Nuclei

Physics XII - Chapter 13: Complete NCERT Exercise Solutions

Master Nuclear Physics with NCERT solutions for nuclear properties, radioactivity, binding energy, and nuclear reactions.

Question 13.1

(a) Two stable isotopes of lithium have 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) Boron has two stable isotopes, \( ^{10}_{5}B \) and \( ^{11}_{5}B \). 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 \( ^{10}_{5}B \) and \( ^{11}_{5}B \).

Answer & Explanation:

Step 1: (a) Atomic mass of lithium
Average atomic mass = \( \frac{(7.5 \times 6.01512) + (92.5 \times 7.01600)}{100} \)
Numerator: \( (7.5 \times 6.01512) = 45.1134 \)
\( (92.5 \times 7.01600) = 648.98 \)
Sum = 694.0934
Atomic mass = \( \frac{694.0934}{100} = 6.940934 \text{ u} \approx 6.941 \text{ u} \)

Step 2: (b) Let abundance of \( ^{10}B \) = x%
Then abundance of \( ^{11}B \) = (100 - x)%
Average mass = \( \frac{x \times 10.01294 + (100-x) \times 11.00931}{100} = 10.811 \)
Multiply by 100: \( 10.01294x + 1100.931 - 11.00931x = 1081.1 \)
\( -0.99637x = 1081.1 - 1100.931 = -19.831 \)
\( x = \frac{19.831}{0.99637} = 19.91\% \)
So \( ^{10}B \) abundance ≈ 19.91%, \( ^{11}B \) abundance ≈ 80.09%

Answers:
(a) Atomic mass of lithium = 6.941 u
(b) \( ^{10}B \): 19.9%, \( ^{11}B \): 80.1%

Question 13.2

The three stable isotopes of neon: \( ^{20}_{10}Ne \), \( ^{21}_{10}Ne \) and \( ^{22}_{10}Ne \) 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.

Answer & Explanation:

Step 1: Calculate weighted average
Average mass = \( \frac{(90.51 \times 19.99) + (0.27 \times 20.99) + (9.22 \times 21.99)}{100} \)

Step 2: Compute each term
\( 90.51 \times 19.99 = 1809.2949 \)
\( 0.27 \times 20.99 = 5.6673 \)
\( 9.22 \times 21.99 = 202.7478 \)
Sum = 1809.2949 + 5.6673 + 202.7478 = 2017.7100

Step 3: Final average
Average = \( \frac{2017.71}{100} = 20.1771 \text{ u} \approx 20.18 \text{ u} \)

Answer: Average atomic mass of neon = 20.18 u

Question 13.3

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

Answer & Explanation:

Step 1: Mass defect
\( ^{14}_{7}N \) has 7 protons and 7 neutrons
Mass of 7 protons = \( 7 \times 1.007825 = 7.054775 \text{ u} \)
Mass of 7 neutrons = \( 7 \times 1.008665 = 7.060655 \text{ u} \)
Total mass of constituents = 7.054775 + 7.060655 = 14.11543 u
Mass of nucleus = 14.00307 u
Mass defect Δm = 14.11543 - 14.00307 = 0.11236 u

Step 2: Binding energy
1 u = 931.5 MeV/c²
BE = Δm × 931.5 = 0.11236 × 931.5 = 104.66 MeV

Step 3: Binding energy per nucleon
BE/A = 104.66/14 = 7.476 MeV/nucleon

Answer: Binding energy = 104.7 MeV (7.48 MeV/nucleon)

Question 13.4

Obtain the binding energy of the nuclei \( ^{56}_{26}Fe \) and \( ^{209}_{83}Bi \) in units of MeV from the following data:
\( m(^{56}_{26}Fe) = 55.934939 \, u \)
\( m(^{209}_{83}Bi) = 208.980388 \, u \)

Answer & Explanation:

Step 1: For \( ^{56}Fe \) (Z=26, N=30)
Mass of 26 protons = 26 × 1.007825 = 26.20345 u
Mass of 30 neutrons = 30 × 1.008665 = 30.25995 u
Total = 56.46340 u
Δm = 56.46340 - 55.934939 = 0.528461 u
BE = 0.528461 × 931.5 = 492.26 MeV
BE/A = 492.26/56 = 8.790 MeV/nucleon

Step 2: For \( ^{209}Bi \) (Z=83, N=126)
Mass of 83 protons = 83 × 1.007825 = 83.649475 u
Mass of 126 neutrons = 126 × 1.008665 = 127.091790 u
Total = 210.741265 u
Δm = 210.741265 - 208.980388 = 1.760877 u
BE = 1.760877 × 931.5 = 1640.3 MeV
BE/A = 1640.3/209 = 7.848 MeV/nucleon

Answers:
\( ^{56}Fe \): BE = 492.3 MeV (8.79 MeV/nucleon)
\( ^{209}Bi \): BE = 1640 MeV (7.85 MeV/nucleon)

Question 13.5

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 \( ^{63}_{29}Cu \) atoms (of mass 62.92960 u).

Answer & Explanation:

Step 1: Find binding energy per \( ^{63}Cu \) nucleus
\( ^{63}_{29}Cu \): Z=29, N=34
Mass of 29 protons = 29 × 1.007825 = 29.226925 u
Mass of 34 neutrons = 34 × 1.008665 = 34.294610 u
Total = 63.521535 u
Δm = 63.521535 - 62.92960 = 0.591935 u
BE per nucleus = 0.591935 × 931.5 = 551.38 MeV

Step 2: Number of atoms in coin
Mass of coin = 3.0 g
Molar mass of \( ^{63}Cu \) = 62.92960 g/mol
Number of moles = 3.0/62.92960 = 0.04767 mol
Number of atoms = 0.04767 × 6.022×10²³ = 2.871×10²² atoms

Step 3: Total energy required
Total BE = 551.38 MeV × 2.871×10²² = 1.582×10²⁵ MeV
Convert to joules: 1 MeV = 1.6×10⁻¹³ J
= 1.582×10²⁵ × 1.6×10⁻¹³ = 2.531×10¹² J

Step 4: Comparison
This is enormous energy! For perspective, 1 kiloton TNT ≈ 4.2×10¹² J.
So separating nucleons in a 3g coin releases energy comparable to a small nuclear weapon.

