Key Fact
Nuclear fission
Heavy nucleus splits into lighter fragments + neutrons + energy. Example: ²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n + ~200 MeV. Basis of atomic energy and nuclear weapons.
-- NCERT Class 12 Physics, Ch. 13, p. 315Nuclear Fission
8 MCQs2 revision cards9-step worked example
Source: NCERT Atoms and NucleiPYQ coverage: NEET 2020, 2022Official key: NTA-verifiedLast updated: 27 Sep 2026
Nuclear Fission, explained for NEET
The confusion this topic punishes is the conceptual swap: fission and fusion both release energy, so it is easy to remember "energy out" and forget which direction the mass number moves. Fission is a heavy nucleus breaking into two lighter fragments. If the question describes light nuclei joining, it is not fission, whatever the energy figure says.
NCERT Class 12 Physics, Chapter 13 (page 315) records the canonical case: a slow neutron absorbed by ²³⁵U produces a compound nucleus that splits into two intermediate-mass fragments plus a few neutrons, releasing about 200 MeV per event. That number is worth holding, because NEET stems frequently give it and ask for reactor power or events per second.
Why energy comes out is settled by the binding-energy-per-nucleon curve, not by any special property of uranium. Uranium sits near 7.6 MeV per nucleon; the fragments land in the A ≈ 100–140 region, near 8.5 MeV per nucleon. Every nucleon in the products is more tightly bound than it was, and that surplus binding appears as kinetic energy of the fragments. Total binding energy increases; total rest mass decreases. The energy released equals the mass defect of the reaction converted through E = mc², with 1 u worth 931.5 MeV.
Two details separate the careless mark from the safe one. First, the released energy is overwhelmingly the kinetic energy of the two fragments, not gamma radiation. Second, fission liberates 2–3 free neutrons, and that is the whole basis of the chain reaction — not an incidental by-product.
Watch out: when a stem gives you the binding energy per nucleon of parent and fragments, the energy released is the difference in total binding energy (B_products − B_parent), not the difference per nucleon. Multiply by the nucleon counts before you subtract.
Can you answer these Nuclear Fission MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Nuclear fission is the process in which
Show answer and why every option is right or wrong
Answer: B. Fission is the splitting of a heavy nucleus into two fragments of roughly comparable, intermediate mass, as described in NCERT Class 12 Physics, Chapter 13, page 315.
Why A is wrong: A is wrong because combining light nuclei into a heavier one is fusion, the opposite process; both release energy, which is exactly why they get swapped.
Why C is wrong: C is wrong because alpha emission is radioactive decay, which removes a small fixed fragment (a helium nucleus) rather than splitting the nucleus into two comparable parts.
Why D is wrong: D is wrong because that describes electron capture, a beta-type decay in which the mass number does not change at all.
MCQ 2Easy RecallPractice
In the fission of a ²³⁵U nucleus by a slow neutron, the number of free neutrons typically liberated is
Show answer and why every option is right or wrong
Answer: A. Each fission event releases on average 2 to 3 neutrons, which is what makes a self-sustaining chain reaction possible (NCERT Class 12 Physics, Chapter 13, page 315).
Why B is wrong: B is wrong because a single emitted neutron would only replace the one absorbed, giving no multiplication and therefore no possibility of a growing chain reaction.
Why C is wrong: C is wrong because with no neutrons emitted the reaction could not propagate at all, contradicting the entire basis of reactor operation.
Why D is wrong: D is wrong because it overstates the yield by roughly a factor of four; the fragments retain almost all the nucleons.
MCQ 3Easy RecallPractice
The energy released in a single fission event of ²³⁵U is approximately
Show answer and why every option is right or wrong
Answer: C. A single ²³⁵U fission releases about 200 MeV, the standard figure quoted in NCERT Class 12 Physics, Chapter 13, page 315.
Why A is wrong: A is wrong because it is the right number with the wrong unit prefix: 200 eV is a millionfold too small and is a chemical-scale, not nuclear-scale, energy.
Why B is wrong: B is wrong because roughly 8 MeV is the binding energy per nucleon, not the energy released by the whole fission event.
