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NCERT Solutions · Class 12 Physics · NCERT Physics Part I and II Part II · Chapter 13

Chapter 13: Nuclei (Modern Physics)

Step-by-step solutions to all 10 exercises of Chapter 13, Nuclei (NCERT Class 12 Physics Part II, 2026-27): binding energy of nitrogen, iron and bismuth, energy to break up a copper coin, ratio of nuclear radii, Q-values of reactions, fission of iron, energy from plutonium fission and deuterium fusion, the Coulomb barrier, and constant nuclear density. All 10 questions are answered, with the key answer highlighted.

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Data: $m_{H} = 1.007825$ u, $m_{n} = 1.008665$ u, 1 u $= 931.5$ MeV/c², 1 MeV $= 1.6 \times 1 0^{- 13}$ J, $N_{A} = 6.023 \times 1 0^{23}$ . Mass defect $Δ m = Z m_{H} + (A - Z) m_{n} - m_{atom}$ ; binding energy $E_{b} = Δ m \times 931.5$ MeV. Q-value: $Q = (mass before - mass after) c^{2}$ ; $Q > 0$ means energy is released. Nuclear radius: $R = R_{0} A^{1/3}$ .

Exercises

13.1

Obtain the binding energy (in MeV) of a nitrogen nucleus

(_{7}^{14} N)

, given

m (_{7}^{14} N) = 14.00307

Solution

N-14 has $Z = 7$ protons and $A - Z = 7$ neutrons.

$Δ m = 7 \times 1.007825 + 7 \times 1.008665 - 14.00307 = 14.11543 - 14.00307 = 0.11236 u$

$E_{b} = 0.11236 \times 931.5 \approx 104.7 MeV$

About 104.7 MeV.

13.2

Obtain the binding energy of the nuclei

_{26}^{56} Fe

and

_{83}^{209} Bi

in units of MeV from the following data:

m (_{26}^{56} Fe) = 55.934939

m (_{83}^{209} Bi) = 208.980388

Solution

Fe-56 (26 protons, 30 neutrons):

$Δ m = 26 \times 1.007825 + 30 \times 1.008665 - 55.934939 = 56.463400 - 55.934939 = 0.528461 u$

$E_{b} = 0.528461 \times 931.5 \approx 492.3$ MeV, i.e. $\frac{492.3}{56} \approx 8.79$ MeV per nucleon.

Bi-209 (83 protons, 126 neutrons):

$Δ m = 83 \times 1.007825 + 126 \times 1.008665 - 208.980388 = 210.741265 - 208.980388 = 1.760877 u$

$E_{b} = 1.760877 \times 931.5 \approx 1640.3$ MeV, i.e. $\frac{1640.3}{209} \approx 7.85$ MeV per nucleon.

Iron is more tightly bound per nucleon than bismuth.

Fe-56: 492.3 MeV (8.79 MeV per nucleon). Bi-209: 1640.3 MeV (7.85 MeV per nucleon).

13.3

A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of

_{29}^{63} Cu

atoms (of mass 62.92960 u).

Solution

One Cu-63 nucleus (29 protons, 34 neutrons):

$Δ m = 29 \times 1.007825 + 34 \times 1.008665 - 62.92960 = 63.521535 - 62.92960 = 0.591935 u$

$E_{b} = 0.591935 \times 931.5 \approx 551.4$ MeV.

Atoms in 3.0 g: $N = \frac{3.0}{62.93} \times 6.023 \times 1 0^{23} \approx 2.87 \times 1 0^{22}$ .

$E = 2.87 \times 1 0^{22} \times 551.4 MeV \approx 1.58 \times 1 0^{25} MeV = 1.58 \times 1 0^{25} \times 1.6 \times 1 0^{- 13} J \approx 2.5 \times 1 0^{12} J$

About $1.6 \times 1 0^{25}$ MeV, i.e. $2.5 \times 1 0^{12}$ J.

13.4

Obtain approximately the ratio of the nuclear radii of the gold isotope

_{79}^{197} Au

and the silver isotope

_{47}^{107} Ag

Solution

$R = R_{0} A^{1/3}$ , so

$\frac{R _{Au}}{R _{Ag}} = (\frac{197}{107})^{1/3} = (1.84)^{1/3} \approx 1.23$

About 1.23.