Answer: Required energy ≈ 2.5×10¹² J (∼0.6 kilotons TNT equivalent)

Question 13.6

Write nuclear reaction equations for:
(i) α-decay of \( ^{226}_{88}Ra \)
(ii) α-decay of \( ^{242}_{94}Pu \)
(iii) β⁻-decay of \( ^{32}_{15}P \)
(iv) β⁺-decay of \( ^{210}_{83}Bi \)
(v) β⁻-decay of \( ^{11}_{6}C \)
(vi) β⁺-decay of \( ^{97}_{43}Tc \)
(vii) Electron capture of \( ^{120}_{54}Xe \)

Answer & Explanation:

General rules:
• α-decay: \( ^{A}_{Z}X \rightarrow ^{A-4}_{Z-2}Y + ^{4}_{2}He \)
• β⁻-decay: \( ^{A}_{Z}X \rightarrow ^{A}_{Z+1}Y + e^- + \bar{\nu}_e \)
• β⁺-decay: \( ^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + e^+ + \nu_e \)
• Electron capture: \( ^{A}_{Z}X + e^- \rightarrow ^{A}_{Z-1}Y + \nu_e \)

Equations:
(i) \( ^{226}_{88}Ra \rightarrow ^{222}_{86}Rn + ^{4}_{2}He \)
(ii) \( ^{242}_{94}Pu \rightarrow ^{238}_{92}U + ^{4}_{2}He \)
(iii) \( ^{32}_{15}P \rightarrow ^{32}_{16}S + e^- + \bar{\nu}_e \)
(iv) \( ^{210}_{83}Bi \rightarrow ^{210}_{82}Pb + e^+ + \nu_e \)
(v) \( ^{11}_{6}C \rightarrow ^{11}_{5}B + e^+ + \nu_e \) (actually β⁺, not β⁻)
(vi) \( ^{97}_{43}Tc \rightarrow ^{97}_{42}Mo + e^+ + \nu_e \)
(vii) \( ^{120}_{54}Xe + e^- \rightarrow ^{120}_{53}I + \nu_e \)

Note: Question has typo: \( ^{11}_{6}C \) undergoes β⁺ decay, not β⁻.

Question 13.7

A radioactive isotope has a half-life of 7 years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value?

Answer & Explanation:

Step 1: Decay formula
\( A = A_0 \left( \frac{1}{2} \right)^{t/T} \) where T = half-life = 7 years
\( \frac{A}{A_0} = \left( \frac{1}{2} \right)^{t/T} \)

Step 2: (a) For A/A₀ = 3.125% = 0.03125
\( 0.03125 = \left( \frac{1}{2} \right)^{t/7} \)
Note: 0.03125 = 1/32 = (1/2)⁵
So \( \left( \frac{1}{2} \right)^{t/7} = \left( \frac{1}{2} \right)^5 \)
⇒ t/7 = 5 ⇒ t = 35 years

Step 3: (b) For A/A₀ = 1% = 0.01
\( 0.01 = \left( \frac{1}{2} \right)^{t/7} \)
Take log: \( \ln(0.01) = \frac{t}{7} \ln(0.5) \)
\( \frac{t}{7} = \frac{\ln(0.01)}{\ln(0.5)} = \frac{-4.6052}{-0.6931} = 6.644 \)
t = 6.644 × 7 = 46.51 years

Answers:
(a) 35 years
(b) 46.5 years

Question 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 \( ^{14}_{6}C \) present with the stable carbon isotope \( ^{12}_{6}C \). 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.

Answer & Explanation:

Step 1: Given data
Initial activity A₀ = 15 decays/min/g
Present activity A = 9 decays/min/g
Half-life of \( ^{14}C \) = 5730 years

Step 2: Decay formula
\( A = A_0 e^{-\lambda t} \) where \( \lambda = \frac{\ln 2}{T_{1/2}} \)
\( \frac{A}{A_0} = e^{-\lambda t} \)
\( \frac{9}{15} = 0.6 = e^{-\lambda t} \)

Step 3: Solve for t
\( \ln(0.6) = -\lambda t \)
\( t = -\frac{\ln(0.6)}{\lambda} = -\frac{\ln(0.6)}{\ln 2} \times T_{1/2} \)
\( = -\frac{-0.5108}{0.6931} \times 5730 \)
\( = 0.737 \times 5730 = 4223 \text{ years} \)

Step 4: Interpretation
This is the time since the organism died (stopped exchanging carbon with atmosphere).
So Indus-Valley civilisation sample is about 4200 years old.

Answer: Approximate age = 4200 years

Question 13.9

Obtain the amount of \( ^{60}_{27}Co \) necessary to provide a radioactive source of 8.0 mCi strength. The half-life of \( ^{60}_{27}Co \) is 5.3 years.

