How to Use This Problem Set
Your full workout for Nuclei, grouped by theme: mass conversions & counting, size & density, binding energy, Q-values, fission & fusion energy budgets, and decay-law problems.
Keep these handy:
- 1 u = 1.66×10−27 kg = 931.5 MeV/c2; 1 MeV = 1.6×10−13 J; 1 year = 3.154×107 s
- Radius/density: R=1.2A1/3 fm; V∝A; ρ≈2.3×1017 kg/m³ (all nuclei)
- Binding: ΔM=[ZmH+(A−Z)mn]−Matom; Eb=ΔM×931.5 MeV; Ebn=Eb/A
- Q-value: Q=[Σmi−Σmf]×931.5 MeV; Q > 0 exothermic
- Decay: N=N0e−λt; T1/2=0.693/λ; τ=1/λ; R=λN; after n half-lives: N0/2n
- Masses: mH = 1.007825 u, mn = 1.008665 u, m(4He) = 4.002603 u
State the condition, then substitute.
Solved Examples - Conversions & Counting
Example 1. Convert 0.5 u into MeV.
Solution: 0.5×931.5=465.8 MeV.
Example 2. Energy equivalent of an electron's mass (0.000548 u)?
Solution: 0.000548×931.5≈0.511 MeV — the famous electron rest energy.
Example 3. Atoms in 10 g of 235U?
Solution: N=23510×6.023×1023=2.56×1022.
Example 4. Express 3.2×10−11 J in MeV.
Solution: 1.6×10−133.2×10−11=200 MeV — one fission's worth.
Example 5. An element's isotopes of mass 6.015 u (7.5%) and 7.016 u (92.5%): mean atomic mass?
Solution: 0.075×6.015+0.925×7.016=6.94 u — lithium.
Solved Examples - Size & Density
Example 6. Radius of 27Al?
Solution: R=1.2×271/3=1.2×3=3.6 fm.
Example 7. Ratio of radii of 208Pb and 26Al… take A = 208, 26.
Solution: (26208)1/3=81/3=2 — lead's nucleus is exactly twice aluminium's radius.
Example 8. Gold vs silver radii (NCERT 13.4)?
Solution: (107197)1/3≈1.23.
Example 9. Nuclear density of 56Fe (mass 55.85 u)? (NCERT Example 13.1)
Solution: ρ=34π(1.2×10−15)3×569.27×10−26=2.29×1017 kg/m³.
Example 10. Mass number of a nucleus with radius 6.0 fm?
Solution: A=(1.26.0)3=53=125.
Solved Examples - Binding Energy
Example 11. Binding energy of 714N (14.00307 u)? (NCERT 13.1)
Solution: ΔM=[7(1.007825)+7(1.008665)]−14.00307=0.11236 u; Eb=104.7 MeV.
Example 12. Ebn of nitrogen-14 from Example 11?
Solution: 104.7/14=7.48 MeV per nucleon.
Example 13. Binding energy of 2656Fe (55.934939 u)? (NCERT 13.2)
Solution: ΔM=[26(1.007825)+30(1.008665)]−55.934939=0.528461 u; Eb=492.3 MeV; Ebn=8.79 MeV.
Example 14. Binding energy of 83209Bi (208.980388 u)? (NCERT 13.2)
Solution: ΔM=[83(1.007825)+126(1.008665)]−208.980388=1.760877 u; Eb=1640 MeV; Ebn=7.85 MeV.
Example 15. Energy to dismantle all nuclei in a 3.0 g copper coin (63Cu, 62.92960 u)? (NCERT 13.3)
Solution: per-nucleus ΔM=[29(1.007825)+34(1.008665)]−62.92960=0.591935 u → 551.4 MeV; atoms =2.87×1022; total ≈ 1.58×1025 MeV =2.5×1012 J.
Example 16. Ebn of 4He (4.002603 u)?
Solution: ΔM=[2(1.007825)+2(1.008665)]−4.002603=0.030377 u; Eb=28.3 MeV; Ebn=7.07 MeV.
Solved Examples - Q-values
Example 17. Q of 1H+3H→2H+2H? (NCERT 13.5i)
Solution: Q=[1.007825+3.016049−2(2.014102)]×931.5=−0.00433×931.5≈−4.03 MeV — endothermic.
Example 18. Q of 12C+12C→20Ne+4He? (NCERT 13.5ii)
Solution: Q=[24.000000−19.992439−4.002603]×931.5=0.004958×931.5≈+4.62 MeV — exothermic.
