2

= [ $m_n(_{Z}^{A}\textrm{X})$ – Zme – $m_n(_{Z-1}^{A}\textrm{Y})$ +( Z – 1 )me – me ] c2

= [$m_n(_{Z}^{A}\textrm{X})$ – $m_n(_{Z-1}^{A}\textrm{Y})$ -2me  ] c2              —–(4)

It can be inferred that if Q2 > 0, then Q1 > 0; Also, if Q1> 0, it does not necessarily mean that Q2 > 0.

In other words, this means that if $\beta ^+$ emission is energetically allowed, then the electron capture process is necessarily allowed, but not vice-versa. This is because the Q-value must be positive for an energetically allowed nuclear reaction.

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