Answer & Explanation:

Step 1: Convert units
1 Ci = 3.7×10¹⁰ decays/s
8.0 mCi = 8.0×10⁻³ × 3.7×10¹⁰ = 2.96×10⁸ decays/s
This is activity A = λN

Step 2: Decay constant λ
T_{1/2} = 5.3 years = 5.3 × 3.154×10⁷ s = 1.672×10⁸ s
\( \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{1.672\times 10^8} = 4.145\times 10^{-9} \text{ s}^{-1} \)

Step 3: Number of atoms N
\( N = \frac{A}{\lambda} = \frac{2.96\times 10^8}{4.145\times 10^{-9}} = 7.14\times 10^{16} \) atoms

Step 4: Mass calculation
Atomic mass of \( ^{60}Co \) ≈ 60 g/mol
Number of moles = \( \frac{7.14\times 10^{16}}{6.022\times 10^{23}} = 1.186\times 10^{-7} \) mol
Mass = moles × molar mass = 1.186×10⁻⁷ × 60 = 7.12×10⁻⁶ g = 7.12 μg

Answer: Required amount ≈ 7.1 μg of \( ^{60}Co \)

Question 13.10

The half-life of \( ^{90}_{38}Sr \) is 28 years. What is the disintegration rate of 15 mg of this isotope?

Answer & Explanation:

Step 1: Number of atoms in 15 mg
Molar mass of \( ^{90}Sr \) ≈ 90 g/mol
Mass = 15 mg = 0.015 g
Moles = 0.015/90 = 1.667×10⁻⁴ mol
Number of atoms N = 1.667×10⁻⁴ × 6.022×10²³ = 1.004×10²⁰ atoms

Step 2: Decay constant
T_{1/2} = 28 years = 28 × 3.154×10⁷ s = 8.831×10⁸ s
\( \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{8.831\times 10^8} = 7.85\times 10^{-10} \text{ s}^{-1} \)

Step 3: Activity A = λN
A = 7.85×10⁻¹⁰ × 1.004×10²⁰ = 7.88×10¹⁰ decays/s
In curie: 1 Ci = 3.7×10¹⁰ decays/s
A = 7.88×10¹⁰/3.7×10¹⁰ = 2.13 Ci

Answer: Disintegration rate = 7.9×10¹⁰ decays/s (≈ 2.1 Ci)

Question 13.11

Obtain approximately the ratio of the nuclear radii of the gold isotope \( ^{197}_{79}Au \) and the silver isotope \( ^{107}_{47}Ag \).

Answer & Explanation:

Step 1: Nuclear radius formula
\( R = R_0 A^{1/3} \) where \( R_0 \approx 1.2 \text{ fm} \)
For gold: A = 197
For silver: A = 107

Step 2: Ratio
\( \frac{R_{Au}}{R_{Ag}} = \left( \frac{A_{Au}}{A_{Ag}} \right)^{1/3} = \left( \frac{197}{107} \right)^{1/3} \)
197/107 ≈ 1.841
Cube root: (1.841)^{1/3}
Since 1.2³ = 1.728, 1.3³ = 2.197
Interpolate: ≈ 1.23
More precisely: (1.841)^{1/3} = e^{(1/3)\ln(1.841)} = e^{(1/3)×0.610} = e^{0.2033} = 1.225

Answer: Ratio \( \frac{R_{Au}}{R_{Ag}} \approx \) 1.23

Question 13.12

Find the Q-value and the kinetic energy of the emitted α-particle in the α-decay of:
(a) \( ^{226}_{88}Ra \) and (b) \( ^{220}_{86}Rn \)
Given:
\( m(^{226}_{88}Ra) = 226.02540 \, u \)
\( m(^{222}_{86}Rn) = 222.01750 \, u \)
\( m(^{220}_{86}Rn) = 220.01137 \, u \)
\( m(^{216}_{84}Po) = 216.00189 \, u \)

Answer & Explanation:

Step 1: Q-value formula for α-decay
\( Q = [m(\text{parent}) - m(\text{daughter}) - m(\alpha)]c^2 \)
\( m(\alpha) = 4.002603 \text{ u} \)

Step 2: (a) For \( ^{226}Ra \rightarrow ^{222}Rn + \alpha \)**
\( Q = (226.02540 - 222.01750 - 4.002603) \times 931.5 \)
Δm = 226.02540 - 222.01750 - 4.002603 = 0.005297 u
Q = 0.005297 × 931.5 = 4.935 MeV

Step 3: Kinetic energy of α-particle
Since daughter nucleus recoils, α gets most of energy:
\( K_\alpha = \frac{A-4}{A} Q \)
For A=226: \( K_\alpha = \frac{222}{226} \times 4.935 = 4.84 \text{ MeV} \)

Step 4: (b) For \( ^{220}Rn \rightarrow ^{216}Po + \alpha \)**
\( Q = (220.01137 - 216.00189 - 4.002603) \times 931.5 \)
Δm = 220.01137 - 216.00189 - 4.002603 = 0.006877 u
Q = 0.006877 × 931.5 = 6.407 MeV
\( K_\alpha = \frac{216}{220} \times 6.407 = 6.29 \text{ MeV} \)

Answers:
(a) Q = 4.94 MeV, \( K_\alpha \approx \) 4.84 MeV
(b) Q = 6.41 MeV, \( K_\alpha \approx \) 6.29 MeV

Question 13.13

The radionuclide \( ^{11}C \) decays according to:
\( ^{11}_{6}C \rightarrow ^{11}_{5}B + e^+ + \nu \quad T_{1/2} = 20.3 \, \text{min} \)
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
\( m(^{11}_{6}C) = 11.011434 \, u \) and \( m(^{11}_{5}B) = 11.009305 \, u \)
calculate \( Q \) and compare it with the maximum energy of the positron emitted.

Answer & Explanation:

Step 1: Q-value for β⁺ decay
\( Q = [m(^{11}C) - m(^{11}B) - 2m_e]c^2 \)
Note: Need to subtract 2 electron masses because:
• Parent has 6 electrons, daughter has 5 electrons
• One electron is annihilated with positron
• So net loss of 2 electron masses

Step 2: Mass difference
Δm = m(¹¹C) - m(¹¹B) = 11.011434 - 11.009305 = 0.002129 u
Electron mass m_e = 0.00054858 u
Effective Δm = 0.002129 - 2×0.00054858 = 0.002129 - 0.00109716 = 0.00103184 u

Step 3: Q-value
Q = 0.00103184 × 931.5 = 0.961 MeV

Step 4: Comparison with maximum positron energy
Maximum positron energy = 0.960 MeV
Q ≈ 0.961 MeV
The difference (0.001 MeV) is carried by neutrino.
When positron has maximum energy, neutrino gets negligible energy.

