How JEE Tests This Chapter

Nuclei contributes 1-2 questions to nearly every JEE Main paper, from tight templates:

  1. Radius/density scalings: R∝A1/3R \propto A^{1/3} ratios, volume ∝ A, density constant; surface-area and volume variants.
  2. Binding-energy arithmetic: mass defect × 931.5; EbnE_{bn} comparisons; energy released from curve values (EbnE_{bn} of parents vs products).
  3. Q-values: compute from masses; exo/endothermic verdicts; kinetic-energy sharing in two-body decays (K∝1/mK \propto 1/m).
  4. Fission/fusion budgets: energy per kg, fissions per second for a given power, fuel-burn durations.
  5. Decay law: N0/2nN_0/2^n chains, activity R=λNR = \lambda N, extracting T1/2T_{1/2} from data pairs, mean-life relations.
  6. Twists: alpha-decay recoil, annihilation energetics, mass-to-power rates (P=m˙c2P = \dot{m}c^2).

All questions below are JEE-style previous-year questions with worked solutions. Years are attached only where attribution is certain; otherwise tags read [JEE Mains] / [JEE Advanced].

Constants: 1 u = 931.5 MeV/c2c^2; R0R_0 = 1.2 fm; 1 MeV = 1.6×10−131.6 \times 10^{-13} J; NA=6.023×1023N_A = 6.023 \times 10^{23}.

JEE PYQ Worked Set A: Size, Density, Binding

PYQ 1. The ratio of the nuclear densities of two nuclei with mass numbers 8 and 64 is: [JEE Mains]

Solution:

  1. Nuclear density is independent of A.
  2. Ratio = 1 : 1 — regardless of the tempting 1 : 8.

PYQ 2. If the radius of 64^{64}Zn is R, the radius of 125^{125}Te in terms of R is: [JEE Mains]

Solution:

  1. R′R=(12564)1/3=54\dfrac{R'}{R} = \left(\dfrac{125}{64}\right)^{1/3} = \dfrac{5}{4}.
  2. R′=1.25RR' = 1.25R.

PYQ 3. The binding energies per nucleon of deuteron and helium-4 are 1.1 MeV and 7.0 MeV. The energy released when two deuterons fuse into 4^4He is: [JEE Mains]

Solution:

  1. Before: binding = 2 × (2 × 1.1) = 4.4 MeV.
  2. After: 4 × 7.0 = 28.0 MeV.
  3. Released: 28.0 - 4.4 = 23.6 MeV.
  4. Takeaway: energy out = binding gained — the template for every 'given EbnE_{bn} values' question.

PYQ 4. A nucleus of mass M emits a photon of frequency ν\nu while at rest. The recoil kinetic energy of the nucleus is: [JEE Mains]

Solution:

  1. Photon momentum p=hνcp = \dfrac{h\nu}{c}; nucleus recoils with the same p.
  2. K=p22M=h2ν22Mc2K = \dfrac{p^2}{2M} = \dfrac{h^2\nu^2}{2Mc^2}.
  3. Takeaway: gamma emission always kicks the emitter — energy hierarchy: photon takes almost all, recoil takes (hν)22Mc2\dfrac{(h\nu)^2}{2Mc^2}.

JEE PYQ Worked Set B: Q-values & Decay Energetics

PYQ 5. In the alpha decay of 226^{226}Ra (Q = 4.87 MeV), the kinetic energy of the alpha particle is closest to: [JEE Advanced pattern]

Solution:

  1. Two-body decay from rest: Kα=Q×A−4A=4.87×222226K_\alpha = Q \times \dfrac{A - 4}{A} = 4.87 \times \dfrac{222}{226}.
  2. Kα≈4.78K_\alpha \approx 4.78 MeV (daughter takes only ~0.09 MeV).
  3. Takeaway: the light particle carries the fraction (heavy mass)/(total) of Q.

PYQ 6. The binding energies per nucleon of X (A = 240) and its two equal fragments (A = 120) are 7.6 and 8.5 MeV. Energy released per fission: [JEE Mains]

Solution:

  1. ΔE=240×(8.5−7.6)=240×0.9=216\Delta E = 240 \times (8.5 - 7.6) = 240 \times 0.9 = 216 MeV.

PYQ 7. A radioactive sample's activity falls to 1/e of its initial value in time t. Then t equals: [JEE Mains]

Solution:

  1. R=R0e−λtR = R_0e^{-\lambda t}; R=R0/eR = R_0/e when λt=1\lambda t = 1.
  2. t=1λ=τt = \dfrac{1}{\lambda} = \tau — the mean life.
  3. Takeaway: 1/e ↔ mean life; 1/2 ↔ half-life. The two flags never mix.

PYQ 8. Two radioactive materials have decay constants 5λ\lambda and λ\lambda. Starting with equal numbers, the ratio of their remaining nuclei becomes 1/e² after time: [JEE Mains]

Solution:

  1. Ratio: N1N2=e−5λt/e−λt=e−4λt\dfrac{N_1}{N_2} = e^{-5\lambda t}/e^{-\lambda t} = e^{-4\lambda t}.
  2. Set e−4λt=e−2e^{-4\lambda t} = e^{-2}: t=12λt = \dfrac{1}{2\lambda}.
  3. Takeaway: ratios of exponentials subtract the exponents — a JEE fixture.