Example 19. Is 56Fe→228Al possible? (NCERT 13.6)
Solution: Q=[55.93494−2(27.98191)]×931.5=−26.9 MeV < 0 — not energetically possible.
Example 20. Q of the deuteron-deuteron → He-3 + n reaction, from masses m(2H) = 2.014102 u, m(3He) = 3.016029 u, mn = 1.008665 u?
Solution: Q=[2(2.014102)−3.016029−1.008665]×931.5=0.003510×931.5≈3.27 MeV. ✔ (Matches NCERT Eq. 13.13b.)
Example 21. Complete and identify: 01n+92235U→54140Xe+X+201n.
Solution: A = 236 - 140 - 2 = 94; Z = 92 - 54 = 38 → X=3894Sr.
Solved Examples - Fission & Fusion Budgets
Example 22. Energy from complete fission of 1 kg of 239Pu at 180 MeV each? (NCERT 13.7)
Solution: N=2391000×6.023×1023=2.52×1024; E=4.53×1026 MeV ≈7.3×1013 J.
Example 23. How long does 2.0 kg of deuterium power a 100 W lamp (3.27 MeV per d-d reaction)? (NCERT 13.8)
Solution: reactions =3.01×1026; E ≈ 1.58×1014 J; t=1.58×1012 s ≈ 5×104 years.
Example 24. Coulomb barrier of two touching deuterons (r = 2.0 fm)? (NCERT 13.9)
Solution: U=4×10−159×109(1.6×10−19)2=5.76×10−14 J ≈ 360 keV.
Example 25. Fissions per second in a 500 MW reactor (200 MeV each)?
Solution: 3.2×10−115×108≈1.6×1019 per second.
Example 26. Verify the p-p cycle total.
Solution: 2(0.42) + 2(1.02) + 2(5.49) + 12.86 = 26.7 MeV. ✔
Example 27. Hydrogen the Sun burns per second (3.8×1026 W)?
Solution: per proton 6.68 MeV = 1.07×10−12 J → 3.6×1038 protons/s → ≈ 6×1011 kg/s.
Solved Examples - Decay Law
Example 28. T1/2 = 20 min. Fraction left after 1 hour?
Solution: 3 half-lives → 81.
Example 29. Decay constant for T1/2 = 1386 s?
Solution: λ=13860.693=5×10−4 s⁻¹.
Example 30. Activity of 1018 atoms with λ=10−4 s⁻¹?
Solution: R=λN=1014 Bq.
Example 31. Activity falls from 8000 to 500 Bq in 36 hours: half-life?
Solution: ratio 16 = 24 → 4 half-lives → T1/2 = 9 hours.
Example 32. Mean life of a nuclide with T1/2 = 693 years?
Solution: τ=0.693T1/2=1000 years.
Example 33. What fraction decays during one mean life?
Solution: remaining =e−1≈37%; decayed ≈ 63%.
Example 34. A sample is 1/64 of its original after 30 days. Find T1/2.
Solution: 1/64=(1/2)6 → 6 half-lives → T1/2 = 5 days.
Solved Examples - Mixed Mastery
Example 35. 92238U alpha-decays. Identify the daughter and compute Q given m(U) = 238.05079 u, m(Th) = 234.04363 u, m(He) = 4.002603 u.
Solution: daughter 90234Th; Q=[238.05079−234.04363−4.002603]×931.5=0.004557×931.5≈4.25 MeV.
Example 36. Mass converted to energy per second by a 1000 MW source?
Solution: m˙=c2P=9×1016109≈1.1×10−8 kg/s ≈ 1 gram per day.
Example 37. The Sun loses how much mass per second (3.8×1026 W)?
Solution: 9×10163.8×1026≈4.2×109 kg/s — four million tonnes a second.
Example 38. Two nuclei have volumes in ratio 8 : 1. Ratio of mass numbers and radii?
Solution: A ratio = 8 : 1 (V ∝ A); radii ratio = 2 : 1.
Example 39. A nucleus at rest emits a 5 MeV alpha. Why is the daughter's recoil energy small, and roughly how big for A = 234?
Solution: momentum conservation: pα=pdaughter; K=2mp2 → Kd=Kα×mdmα≈5×2344≈0.085 MeV — the heavy partner takes the momentum but little energy.
Example 40. How much 235U (200 MeV/fission) equals the chemical energy of 1000 kg of coal (107 J/kg)?
Solution: need 1010 J; per kg U ≈ 8.2×1013 J → m=8.2×10131010 kg ≈ 0.12 g. A coal truck versus a dust grain.