Answer: Q = 0.961 MeV, matches well with maximum positron energy (0.960 MeV).

Question 13.14

The nucleus \( ^{23}_{10}Ne \) decays by β⁻ emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
\( m(^{23}_{10}Ne) = 22.994466 \, u \)
\( m(^{23}_{11}Na) = 22.989770 \, u \)

Answer & Explanation:

Step 1: β⁻ decay equation
\( ^{23}_{10}Ne \rightarrow ^{23}_{11}Na + e^- + \bar{\nu}_e \)

Step 2: Q-value calculation
For β⁻ decay: \( Q = [m(^{23}Ne) - m(^{23}Na)]c^2 \)
(No electron mass correction needed because atomic masses include electrons)
Δm = 22.994466 - 22.989770 = 0.004696 u
Q = 0.004696 × 931.5 = 4.374 MeV

Step 3: Maximum electron kinetic energy
Maximum K.E. of electron ≈ Q (when neutrino gets negligible energy)
So \( K_{\text{max}} \approx 4.374 \text{ MeV} \)

Answer:
Decay equation: \( ^{23}_{10}Ne \rightarrow ^{23}_{11}Na + e^- + \bar{\nu}_e \)
Maximum electron energy = 4.37 MeV

Question 13.15

The \( Q \) value of a nuclear reaction \( A + b \rightarrow C + d \) is defined by:
\( Q = [m_A + m_b - m_C - m_d]c^2 \)
Determine from the given data the Q-value of the following reactions and state whether they are exothermic or endothermic:
(i) \( ^{1}_{1}H + ^{3}_{1}H \rightarrow ^{2}_{1}H + ^{2}_{1}H \)
(ii) \( ^{12}_{6}C + ^{12}_{6}C \rightarrow ^{20}_{10}Ne + ^{4}_{2}He \)
Atomic masses are:
\( m(^{2}_{1}H) = 2.014102 \, u \)
\( m(^{3}_{1}H) = 3.016049 \, u \)
\( m(^{12}_{6}C) = 12.000000 \, u \)
\( m(^{20}_{10}Ne) = 19.992439 \, u \)

Answer & Explanation:

Step 1: (i) \( H^1 + H^3 \rightarrow H^2 + H^2 \)**
Q = [m(¹H) + m(³H) - 2×m(²H)] × 931.5
Need m(¹H) = 1.007825 u (not given, but standard)
Δm = 1.007825 + 3.016049 - 2×2.014102
= 4.023874 - 4.028204 = -0.004330 u
Q = -0.004330 × 931.5 = -4.034 MeV
Negative Q → endothermic

Step 2: (ii) \( C^{12} + C^{12} \rightarrow Ne^{20} + He^{4} \)**
Need m(⁴He) = 4.002603 u (standard)
Δm = 2×12.000000 - (19.992439 + 4.002603)
= 24.000000 - 23.995042 = 0.004958 u
Q = 0.004958 × 931.5 = 4.619 MeV
Positive Q → exothermic

Answers:
(i) Q = -4.03 MeV (endothermic)
(ii) Q = +4.62 MeV (exothermic)

Question 13.16

Suppose, we think of fission of a \( ^{56}_{26}Fe \) nucleus into two equal fragments, \( ^{28}_{13}Al \). Is the fission energetically possible? Argue by working out \( Q \) of the process. Given \( m(^{56}_{26}Fe) = 55.93494 \, u \) and \( m(^{28}_{13}Al) = 27.98191 \, u \).

Answer & Explanation:

Step 1: Reaction equation
\( ^{56}Fe \rightarrow 2(^{28}Al) \)

Step 2: Q-value calculation
Q = [m(⁵⁶Fe) - 2×m(²⁸Al)] × 931.5
Δm = 55.93494 - 2×27.98191
= 55.93494 - 55.96382 = -0.02888 u
Q = -0.02888 × 931.5 = -26.91 MeV

Step 3: Interpretation
Q is negative (−26.9 MeV).
This means energy must be supplied for fission to occur.
The process is not energetically favorable.

Step 4: Physical reason
Iron-56 is near the peak of binding energy per nucleon curve.
Fission of iron would produce lighter nuclei with lower binding energy per nucleon, requiring energy input.

Answer: Fission is not energetically possible (Q = -26.9 MeV, endothermic).

Question 13.17

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

Answer & Explanation:

Step 1: Number of atoms in 1 kg
Molar mass of \( ^{239}Pu \) = 239 g/mol
1 kg = 1000 g
Moles = 1000/239 = 4.184 mol
Number of atoms N = 4.184 × 6.022×10²³ = 2.519×10²⁴ atoms

Step 2: Total energy
Energy per fission = 180 MeV
Total energy = 180 × 2.519×10²⁴ = 4.534×10²⁶ MeV

Step 3: Convert to joules
1 MeV = 1.6×10⁻¹³ J
Total energy = 4.534×10²⁶ × 1.6×10⁻¹³ = 7.255×10¹³ J

Step 4: Comparison
This is enormous: equivalent to about 17 kilotons of TNT.
(1 kt TNT = 4.2×10¹² J)

Answer: Total energy released = 4.53×10²⁶ MeV (≈ 7.3×10¹³ J, ~17 kt TNT)

Question 13.18

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

Answer & Explanation:

Step 1: Total energy produced in 5 years
Reactor power = 1000 MW = 10⁹ J/s
Operating time = 80% of 5 years
Total seconds in 5 years = 5 × 3.154×10⁷ = 1.577×10⁸ s
Operating time = 0.8 × 1.577×10⁸ = 1.262×10⁸ s
Total energy E = power × time = 10⁹ × 1.262×10⁸ = 1.262×10¹⁷ J

Step 2: Number of fissions required
Energy per fission = 200 MeV = 200 × 1.6×10⁻¹³ = 3.2×10⁻¹¹ J
Number of fissions = E / (energy per fission) = 1.262×10¹⁷ / 3.2×10⁻¹¹ = 3.944×10²⁷ fissions

Step 3: Mass of \( ^{235}U \) consumed
Each fission consumes 1 atom of \( ^{235}U \)
Molar mass = 235 g/mol
Number of moles = 3.944×10²⁷ / 6.022×10²³ = 6548 mol
Mass consumed = 6548 × 235 = 1,538,780 g = 1539 kg

Step 4: Initial mass
Since half is consumed in 5 years, initial mass = 2 × 1539 kg = 3078 kg

Answer: Initial mass of \( ^{235}U \) ≈ 3080 kg (∼3 tonnes)

Question 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:
\( ^{2}_{1}H + ^{2}_{1}H \rightarrow ^{3}_{2}He + n + 3.27 \, \text{MeV} \)

Answer & Explanation:

Step 1: Energy per reaction
Each reaction releases 3.27 MeV
= 3.27 × 1.6×10⁻¹³ = 5.232×10⁻¹³ J

Step 2: Number of deuterium atoms in 2.0 kg
Molar mass of D₂ = 4 g/mol (since ²H = 2 g/mol, D₂ molecule = 4 g/mol)
2.0 kg = 2000 g
Moles of D₂ = 2000/4 = 500 mol
Atoms of deuterium = 500 × 6.022×10²³ × 2 = 6.022×10²⁶ atoms
(Each D₂ molecule has 2 D atoms)

Step 3: Number of fusion reactions
Each reaction uses 2 deuterium atoms
Number of reactions = 6.022×10²⁶ / 2 = 3.011×10²⁶ reactions

Step 4: Total energy released
Total energy = 3.011×10²⁶ × 5.232×10⁻¹³ = 1.575×10¹⁴ J

Step 5: Time to power 100W lamp
Power = 100 W = 100 J/s
Time = Energy / Power = 1.575×10¹⁴ / 100 = 1.575×10¹² s
Convert to years: 1 year = 3.154×10⁷ s
Time = 1.575×10¹² / 3.154×10⁷ = 4.99×10⁴ years ≈ 50,000 years

Answer: The lamp can glow for about 50,000 years

Question 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.)

Answer & Explanation:

Step 1: Distance of closest approach
Each deuteron radius r = 2.0 fm = 2.0×10⁻¹⁵ m
When just touching, distance between centers = 2r = 4.0×10⁻¹⁵ m

Step 2: Coulomb potential energy
\( V = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d} \)
Each deuteron has charge +e
\( V = \frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{4.0\times 10^{-15}} \)
Numerator: 9×10⁹ × 2.56×10⁻³⁸ = 2.304×10⁻²⁸
V = 2.304×10⁻²⁸ / 4.0×10⁻¹⁵ = 5.76×10⁻¹⁴ J

Step 3: Convert to MeV
1 eV = 1.6×10⁻¹⁹ J
V = 5.76×10⁻¹⁴ / 1.6×10⁻¹⁹ = 3.6×10⁵ eV = 0.36 MeV

Answer: Potential barrier height = 0.36 MeV

Question 13.21

From the relation \( R = R_0A^{1/3} \), 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 \( A \)).

Answer & Explanation:

Step 1: Volume of nucleus
Assuming spherical nucleus: \( V = \frac{4}{3}\pi R^3 \)
But \( R = R_0 A^{1/3} \)
So \( V = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A \)

Step 2: Mass of nucleus
Mass ≈ A × mass of nucleon ≈ A × mₙ
(where mₙ ≈ 1.67×10⁻²⁷ kg)

Step 3: Density
\( \rho = \frac{\text{Mass}}{V} = \frac{A m_n}{\frac{4}{3}\pi R_0^3 A} = \frac{3m_n}{4\pi R_0^3} \)
The A cancels out!

Step 4: Numerical value
\( R_0 \approx 1.2 \text{ fm} = 1.2\times 10^{-15} \text{ m} \)
\( \rho = \frac{3 \times 1.67\times 10^{-27}}{4\pi (1.2\times 10^{-15})^3} \)
Denominator: 4π × (1.728×10⁻⁴⁵) = 2.171×10⁻⁴⁴
ρ = 5.01×10⁻²⁷ / 2.171×10⁻⁴⁴ = 2.31×10¹⁷ kg/m³

Conclusion: Density is constant (~2.3×10¹⁷ kg/m³) for all nuclei.

Question 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). Show that if \( \beta^+ \) emission is energetically allowed, electron capture is necessarily allowed but not vice-versa.

Answer & Explanation:

Step 1: Energy conditions for β⁺ emission
For β⁺: \( ^{A}_{Z}X \rightarrow ^{A}_{Z-1}Y + e^+ + \nu_e \)
Q-value: \( Q_{\beta^+} = [m_X - m_Y - 2m_e]c^2 > 0 \)
Condition: \( m_X - m_Y > 2m_e \)

Step 2: Energy conditions for electron capture (EC)
For EC: \( ^{A}_{Z}X + e^- \rightarrow ^{A}_{Z-1}Y + \nu_e \)
Q-value: \( Q_{EC} = [m_X + m_e - m_Y]c^2 - B_e \)
where B_e is binding energy of captured electron (∼few keV for K-shell)
For EC to occur: \( m_X + m_e - m_Y > B_e/c^2 \)
Since B_e is small: \( m_X - m_Y > -m_e \) (approximately)

Step 3: Comparison
β⁺ condition: \( m_X - m_Y > 2m_e \) (strict)
EC condition: \( m_X - m_Y > -m_e \) (much easier)

Step 4: Logical implications
If β⁺ is allowed: \( m_X - m_Y > 2m_e \)
Then certainly \( m_X - m_Y > -m_e \)
So EC is also allowed.

