Key Fact
Nuclear fusion
Light nuclei combine to form heavier nucleus + energy. Sun's energy from ⁴ ¹H → ⁴He + 2e⁺ + 2ν + 26.7 MeV. Requires extreme T to overcome Coulomb repulsion.
-- NCERT Class 12 Physics, Ch. 13, p. 316Nuclear Fusion
8 MCQs2 revision cards9-step worked example
Source: NCERT Atoms and NucleiPYQ coverage: NEET 2023Official key: NTA-verifiedLast updated: 27 Sep 2026
Nuclear Fusion, explained for NEET
Fusion questions punish one habit: reading "light nuclei combine" and stopping there. The exam asks why it needs a hundred million kelvin, and a student who only memorised the definition has nothing to say.
Two light nuclei carry positive charge. Bringing them within range of the strong nuclear force — a few femtometres — means pushing them through their mutual Coulomb repulsion. For two protons that barrier is of order 400 keV. Nothing supplies this at room temperature. Only thermal kinetic energy at temperatures around 10⁹ K gives a significant fraction of nuclei enough speed, which is why the process is called thermonuclear fusion (NCERT Class 12 Physics, Chapter 13, page 317).
Why energy comes out is the other half. On the binding-energy-per-nucleon curve, very light nuclei sit low. Fuse them and the product sits higher, so the products are more tightly bound and their combined rest mass is less than that of the reactants. That lost mass appears as kinetic energy of the products, E = Δm·c², with 1 u ≡ 931.5 MeV.
The proton–proton cycle powering the Sun nets four protons into one ⁴He nucleus plus two positrons and two neutrinos, releasing 26.7 MeV. The Sun's core, at about 1.5 × 10⁷ K, is cooler than the naive barrier estimate suggests it must be — the reaction proceeds anyway because of the high-energy tail of the speed distribution and the enormous number of nuclei available.
Watch out for two swaps. First, fusion releases more energy per unit mass than fission, but far less per event — a single fission gives ~200 MeV against fusion's ~26.7 MeV for the whole p–p cycle. Second, "light nuclei" means the rising part of the curve, up to about iron; past that peak, combining nuclei absorbs energy rather than releasing it.
Can you answer these Nuclear Fusion 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
Thermonuclear fusion requires extremely high temperature principally in order to
Show answer and why every option is right or wrong
Answer: A. A is correct. Both reacting nuclei are positively charged, so they repel; only large thermal kinetic energy brings them close enough for the short-range strong force to take over. This is why the process is called thermonuclear (NCERT Class 12 Physics, Chapter 13, page 317).
Why B is wrong: B is wrong because fusion joins intact nuclei; dismantling them into free nucleons would cost energy, not enable the reaction.
Why C is wrong: C is wrong because the strong nuclear force acts between nucleons, not electrons, and ionisation is incidental rather than the reason for the temperature requirement.
Why D is wrong: D is wrong because the mass defect of a given reaction is fixed by the rest masses of reactants and products; temperature changes the reaction rate, not the energy released per event.
MCQ 2Easy RecallPractice
In the proton–proton cycle that powers the Sun, the net result is the conversion of
Show answer and why every option is right or wrong
Answer: B. B is correct. The p–p cycle nets 4 ¹H → ⁴He + 2e⁺ + 2ν, liberating approximately 26.7 MeV in total (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because the first step alone (two protons to a deuteron) releases only a fraction of an MeV; 26.7 MeV is the figure for the complete cycle ending in ⁴He.
Why C is wrong: C is wrong because it reverses the reaction; the ⁴He nucleus is more tightly bound than four separate protons, so forming it releases energy.
Why D is wrong: D is wrong on both counts: the end product is ⁴He, not two separate deuterons, and ~200 MeV is the fission figure, not a fusion one.
MCQ 3Easy RecallPractice
For fusion of two nuclei to release energy, the product nucleus must have
Show answer and why every option is right or wrong
Answer: B. B is correct. The released energy is the rest-mass deficit converted by E = Δm·c²; a product lighter than the reactants combined is exactly the condition for exoergic fusion (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because nucleon number is conserved in fusion — the product's mass number is simply the sum, which says nothing about whether energy is released.
Why C is wrong: C is wrong because it is inverted: energy is released when the product is more tightly bound, i.e. has a larger binding energy per nucleon.
Why D is wrong: D is wrong because if rest masses balanced exactly there would be no mass defect and no net energy release; the input kinetic energy only gets the nuclei together.