Why D is wrong: D is wrong because 13.6 eV is the ionisation energy of hydrogen, an atomic quantity that has nothing to do with nuclear fission.
MCQ 4Direct ApplicationPractice
A nucleus of mass number 90 has a binding energy per nucleon of 8.70 MeV. Its total binding energy is closest to
Show answer and why every option is right or wrong
Answer: A. Total binding energy = (binding energy per nucleon) × (mass number) = 8.70 × 90 = 783 MeV = 7.83 × 10² MeV, a single application of B = (B/A) × A (NCERT Class 12 Physics, Chapter 13, page 315).
Why B is wrong: B is wrong because it divides 8.70 by 90 (giving 0.0967 MeV) instead of multiplying, inverting the relationship between per-nucleon and total binding energy.
Why C is wrong: C is wrong because it reports the per-nucleon value unchanged, as though multiplying by the mass number were unnecessary.
Why D is wrong: D is wrong because it multiplies by 900 instead of 90, misplacing a decimal in the mass number.
MCQ 5Direct ApplicationPractice
In a fission reaction the total rest mass of the products is less than that of the reactants by 0.2100 u. Taking 1 u ≡ 931.5 MeV, the energy released is closest to
Show answer and why every option is right or wrong
Answer: A. E = Δm × 931.5 MeV/u = 0.2100 × 931.5 ≈ 195.6 MeV, i.e. about 1.96 × 10² MeV (NCERT Class 12 Physics, Chapter 13, page 315).
Why B is wrong: B is wrong because it divides 931.5 by 0.2100 instead of multiplying, inverting the conversion.
Why C is wrong: C is wrong because it is the energy equivalent of 1 u itself; the mass defect of 0.2100 u was never applied.
Why D is wrong: D is wrong because it quotes the mass defect in u as though it were already an energy in MeV, skipping the conversion entirely.
MCQ 6Direct ApplicationPractice
A reactor produces 3.00 × 10⁸ W of thermal power. Taking the energy released per fission as 2.00 × 10² MeV and 1 MeV = 1.60 × 10⁻¹³ J, the number of fission events per second is closest to
Show answer and why every option is right or wrong
Answer: C. Energy per fission = 2.00 × 10² × 1.60 × 10⁻¹³ = 3.20 × 10⁻¹¹ J; rate = 3.00 × 10⁸ ÷ 3.20 × 10⁻¹¹ ≈ 9.38 × 10¹⁸ per second (NCERT Class 12 Physics, Chapter 13, page 315).
Why A is wrong: A is wrong because it is 1000 times too small, as if the power had been converted to kilowatts (3.00 × 10⁵ kW) before dividing by the energy per fission in joules.
Why B is wrong: B is wrong because it divides the power by 2.00 × 10² directly, treating the per-fission energy as though it were already in joules.
Why D is wrong: D is wrong because it stops at the energy released per fission in joules and never divides the power by it.
MCQ 7CalculationPractice
A nucleus of mass number 236 with binding energy per nucleon 7.60 MeV fissions into two fragments, each of mass number 118 and binding energy per nucleon 8.50 MeV. The energy released is
Show answer and why every option is right or wrong
Answer: B. Total binding energy of products = 236 × 8.50 = 2006 MeV; of the parent = 236 × 7.60 = 1793.6 MeV; the released energy is the increase, 2006 − 1793.6 ≈ 2.12 × 10² MeV (NCERT Class 12 Physics, Chapter 13, page 315).
Why A is wrong: A is wrong because it subtracts the binding energies per nucleon (8.50 − 7.60) without multiplying by the 236 nucleons involved.
Why C is wrong: C is wrong because it reports the parent's total binding energy instead of the difference between products and parent.
Why D is wrong: D is wrong because it adds the two total binding energies rather than subtracting them, which no conservation argument supports.
MCQ 8CalculationPractice
A slow neutron is absorbed by ²³⁵₉₂U, forming a compound nucleus that splits into ¹⁴¹₅₆Ba, a second fragment, and three free neutrons. The second fragment is
Show answer and why every option is right or wrong
Answer: A. Mass number: 235 + 1 = 236, so the fragment has A = 236 − 141 − 3 = 92; charge: 92 + 0 = 92, so Z = 92 − 56 = 36, giving ⁹²₃₆Kr (NCERT Class 12 Physics, Chapter 13, page 315).