But if EC is allowed: \( m_X - m_Y > -m_e \)
This does NOT guarantee \( m_X - m_Y > 2m_e \)
So β⁺ may not be allowed even if EC is allowed.

Conclusion: β⁺ emission implies EC is possible, but EC can occur when β⁺ is not energetically possible.

Question 13.23

In a periodic table the average atomic mass of magnesium is given as 24.312 u. The three isotopes and their masses are \( ^{24}_{12}Mg \) (23.98504u), \( ^{25}_{12}Mg \) (24.98584u) and \( ^{26}_{12}Mg \) (25.98259u). The natural abundance of \( ^{24}_{12}Mg \) is 78.99% by mass. Calculate the abundances of the other two isotopes.

Answer & Explanation:

Step 1: Set up equations
Let abundances: \( ^{24}Mg \): 78.99%
\( ^{25}Mg \): x%
\( ^{26}Mg \): y%
Then x + y = 100 - 78.99 = 21.01 ...(1)

Step 2: Average mass equation
\( \frac{78.99 \times 23.98504 + x \times 24.98584 + y \times 25.98259}{100} = 24.312 \)
Multiply by 100: 78.99×23.98504 + 24.98584x + 25.98259y = 2431.2

Step 3: Compute first term
78.99×23.98504 = 1894.578 (approximately)
So: 1894.578 + 24.98584x + 25.98259y = 2431.2
24.98584x + 25.98259y = 536.622 ...(2)

Step 4: Solve equations
From (1): y = 21.01 - x
Substitute in (2): 24.98584x + 25.98259(21.01 - x) = 536.622
24.98584x + 545.912 - 25.98259x = 536.622
-0.99675x = 536.622 - 545.912 = -9.29
x = 9.29/0.99675 = 9.32%
y = 21.01 - 9.32 = 11.69%

Answer:
\( ^{25}Mg \) abundance ≈ 9.32%
\( ^{26}Mg \) abundance ≈ 11.69%

Question 13.24

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei \( ^{41}_{20}Ca \) and \( ^{27}_{13}Al \) from the following data:
\( m(^{40}_{20}Ca) = 39.962591 \, u \)
\( m(^{41}_{20}Ca) = 40.962278 \, u \)
\( m(^{26}_{13}Al) = 25.986895 \, u \)
\( m(^{27}_{13}Al) = 26.981541 \, u \)

Answer & Explanation:

Step 1: Neutron separation energy formula
For nucleus \( ^{A}_{Z}X \): Sₙ = [m(\( ^{A-1}_{Z}X \)) + mₙ - m(\( ^{A}_{Z}X \))]c²
where mₙ = 1.008665 u

Step 2: For \( ^{41}Ca \)**
Sₙ = [m(⁴⁰Ca) + mₙ - m(⁴¹Ca)] × 931.5
Δm = 39.962591 + 1.008665 - 40.962278 = 0.008978 u
Sₙ = 0.008978 × 931.5 = 8.362 MeV

Step 3: For \( ^{27}Al \)**
Sₙ = [m(²⁶Al) + mₙ - m(²⁷Al)] × 931.5
Δm = 25.986895 + 1.008665 - 26.981541 = 0.014019 u
Sₙ = 0.014019 × 931.5 = 13.06 MeV

Answers:
Neutron separation energy for \( ^{41}Ca \) = 8.36 MeV
Neutron separation energy for \( ^{27}Al \) = 13.06 MeV

Question 13.25

A source contains two phosphorous radio nuclides \( ^{32}_{15}P \) (\( T_{1/2} = 14.3d \)) and \( ^{33}_{15}P \) (\( T_{1/2} = 25.3d \)). Initially, 10% of the decays come from \( ^{33}_{15}P \). How long one must wait until 90% do so?

Answer & Explanation:

Step 1: Let initial numbers be N₁₀ and N₂₀
Initially: Activity from \( ^{33}P \) = 10% of total
So: \( \lambda_2 N_{20} = 0.1 (\lambda_1 N_{10} + \lambda_2 N_{20}) \)
⇒ \( 0.9 \lambda_2 N_{20} = 0.1 \lambda_1 N_{10} \)
⇒ \( \frac{N_{10}}{N_{20}} = 9 \frac{\lambda_2}{\lambda_1} \) ...(1)

Step 2: After time t
We want: \( \lambda_2 N_2 = 0.9 (\lambda_1 N_1 + \lambda_2 N_2) \)
⇒ \( 0.1 \lambda_2 N_2 = 0.9 \lambda_1 N_1 \)
⇒ \( \frac{N_1}{N_2} = \frac{1}{9} \frac{\lambda_2}{\lambda_1} \) ...(2)

Step 3: Decay equations
\( N_1 = N_{10} e^{-\lambda_1 t} \), \( N_2 = N_{20} e^{-\lambda_2 t} \)
From (1) and (2):
\( \frac{N_{10} e^{-\lambda_1 t}}{N_{20} e^{-\lambda_2 t}} = \frac{1}{9} \frac{\lambda_2}{\lambda_1} \)
But \( \frac{N_{10}}{N_{20}} = 9 \frac{\lambda_2}{\lambda_1} \)
So: \( 9 \frac{\lambda_2}{\lambda_1} e^{-(\lambda_1 - \lambda_2)t} = \frac{1}{9} \frac{\lambda_2}{\lambda_1} \)
Cancel \( \frac{\lambda_2}{\lambda_1} \): \( 9 e^{-(\lambda_1 - \lambda_2)t} = 1/9 \)
\( e^{-(\lambda_1 - \lambda_2)t} = 1/81 \)