JEE PYQ Worked Set C: Power, Fuel & Multi-step

PYQ 9. A 100 MW reactor consumes half its fuel in 5 years via 235^{235}U fission (200 MeV each). The initial fuel load is about: [JEE Advanced pattern]

Solution:

  1. Energy in 5 years: E=108×5×3.154×107=1.58×1016E = 10^8 \times 5 \times 3.154 \times 10^7 = 1.58 \times 10^{16} J.
  2. Fissions: 1.58×10163.2×10−11=4.9×1026\dfrac{1.58 \times 10^{16}}{3.2 \times 10^{-11}} = 4.9 \times 10^{26} — this consumed HALF the fuel.
  3. Initial atoms: 9.9×10269.9 \times 10^{26}; mass =9.9×10266.02×1023×235= \dfrac{9.9 \times 10^{26}}{6.02 \times 10^{23}} \times 235 g ≈ 386 kg.
  4. Takeaway: power → energy → fissions → atoms → kilograms: the five-step reactor chain.

PYQ 10. The half-life of a nuclide is 10 hours. The fraction remaining after 4 hours is: [JEE Mains]

Solution:

  1. Non-integer half-lives: NN0=2−t/T1/2=2−0.4\dfrac{N}{N_0} = 2^{-t/T_{1/2}} = 2^{-0.4}.
  2. 2−0.4=e−0.4×0.693=e−0.277≈0.7582^{-0.4} = e^{-0.4 \times 0.693} = e^{-0.277} \approx 0.758.
  3. Answer: ≈ 76% remains.
  4. Takeaway: when t isn't a whole number of half-lives, use 2−t/T2^{-t/T} — no shortcuts.

PYQ 11. If 200 MeV is released per fission, the number of fissions per second for 1 W of power is: [JEE Mains]

Solution:

  1. Per fission: 3.2×10−113.2 \times 10^{-11} J.
  2. Rate =13.2×10−11≈3.125×1010= \dfrac{1}{3.2 \times 10^{-11}} \approx 3.125 \times 10^{10} per second.

PYQ 12. A star converts all its helium (4×10304 \times 10^{30} kg… illustrative) to carbon via 3 4He→12C+7.273\,^4He \to {}^{12}C + 7.27 MeV. Energy released per kg of helium: [JEE Advanced pattern]

Solution:

  1. Per reaction: 3 heliums (12 u) yield 7.27 MeV → per kg: 100012×1.66×10−27\dfrac{1000}{12 \times 1.66 \times 10^{-27}}… cleaner: reactions per kg =13×4×1.66×10−27=5.0×1025= \dfrac{1}{3 \times 4 \times 1.66 \times 10^{-27}} = 5.0 \times 10^{25}.
  2. Energy: 5.0×1025×7.27×1.6×10−13≈5.8×10135.0 \times 10^{25} \times 7.27 \times 1.6 \times 10^{-13} \approx 5.8 \times 10^{13} J per kg.
  3. Takeaway: helium burning yields ~101310^{13}-101410^{14} J/kg — same order as fission, an order below hydrogen fusion.

PYQ 13. The mass defect of a reaction is 0.02866 u. The energy released per event, and per mole of events, are: [JEE Mains]

Solution:

  1. Per event: 0.02866×931.5≈26.70.02866 \times 931.5 \approx 26.7 MeV (the p-p cycle's number!).
  2. Per mole: 26.7×1.6×10−13×6.02×1023≈2.6×101226.7 \times 1.6 \times 10^{-13} \times 6.02 \times 10^{23} \approx 2.6 \times 10^{12} J.
  3. Takeaway: Avogadro converts nuclear per-event MeV into terajoules per mole — twelve orders in one hop.

PYQ 14. At time t = 0, a sample has N0N_0 nuclei of half-life T. The number DECAYED between t = T and t = 2T is: [JEE Mains]

Solution:

  1. At T: N0/2N_0/2 remain; at 2T: N0/4N_0/4 remain.
  2. Decayed in the interval: N02−N04=N04\dfrac{N_0}{2} - \dfrac{N_0}{4} = \dfrac{N_0}{4}.
  3. Takeaway: each successive half-life kills half of WHAT REMAINS — absolute decay counts shrink geometrically.

PYQ 15. A nucleus with EbnE_{bn} = 7.8 MeV (A = 220) splits into fragments with EbnE_{bn} = 8.4 MeV. If the fragments' mass numbers are 110 each, the total kinetic energy released is: [JEE Mains]

Solution:

  1. ΔE=220×(8.4−7.8)=220×0.6=132\Delta E = 220 \times (8.4 - 7.8) = 220 \times 0.6 = 132 MeV.
  2. Takeaway: always (total nucleons) × (per-nucleon gain) — fragment symmetry is irrelevant to the total.

PYQ 16. The activity of a sample is R1R_1 at time t1t_1 and R2R_2 at t2t_2 (t2>t1t_2 > t_1). The number of nuclei that decayed in the interval is proportional to: [JEE Advanced]

Solution:

  1. Decayed count =N1−N2=R1−R2λ= N_1 - N_2 = \dfrac{R_1 - R_2}{\lambda}.
  2. With λ=0.693T1/2\lambda = \dfrac{0.693}{T_{1/2}}: decayed =(R1−R2)T1/20.693∝(R1−R2)T1/2= \dfrac{(R_1 - R_2)T_{1/2}}{0.693} \propto (R_1 - R_2)T_{1/2}.
  3. Takeaway: activity differences convert to population differences through 1/λ — a favourite JEE Advanced twist.