MCQ 4Direct ApplicationPractice
Two ²H nuclei fuse to form ³He and a free neutron. The total rest mass of the products is less than that of the reactants by 3.50 × 10⁻³ u. Taking 1 u ≡ 931.5 MeV, the energy released is closest to
Show answer and why every option is right or wrong
Answer: C. C is correct. Q = Δm × 931.5 MeV/u = (3.50 × 10⁻³)(931.5) = 3.26 MeV (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because it multiplies the mass defect by 0.9315 rather than 931.5 — a factor-of-1000 slip in the conversion constant.
Why B is wrong: B is wrong because it divides by the mass defect instead of multiplying by it.
Why D is wrong: D is wrong because it quotes the conversion constant itself (931.5 MeV for 1 u) without scaling it by the actual mass defect.
MCQ 5Direct ApplicationPractice
The binding energy per nucleon of a ⁴He nucleus is 7.10 MeV. Its total binding energy is closest to
Show answer and why every option is right or wrong
Answer: B. B is correct. Total binding energy = (binding energy per nucleon) × (mass number) = 7.10 × 4 = 28.4 MeV, a single application of B = (B/A) × A (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because it divides 7.10 by 4 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 4 were unnecessary.
Why D is wrong: D is wrong because it multiplies by 40 instead of 4, misplacing a decimal in the mass number.
MCQ 6Direct ApplicationPractice
Two nuclei each of charge +e are brought to a centre-to-centre separation of 4.0 × 10⁻¹⁵ m, at which the strong force can act. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻² and e = 1.6 × 10⁻¹⁹ C, the electrostatic potential energy of the pair at this separation, expressed in keV, is closest to
Show answer and why every option is right or wrong
Answer: A. A is correct. U = (9.0 × 10⁹)(1.6 × 10⁻¹⁹)²/(4.0 × 10⁻¹⁵) = 5.76 × 10⁻¹⁴ J; dividing by 1.6 × 10⁻¹⁹ J/eV gives 3.6 × 10⁵ eV = 3.6 × 10² keV — the order-of-magnitude Coulomb barrier quoted for proton–proton fusion (NCERT Class 12 Physics, Chapter 13, page 317).
Why B is wrong: B is wrong because it reports the energy in joules while labelling it keV; the conversion to electron-volts has been omitted.
Why C is wrong: C is wrong because it gives the value in eV but labels it keV, a factor-of-1000 slip at the final step.
Why D is wrong: D is wrong because it doubles U, counting the pair's energy once for each nucleus; the potential energy belongs to the pair and is counted once, U = ke²/r.
MCQ 7CalculationPractice
In the net p–p reaction 4 ¹H → ⁴He + 2e⁺ + 2ν, the mass of ¹H is 1.00783 u, the mass of ⁴He is 4.00260 u and the electron (positron) mass is 0.00055 u. Neglecting the neutrino masses and taking 1 u ≡ 931.5 MeV, the total energy released — counting the annihilation of the two positrons with two of the star's electrons — is closest to
Show answer and why every option is right or wrong
Answer: D. D is correct. The quoted masses are ATOMIC masses, so each ¹H carries one electron and the ⁴He carries two. When the two positrons annihilate with two electrons, every electron on both sides is accounted for and the whole balance reduces to Δm = 4M(¹H) − M(⁴He) = 4(1.00783) − 4.00260 = 0.02872 u, giving 0.02872 × 931.5 = 26.7 MeV. That is why the textbook figure for the p–p cycle is 26.7 MeV: it is the total including annihilation, which is what this question asks for. Note that the neutrinos do not add to this — they carry energy AWAY, so the energy actually deposited in the Sun is slightly less (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because it subtracts only the two positron masses, 0.02872 − 0.00110 = 0.02762 u → 25.7 MeV, forgetting that the four ¹H atomic masses also carry two more electrons than the ⁴He atom; the question in any case asks for the total including annihilation.
Why B is wrong: B is wrong because 6.69 MeV is the energy per hydrogen atom consumed, 26.7/4; the question asks for the whole reaction, which uses four.
Why C is wrong: C is wrong for a subtle and instructive reason: 24.7 MeV is the energy released BEFORE the positrons annihilate. Stopping at Δm = 4M(¹H) − M(⁴He) − 4mₑ = 0.02872 − 0.00220 = 0.02652 u gives 24.7 MeV, and the two annihilations then release a further 4mₑc² = 4 × 0.511 = 2.04 MeV, which brings the total to 26.7 MeV. The question asks for the total.