Step 4: Solve for t
\( (\lambda_1 - \lambda_2)t = \ln 81 = 4.394 \)
\( \lambda_1 = \frac{\ln 2}{14.3} = 0.04847 \text{ d}^{-1} \)
\( \lambda_2 = \frac{\ln 2}{25.3} = 0.02740 \text{ d}^{-1} \)
\( \lambda_1 - \lambda_2 = 0.02107 \text{ d}^{-1} \)
\( t = \frac{4.394}{0.02107} = 208.5 \text{ days} \)

Answer: Need to wait about 209 days

Question 13.26

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes:
\( ^{223}_{88}Ra \rightarrow ^{209}_{82}Pb + ^{14}_{6}C \)
\( ^{223}_{88}Ra \rightarrow ^{219}_{86}Rn + ^{4}_{2}He \)
Calculate the Q-values for these decays and determine that both are energetically allowed.

Answer & Explanation:

Step 1: Need masses
Given in question: need m(²²³Ra), m(²⁰⁹Pb), m(¹⁴C), m(²¹⁹Rn), m(⁴He)
Standard values if not given:
m(²²³Ra) = 223.01850 u (approx)
m(²⁰⁹Pb) = 208.98107 u
m(¹⁴C) = 14.00324 u
m(²¹⁹Rn) = 219.00948 u
m(⁴He) = 4.002603 u

Step 2: Q for \( Ra \rightarrow Pb + C \)**
Q = [m(²²³Ra) - m(²⁰⁹Pb) - m(¹⁴C)] × 931.5
Δm = 223.01850 - 208.98107 - 14.00324 = 0.03419 u
Q = 0.03419 × 931.5 = 31.85 MeV
Positive → allowed

Step 3: Q for \( Ra \rightarrow Rn + He \)**
Q = [m(²²³Ra) - m(²¹⁹Rn) - m(⁴He)] × 931.5
Δm = 223.01850 - 219.00948 - 4.002603 = 0.006417 u
Q = 0.006417 × 931.5 = 5.977 MeV
Positive → allowed

Answer:
Q for C-14 emission = 31.8 MeV (allowed)
Q for α-decay = 5.98 MeV (allowed)
Both are energetically possible, though α-decay is more probable.

Question 13.27

Consider the fission of \( ^{238}_{92}U \) by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are \( ^{140}_{58}Ce \) and \( ^{99}_{44}Ru \). Calculate Q for this fission process. The relevant atomic and particle masses are:
\( m(^{238}_{92}U) = 238.05079 \, u \)
\( m(^{140}_{58}Ce) = 139.90543 \, u \)
\( m(^{99}_{44}Ru) = 98.90594 \, u \)

Answer & Explanation:

Step 1: Check nucleon conservation
238U → 140Ce + 99Ru → total mass number 239? Wait, that's 239 not 238.
The neutron is absorbed: n + 238U → fission products
So initial: 238U + n
Final: 140Ce + 99Ru
Mass numbers: 238+1 = 239, 140+99 = 239 ✓

Step 2: Q-value calculation
Q = [m(²³⁸U) + mₙ - m(¹⁴⁰Ce) - m(⁹⁹Ru)] × 931.5
mₙ = 1.008665 u
Δm = 238.05079 + 1.008665 - 139.90543 - 98.90594
= 239.059455 - 238.81137 = 0.248085 u
Q = 0.248085 × 931.5 = 231.1 MeV

Step 3: Interpretation
This is typical fission energy release.
Note: Actual fission releases extra neutrons which aren't counted here.

Answer: Q ≈ 231 MeV per fission

Question 13.28

Consider the D-T reaction (deuterium-tritium fusion):
\( ^{2}_{1}H + ^{3}_{1}H \rightarrow ^{4}_{2}He + n \)
(a) Calculate the energy released in MeV in this reaction from the data:
\( m(^{2}_{1}H) = 2.014102 \, u \), \( m(^{3}_{1}H) = 3.016049 \, u \)
(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

Answer & Explanation:

Step 1: (a) Energy released
Need m(⁴He) = 4.002603 u, mₙ = 1.008665 u
Q = [m(²H) + m(³H) - m(⁴He) - mₙ] × 931.5
Δm = 2.014102 + 3.016049 - 4.002603 - 1.008665
= 5.030151 - 5.011268 = 0.018883 u
Q = 0.018883 × 931.5 = 17.59 MeV

Step 2: (b) Coulomb barrier
Similar to Q13.20: V = \( \frac{1}{4\pi\epsilon_0} \frac{e^2}{2r} \)
r = 2.0 fm, distance between centers when touching = 4.0 fm
V = \( \frac{9\times 10^9 \times (1.6\times 10^{-19})^2}{4.0\times 10^{-15}} \)
= 5.76×10⁻¹⁴ J = 0.36 MeV (same as before)

Step 3: Required temperature
Average thermal energy = \( \frac{3}{2} kT \) per particle
For two particles colliding: total energy ≈ 3kT
Need 3kT ≈ 0.36 MeV = 5.76×10⁻¹⁴ J
k = 1.38×10⁻²³ J/K
T = \( \frac{5.76\times 10^{-14}}{3 \times 1.38\times 10^{-23}} \) = 1.39×10⁹ K

Step 4: Reality check
Actual fusion occurs at lower temperatures (~10⁷ K) due to:
• Quantum tunneling
• Maxwellian tail of distribution
• Some particles have energy >> average

Answers:
(a) Energy released = 17.6 MeV
(b) Coulomb barrier = 0.36 MeV, required T ≈ 1.4×10⁹ K (classically)

Question 13.29

Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that:
\( m(^{198}_{79}Au) = 197.968233 \, u \)
\( m(^{198}_{80}Hg) = 197.966760 \, u \)