Why B is wrong: B is wrong because it omits the three emitted neutrons from the mass-number balance, leaving A too large by 3.
Why C is wrong: C is wrong because its atomic number 34 does not balance the charge: 56 + 34 = 90, not the required 92.
Why D is wrong: D is wrong because it subtracts the three neutrons twice over, giving A = 89 instead of 92.
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Nuclear Fission: quick recall before you leave
How do you solve a Nuclear Fission question? A worked example
- 1
Given
A ²³⁵U nucleus absorbs a slow neutron and undergoes fission. The total rest mass of the reactants exceeds that of the products by Δm = 0.2150 u.
Conversion factor: 1 u ≡ 931.5 MeV (exact by definition of the conversion used in NEET).
A reactor running on this reaction delivers a thermal power of 2.00 × 10⁹ W.
1 MeV = 1.60 × 10⁻¹³ J. - 2
Required
(a) The energy released per fission event, in MeV.
(b) The number of fission events occurring per second. - 3
Concept
Fission converts a heavy nucleus into two intermediate-mass fragments. The products are collectively lighter than the reactants; that missing rest mass reappears as energy, almost all of it as kinetic energy of the fragments. Power is energy per unit time, so dividing the reactor's power by the energy of one event gives the event rate.
- 4
Formula
E = Δm·c², used in the practical form E (MeV) = Δm (u) × 931.5.
Rate N = P / E_per_event, with E_per_event in joules. - 5
Substitution
E = 0.2150 × 931.5 MeV
E_joules = E × 1.60 × 10⁻¹³ J/MeV
N = (2.00 × 10⁹ W) / E_joules - 6
Calculation
E = 0.2150 × 931.5 = 200.27 MeV ≈ 2.00 × 10² MeV.
E_joules = 200.27 × 1.60 × 10⁻¹³ = 3.204 × 10⁻¹¹ J.
N = 2.00 × 10⁹ / 3.204 × 10⁻¹¹ = 6.242 × 10¹⁹ s⁻¹.
The factor 931.5 MeV/u is a defined conversion constant and 1.60 × 10⁻¹³ J/MeV is likewise a unit conversion; neither is a measured quantity, so neither limits the significant-figure count. The precision is set by Δm (4 s.f.) and P (3 s.f.). - 7
Final answer
(a) E ≈ 2.00 × 10² MeV per fission.
(b) N ≈ 6.24 × 10¹⁹ fissions per second, limited to three significant figures by the power. - 8
Common trap
The pattern's recorded distractor is the fission–fusion conceptual swap. A stem that says "two light nuclei combine, releasing energy" is fusion even though the arithmetic that follows is identical; read the direction of the mass number before reaching for the calculator. The second trap here is unit-mixing at step (b): dividing watts by an energy still expressed in MeV gives a number roughly 10¹³ times too small. Convert to joules first.
- 9
Similar NEET-style question
In a fission reaction the products are lighter than the reactants by 0.1900 u. A power station using this reaction operates at 5.00 × 10⁸ W thermal. Taking 1 u ≡ 931.5 MeV and 1 MeV = 1.60 × 10⁻¹³ J, find the energy released per event in MeV and the number of events per second. (Answer: ≈ 1.77 × 10² MeV; ≈ 1.77 × 10¹⁹ s⁻¹.)
What to remember before solving Nuclear Fission questions
More in Atoms and Nuclei: 3 exam traps and mistakes · 6 formulas · 4 question patterns from its other lessons.
Nuclear Fission questions from past NEET papers
2 questions from NEET 2020, 2022. Answers verified against NTA official keys. — click to collapse
NEET 2022
125 : 16
21 : 1
34 : 5
45 : 4
NTA Answer: Option 4(final)
NEET 2020
1103 36Kr
2144 56Ba
391 40Zr
4101 36Kr
NTA Answer: Option 2(final)
How does NEET ask about Nuclear Fission?
1 recurring pattern from past papers — click to collapse
Identify fission vs fusion; compute energy released from mass defect.
InterpretationMedium
Common distractors
confuses fission fusion
Conceptual swap
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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