13.5

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

Q = [m_{A} + m_{b} - m_{C} - m_{d}] c^{2}

, 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 \to_{1}^{2} H +_{1}^{2} H

(ii)

_{6}^{12} C +_{6}^{12} C \to_{10}^{20} Ne +_{2}^{4} He

. Atomic masses are given to be

m (_{1}^{2} H) = 2.014102

m (_{1}^{3} H) = 3.016049

m (_{6}^{12} C) = 12.000000

m (_{10}^{20} Ne) = 19.992439

Solution

The electrons balance on both sides, so atomic masses can be used. Also $m (_{1}^{1} H) = 1.007825$ u and $m (_{2}^{4} He) = 4.002603$ u.

(i) Before: $1.007825 + 3.016049 = 4.023874$ u. After: $2 \times 2.014102 = 4.028204$ u.

$Q = (4.023874 - 4.028204) \times 931.5 = - 0.004330 \times 931.5 \approx - 4.03$ MeV: endothermic.

(ii) Before: $2 \times 12.000000 = 24.000000$ u. After: $19.992439 + 4.002603 = 23.995042$ u.

$Q = 0.004958 \times 931.5 \approx + 4.62$ MeV: exothermic.

(i) Q ≈ −4.03 MeV, endothermic. (ii) Q ≈ +4.62 MeV, exothermic.

13.6

Suppose, we think of fission of a

_{26}^{56} Fe

nucleus into two equal fragments,

_{13}^{28} Al

. Is the fission energetically possible? Argue by working out Q of the process. Given

m (_{26}^{56} Fe) = 55.93494

u and

m (_{13}^{28} Al) = 27.98191

Solution

$Q = [55.93494 - 2 \times 27.98191] \times 931.5 = (- 0.02888) \times 931.5 \approx - 26.9 MeV$

Q is negative: energy would have to be supplied. So the fission of Fe-56 into two Al-28 nuclei is not energetically possible (iron is near the peak of binding energy per nucleon).

Q ≈ −26.9 MeV, so the fission is not possible on its own.

13.7

The fission properties of

_{94}^{239} Pu

are very similar to those of

_{92}^{235} 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

_{94}^{239} Pu

undergo fission?

Solution

Number of atoms in 1 kg:

$N = \frac{1000}{239} \times 6.023 \times 1 0^{23} \approx 2.52 \times 1 0^{24}$

$E = 2.52 \times 1 0^{24} \times 180 \approx 4.5 \times 1 0^{26} MeV$

About $4.5 \times 1 0^{26}$ MeV.

13.8

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

_{1}^{2} H +_{1}^{2} H \to_{2}^{3} He + n + 3.27

MeV.

Solution

Deuterium atoms in 2.0 kg: $N = \frac{2000}{2} \times 6.023 \times 1 0^{23} = 6.02 \times 1 0^{26}$ .

Each reaction uses 2 deuterons, so there are $3.01 \times 1 0^{26}$ reactions.

$E = 3.01 \times 1 0^{26} \times 3.27 MeV \approx 9.85 \times 1 0^{26} MeV = 9.85 \times 1 0^{26} \times 1.6 \times 1 0^{- 13} \approx 1.58 \times 1 0^{14} J$

$t = \frac{E}{P} = \frac{1.58 \times 1 0 ^{14}}{100} = 1.58 \times 1 0^{12} s = \frac{1.58 \times 1 0 ^{12}}{3.154 \times 1 0 ^{7}} years \approx 5 \times 1 0^{4} years$

About $1.6 \times 1 0^{12}$ s, roughly 50,000 years.

13.9

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.)

Solution

When they just touch, their centres are $r = 2 \times 2.0 = 4.0$ fm $= 4.0 \times 1 0^{- 15}$ m apart.

$U = \frac{k e ^{2}}{r} = \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{4.0 \times 1 0 ^{- 15}} = 5.76 \times 1 0^{- 14} J = \frac{5.76 \times 1 0 ^{- 14}}{1.6 \times 1 0 ^{- 13}} MeV = 0.36 MeV$

About 360 keV (0.36 MeV).

13.10

From the relation

R = R_{0} A^{1/3}

, where

R_{0}

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).

Solution

Mass of a nucleus $\approx A m$ , where $m$ is the mass of a nucleon. Its volume:

$V = \frac{4}{3} π R^{3} = \frac{4}{3} π R_{0}^{3} A$

$ρ = \frac{A m}{\frac{4}{3} π R _{0}^{3} A} = \frac{3 m}{4 π R _{0}^{3}}$

$A$ cancels, so the density is the same for all nuclei (about $2.3 \times 1 0^{17}$ kg m⁻³).

$ρ = 3 m / (4 π R_{0}^{3})$ , which does not depend on A.

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