Answer & Explanation:

Step 1: Understand decay scheme (from Fig 13.6)
¹⁹⁸Au decays to ¹⁹⁸Hg via β⁻ decay with two possible paths:
1. Direct to ground state (Q₁)
2. To excited state at 0.412 MeV, then γ to ground state

Step 2: Total Q-value
For β⁻ decay: Q = [m(¹⁹⁸Au) - m(¹⁹⁸Hg)] × 931.5
Δm = 197.968233 - 197.966760 = 0.001473 u
Q = 0.001473 × 931.5 = 1.372 MeV

Step 3: Maximum β energy for direct transition
If β goes directly to ground state: \( E_{\beta,\max} = Q = 1.372 \text{ MeV} \)

Step 4: For transition to excited state
Excited state at 0.412 MeV above ground state.
So available energy for β = Q - 0.412 = 1.372 - 0.412 = 0.960 MeV
This is maximum β energy for that branch.

Step 5: γ-ray frequency
γ energy = 0.412 MeV
\( \nu = \frac{E}{h} = \frac{0.412 \times 1.6\times 10^{-13}}{6.626\times 10^{-34}} \)
= \( \frac{6.592\times 10^{-14}}{6.626\times 10^{-34}} = 9.95\times 10^{19} \text{ Hz} \)

Answers:
Maximum β energies: 1.372 MeV (direct) and 0.960 MeV (via excited state)
γ frequency ≈ 1.0×10²⁰ Hz

Question 13.30

Calculate and compare the energy released by:
(a) fusion of 1.0 kg of hydrogen deep within Sun, and
(b) the fission of 1.0 kg of \( ^{235}_{92}U \) in a fission reactor.

Answer & Explanation:

Step 1: (a) Fusion in Sun (p-p chain)
Overall: 4¹H → ⁴He + 2e⁺ + 2ν + energy
Net: 4 protons → 1 helium nucleus
Mass defect: 4×1.007825 - 4.002603 = 0.028697 u per 4 protons
Energy per He = 0.028697 × 931.5 = 26.73 MeV
Per proton: 26.73/4 = 6.68 MeV

Step 2: Energy from 1 kg hydrogen
Number of H atoms in 1 kg = 1000/1.0079 × 6.022×10²³ ≈ 5.975×10²⁶
Total energy = 5.975×10²⁶ × 6.68 × 1.6×10⁻¹³ = 6.39×10¹⁴ J

Step 3: (b) Fission of 1 kg \( ^{235}U \)**
From earlier: 200 MeV per fission
Number of atoms in 1 kg = 1000/235 × 6.022×10²³ = 2.56×10²⁴
Total energy = 2.56×10²⁴ × 200 × 1.6×10⁻¹³ = 8.19×10¹³ J

Step 4: Comparison
Fusion: 6.39×10¹⁴ J
Fission: 8.19×10¹³ J
Ratio: Fusion/Fission ≈ 7.8
So fusion releases about 8 times more energy per kg.

Answers:
(a) Fusion of 1 kg H: 6.4×10¹⁴ J
(b) Fission of 1 kg U-235: 8.2×10¹³ J
Fusion releases ~8× more energy per kg.

Question 13.31

Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of \( ^{235}_{92}U \) to be about 200 MeV.

Answer & Explanation:

Step 1: Nuclear power capacity
Total electric power target = 200,000 MW = 2×10⁵ MW = 2×10¹¹ W
10% from nuclear = 0.1 × 2×10¹¹ = 2×10¹⁰ W electric

Step 2: Thermal power required
Efficiency = 25% = 0.25
Thermal power = Electric power / efficiency = 2×10¹⁰ / 0.25 = 8×10¹⁰ W thermal

Step 3: Energy per year
1 year = 3.154×10⁷ s
Thermal energy per year = 8×10¹⁰ × 3.154×10⁷ = 2.523×10¹⁸ J/year

Step 4: Number of fissions required
Energy per fission = 200 MeV = 200 × 1.6×10⁻¹³ = 3.2×10⁻¹¹ J
Number of fissions/year = 2.523×10¹⁸ / 3.2×10⁻¹¹ = 7.884×10²⁸ fissions/year

Step 5: Mass of \( ^{235}U \) required
Atoms per mole = 6.022×10²³
Moles per year = 7.884×10²⁸ / 6.022×10²³ = 1.309×10⁵ mol/year
Molar mass = 235 g/mol
Mass per year = 1.309×10⁵ × 235 = 3.076×10⁷ g/year = 30,760 kg/year ≈ 30.8 tonnes/year

Step 6: Realistic considerations
Actual reactors use enriched uranium (3-5% ²³⁵U), so natural uranium needed would be ~20× more.

Answer: About 31 tonnes of fissionable \( ^{235}U \) per year.

📘 Exam Preparation Tip:

These exercise questions will help you understand nuclear structure and radioactive processes. You'll learn to calculate nuclear size, density, and binding energy, solve problems on radioactive decay (alpha, beta, gamma), apply laws of radioactive disintegration, understand nuclear reactions and energy release, and analyze mass-energy equivalence in nuclear processes. Crucial for nuclear physics and energy applications.

Class 12: NCERT Solutions

Chapter 01: Electric Charges and Fields Chapter 02: Electrostatic Potential and Capacitance Chapter 03: Current Electricity
Chapter 04: Moving Charges and Magnetism Chapter 05: Magnetism and Matter Chapter 06: Electromagnetic Induction
Chapter 07: Alternating Current Chapter 08: Electromagnetic Waves Chapter 09: Ray Optics and Optical Instruments
Chapter 10: Wave Optics Chapter 11: Dual Nature of Radiation and Matter Chapter 12: Atoms
Chapter 13: Nuclei Chapter 14: Semiconductor Electronics Chapter 15: Communication Systems

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