Nuclei is a conceptual and high-weightage chapter in Class 12 Physics that explains the structure and properties of atomic nuclei. This chapter covers important topics such as nuclear size and density, mass–energy equivalence, binding energy and binding energy per nucleon, nuclear forces, radioactivity, radioactive decay law, half-life and mean life, nuclear fission and fusion, which are frequently asked in CBSE board exams and competitive exams like JEE and NEET.
NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear derivations, graphs, and solved numericals, helping students understand nuclear physics concepts clearly and score well in board examinations.
NCERT Solutions for Class 12 Physics Chapter 13 – Nuclei
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Calculate and compare the energy released by(a) fusion of1.0 kg of hydrogen deep within the sun and (b) the fissionof 1.0 kg of 235Uin a fission reactor.
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(a) Consider the D-T reaction (deuterium-tritium–fusion) given in eqn:12H+13H–>24He+01n + QCalculate the energy released in MeV in this reaction from the data:m (12H) = 2.014102 u; m (13H) = 3.016049 u; m (24He) = 4.002603 u; mn= 1.00867 u(b) Consider the radius of both deuterium and tritium to be approximately2.0 fm. What is the kinetic energy needed to overcome the Coulombrepulsion between the two nuclei? To what temperature must the gasesbe heated to initiate the reaction?
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A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to
(a) 3.125 % of its original activity (b) 1% of original value ?
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Suppose India had a target of producing by 2020 AD, 2 × 105 MW of electric power, ten percent of which was to be obtained from nuclear power plant. 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 did our country need per year by 2020? Take the heat energy per fission of U235 to be about 200 MeV.
Q. Suppose India had a target of producing by 2020 AD, 2 × 105 MW of electric power, ten percent of which was to be obtained from nuclear power plant. 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 did our country need per year by 2020? Take the heat energy per fission of U235 to be about 200 MeV.
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Total power target = 2 × 105 MW = 2 × 1011 W
Nuclear power target is 10% of total power target = 2 × 1010 W
As efficiency is defined as,
η = (Total useful power) / (Total power generated)
Hence Total power generated = (Total useful power) / η
= (2 × 1010) / 0.25 = 8 × 1010 W
Total energy required for the year 2020 is
E = P × t = 8 × 1010 × 366 × 24 × 60 × 60
(as 2020 is a leap year)
E = 2.265 × 1018 J
1 fission of U92235 produces 200 MeV of energy
= 200 × 1.6 × 10−13 J = 3.2 × 10−11 J
Hence, no. of fissions required for the generation of energy E
= (2.265 × 1018) / (3.2 × 10−11)
= 7.90 × 1028
As 6.023 × 1023 atoms of U92235 have mass = 235 g
Thus mass required to produce 7.90 × 1028 atoms
= (235 × 7.90 × 1028) / (6.023 × 1023) g
≃ 3.08 × 104 kg
Hence, mass of uranium needed per year ≃ 3.08 × 104 kg.
subject: Physics | sub subject: Physics : Part - II | chapter: Nuclei | Difficulty Level: Medium
Q. (a) Consider the D-T reaction (deuterium-tritium–fusion) given in eqn:
12H + 13H → 24He + 01n + Q
Calculate the energy released in MeV in this reaction from the data:
m(12H) = 2.014102 u; m(13H) = 3.016049 u;
m(24He) = 4.002603 u; mn = 1.00867 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 gases be heated to initiate the reaction?
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(a) From the equation given in the question
Q = [mN(12H) + mN(13H) − mN(24He) − mn] × 931 MeV
If m represents the atomic mass, then
mN(12H) = m(12H) − me
mN(13H) = m(13H) − me
mN(24He) = m(24He) − 2me
Q = [2.014102 + 3.016049 − 4.002603 − 1.00867] × 931 MeV
= 17.575 MeV = 17.58 MeV.