MCQ 8CalculationPractice
Using the binding-energy-per-nucleon curve, consider fusing two ⁵⁶Fe nuclei (binding energy per nucleon 8.8 MeV, the curve's maximum) into a single nucleus of mass number 112, for which the binding energy per nucleon is 8.5 MeV. The reaction
Show answer and why every option is right or wrong
Answer: B. B is correct. Total binding energy of reactants = 112 × 8.8 = 985.6 MeV; of the product = 112 × 8.5 = 952 MeV. Binding energy has fallen by 33.6 ≈ 3.4 × 10¹ MeV, so that much energy must be supplied — fusion is exoergic only on the rising part of the curve, below iron (NCERT Class 12 Physics, Chapter 13, page 317).
Why A is wrong: A is wrong because it has the right magnitude but the wrong sign; beyond the iron peak, fusing nuclei costs energy rather than releasing it.
Why C is wrong: C is wrong because the energy exchanged is the change in total binding energy, not the product's total binding energy.
Why D is wrong: D is wrong because conservation of nucleon number says nothing about the energy balance, which is set by how tightly those nucleons are bound before and after.
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Nuclear Fusion: quick recall before you leave
How do you solve a Nuclear Fusion question? A worked example
Pattern: P.PHY.U18.NUCLEAR_FISSION_FUSION — identify fusion and compute the energy released from the mass defect.
- 1
Given
Deuterium–tritium reaction: ²H + ³H → ⁴He + n.
m(²H) = 2.01410 u
m(³H) = 3.01605 u
m(⁴He) = 4.00260 u
m(n) = 1.00866 u
1 u ≡ 931.5 MeV (exact by definition of the working conversion) - 2
Required
The energy released, in MeV, and confirmation that this is a fusion rather than a fission process.
- 3
Concept
Two light nuclei combine into a heavier one that lies higher on the binding-energy-per-nucleon curve. The products are more tightly bound, so their combined rest mass is smaller than the reactants'. The missing rest mass emerges as kinetic energy of the ⁴He nucleus and the neutron.
- 4
Formula
Δm = (sum of reactant masses) − (sum of product masses)
Q = Δm × 931.5 MeV/u - 5
Substitution
Reactants: 2.01410 + 3.01605 = 5.03015 u
Products: 4.00260 + 1.00866 = 5.01126 u
Δm = 5.03015 − 5.01126 - 6
Calculation
Δm = 0.01889 u
Q = 0.01889 × 931.5 = 17.596... MeV
The conversion factor 931.5 MeV/u is a defined equivalence, not a measured quantity, so it does not limit the significant figures. The mass defect carries four significant figures (0.01889 u), set by the fifth-decimal-place precision of the tabulated masses, so the answer is quoted to four. - 7
Final answer
Q = 17.60 MeV, released.
Nucleon number is conserved (2 + 3 = 4 + 1 = 5), charge is conserved (1 + 1 = 2 + 0 = 2), and two light nuclei have combined into a heavier one — this is fusion. The mass defect is positive, so energy is released. - 8
Common trap
Candidates who recognise "nuclear reaction, mass defect, ~10 MeV" sometimes answer from the fission template and reach for 200 MeV, or label the process fission because a neutron appears among the products. A free neutron on the product side is not a signature of fission. Read the direction: here two small nuclei go in and one larger nucleus comes out, so it is fusion. The 17.6 MeV figure is also worth holding alongside fission's ~200 MeV — per event, D–T fusion delivers roughly a tenth as much, even though per kilogram of fuel it delivers several times more.
- 9
Similar NEET-style question
Two ²H nuclei fuse by the route ²H + ²H → ³H + ¹H. Given m(²H) = 2.01410 u, m(³H) = 3.01605 u and m(¹H) = 1.00783 u, with 1 u ≡ 931.5 MeV, find the energy released and state whether the reaction is exoergic. *(Answer: Δm = 4.02820 − 4.02388 = 0.00432 u, so Q = 4.02 MeV, released — exoergic.)*
What to remember before solving Nuclear Fusion questions
More in Atoms and Nuclei: 3 exam traps and mistakes · 6 formulas · 4 question patterns from its other lessons.
Nuclear Fusion questions from past NEET papers
1 question from NEET 2023. Answers verified against NTA official keys. — click to collapse
NEET 2023
140 minutes
260 minutes
380 minutes
420 minutes
NTA Answer: Option 3(final)
How does NEET ask about Nuclear Fusion?
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
NCERT refs: Class 12 Physics Chapter 13, p.317
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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