(b) Repulsive potential energy between two nuclei is
= (1 / 4π ε0) × (q1q2) / (2r)
= (1 / 4π ε0) × e2 / (2r)
Thus,
P.E = [9×109(1.6×10−19)2] / [2×2×10−15] J
= 5.76×10−14 J
As repulsive P.E = K.E and K.E = 2(3kT/2) = 3kT
Hence,
T = K.E / (3k) = (5.76×10−14) / (3×1.38×10−23)
T = 1.39×109 K
subject: Physics | sub subject: Physics : Part - II | chapter: Nuclei | Difficulty Level: Medium
Q. Obtain the maximum kinetic energy of β particle and the radiation frequencies of decays in the following decay scheme shown in the fig 13.6. You are given that mAu198 = 197.968233 u, mHg198 = 197.966760 u.
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For γ1, the frequency
ν1 = E1/h = (1.088×1.6×10−13) / (6.626×10−34)
ν1 = 2.63×1020 Hz
For γ2, the frequency
ν2 = E2/h = (0.412×1.6×10−13) / (6.626×10−34)
ν2 = 9.95×1019 Hz
For γ3, the energy
E3 = 1.088 − 0.412 = 0.676 MeV
= 0.676×1.6×10−13 J = 1.082×10−13 J
ν3 = E3/h = (1.082×10−13) / (6.626×10−34)
ν3 = 1.63×1020 Hz
For β1− decay, maximum kinetic energy
= (mAu − mHg)×931 − 1.088 MeV
= (197.968233 − 197.966760)×931 − 1.088
= 1.372 − 1.088 = 0.283 MeV
For β2− decay, maximum kinetic energy
= (mAu − mHg)×931 − 0.412 MeV
= 1.372 − 0.412 = 0.96 MeV
subject: Physics | sub subject: Physics : Part - II | chapter: Nuclei | Difficulty Level: Medium
Q. Calculate and compare the energy released by (a) fusion of 1.0 kg of hydrogen deep within the sun and (b) the fission of 1.0 kg of 235U in a fission reactor.
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(a) In sun fusion takes place according to the equation
411H → 24He + 2+10e + 26 MeV
So, 4 hydrogen atoms combine to produce 26 MeV of energy.
1 g of hydrogen contains 6.02×1023 atoms
Hence, 1000 g contains 6.02×1026 atoms
Energy released
E = (26 MeV × 6.02×1026) / 4 = 39×1026 MeV
(b) Fission of one 235U nucleus gives energy = 200 MeV
235 g of 235U has 6.02×1023 atoms
1 kg (1000 g) has (6.02×1023 × 1000) / 235 = 2.56×1024 atoms
Total energy released
E′ = 200 × 2.56×1024 = 5.1×1026 MeV
For comparison:
E / E′ = (3.913×1027) / (5.12×1026) = 7.64
subject: Physics | sub subject: Physics : Part - II | chapter: Nuclei | Difficulty Level: Medium
Q. The fission properties of Pu are very similar to those of U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all atoms in 1 kg of pure Pu239 undergo fission?
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Number of atoms present in 239 g of Pu = 6.023×1023
Number of atoms present in 1000 g of Pu = (6.023×1023 × 1000) / 239
= 2.52×1024 atoms
Average energy released per fission = 180 MeV
Total energy released, Q = 180 × 2.52×1024 MeV
Q = 4.536×1026 MeV
subject: Physics | sub subject: Physics : Part - II | chapter: Nuclei | Difficulty Level: Medium
Q.The three stable isotopes of neon: 2010Ne, 2110Ne and 2210Ne have respective abundance of 90.51%, 0.27% and 9.22%. The atomic masses are 19.99 u, 20.99 u and 21.99 u. Obtain the average atomic mass of neon.
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Average atomic mass = (90.51 × 19.99 + 0.27 × 20.99 + 9.22 × 21.99) / 100
= (1809.29 + 5.67 + 202.75) / 100
= 20.18 u
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FAQs: Class 12 Physics Chapter 13 – Nuclei
Q1. Is Nuclei an important chapter for exams?
Yes, it is a high-weightage chapter for CBSE, JEE, and NEET.
Q2. Which topics are most important in this chapter?
Binding energy, radioactive decay law, half-life, and nuclear fission & fusion.
Q3. Are numericals asked from this chapter?
Yes, radioactive decay and binding energy numericals are very common.
Q4. Are derivations important here?
Yes, derivations related to radioactive decay law and mass–energy relation are frequently asked.
Q5. How do NCERT Solutions help?
They provide NCERT-aligned, exam-ready explanations with solved numericals and graphs.