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 = 1.66×10271.66 \times 10^{-27} kg = 931.5 MeV/c2c^2; 1 MeV = 1.6×10131.6 \times 10^{-13} J; 1 year = 3.154×1073.154 \times 10^7 s
  • Radius/density: R=1.2A1/3R = 1.2A^{1/3} fm; VAV \propto A; ρ2.3×1017\rho \approx 2.3 \times 10^{17} kg/m³ (all nuclei)
  • Binding: ΔM=[ZmH+(AZ)mn]Matom\Delta M = [Zm_H + (A-Z)m_n] - M_{atom}; Eb=ΔM×931.5E_b = \Delta M \times 931.5 MeV; Ebn=Eb/AE_{bn} = E_b/A
  • Q-value: Q=[ΣmiΣmf]×931.5Q = [\Sigma m_i - \Sigma m_f] \times 931.5 MeV; Q > 0 exothermic
  • Decay: N=N0eλtN = N_0e^{-\lambda t}; T1/2=0.693/λT_{1/2} = 0.693/\lambda; τ=1/λ\tau = 1/\lambda; R=λNR = \lambda N; after n half-lives: N0/2nN_0/2^n
  • Masses: mHm_H = 1.007825 u, mnm_n = 1.008665 u, m(4He)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=465.80.5 \times 931.5 = 465.8 MeV.

Example 2. Energy equivalent of an electron's mass (0.000548 u)?

Solution: 0.000548×931.50.5110.000548 \times 931.5 \approx 0.511 MeV — the famous electron rest energy.

Example 3. Atoms in 10 g of 235^{235}U?

Solution: N=10235×6.023×1023=2.56×1022N = \dfrac{10}{235} \times 6.023 \times 10^{23} = 2.56 \times 10^{22}.

Example 4. Express 3.2×10113.2 \times 10^{-11} J in MeV.

Solution: 3.2×10111.6×1013=200\dfrac{3.2 \times 10^{-11}}{1.6 \times 10^{-13}} = 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=6.940.075 \times 6.015 + 0.925 \times 7.016 = 6.94 u — lithium.

Solved Examples - Size & Density

Example 6. Radius of 27^{27}Al?

Solution: R=1.2×271/3=1.2×3=3.6R = 1.2 \times 27^{1/3} = 1.2 \times 3 = 3.6 fm.

Example 7. Ratio of radii of 208^{208}Pb and 26^{26}Al… take A = 208, 26.

Solution: (20826)1/3=81/3=2\left(\dfrac{208}{26}\right)^{1/3} = 8^{1/3} = 2 — lead's nucleus is exactly twice aluminium's radius.

Example 8. Gold vs silver radii (NCERT 13.4)?

Solution: (197107)1/31.23\left(\dfrac{197}{107}\right)^{1/3} \approx 1.23.

Example 9. Nuclear density of 56^{56}Fe (mass 55.85 u)? (NCERT Example 13.1)

Solution: ρ=9.27×10264π3(1.2×1015)3×56=2.29×1017\rho = \dfrac{9.27 \times 10^{-26}}{\frac{4\pi}{3}(1.2 \times 10^{-15})^3 \times 56} = 2.29 \times 10^{17} kg/m³.

Example 10. Mass number of a nucleus with radius 6.0 fm?

Solution: A=(6.01.2)3=53=125A = \left(\dfrac{6.0}{1.2}\right)^3 = 5^3 = 125.

Solved Examples - Binding Energy

Example 11. Binding energy of 714^{14}_{7}N (14.00307 u)? (NCERT 13.1)

Solution: ΔM=[7(1.007825)+7(1.008665)]14.00307=0.11236\Delta M = [7(1.007825) + 7(1.008665)] - 14.00307 = 0.11236 u; Eb=104.7E_b = 104.7 MeV.

Example 12. EbnE_{bn} of nitrogen-14 from Example 11?

Solution: 104.7/14=7.48104.7/14 = 7.48 MeV per nucleon.

Example 13. Binding energy of 2656^{56}_{26}Fe (55.934939 u)? (NCERT 13.2)

Solution: ΔM=[26(1.007825)+30(1.008665)]55.934939=0.528461\Delta M = [26(1.007825) + 30(1.008665)] - 55.934939 = 0.528461 u; Eb=492.3E_b = 492.3 MeV; Ebn=8.79E_{bn} = 8.79 MeV.

Example 14. Binding energy of 83209^{209}_{83}Bi (208.980388 u)? (NCERT 13.2)

Solution: ΔM=[83(1.007825)+126(1.008665)]208.980388=1.760877\Delta M = [83(1.007825) + 126(1.008665)] - 208.980388 = 1.760877 u; Eb=1640E_b = 1640 MeV; Ebn=7.85E_{bn} = 7.85 MeV.

Example 15. Energy to dismantle all nuclei in a 3.0 g copper coin (63^{63}Cu, 62.92960 u)? (NCERT 13.3)

Solution: per-nucleus ΔM=[29(1.007825)+34(1.008665)]62.92960=0.591935\Delta M = [29(1.007825) + 34(1.008665)] - 62.92960 = 0.591935 u → 551.4 MeV; atoms =2.87×1022= 2.87 \times 10^{22}; total ≈ 1.58×10251.58 \times 10^{25} MeV =2.5×1012= 2.5 \times 10^{12} J.

Example 16. EbnE_{bn} of 4^{4}He (4.002603 u)?

Solution: ΔM=[2(1.007825)+2(1.008665)]4.002603=0.030377\Delta M = [2(1.007825) + 2(1.008665)] - 4.002603 = 0.030377 u; Eb=28.3E_b = 28.3 MeV; Ebn=7.07E_{bn} = 7.07 MeV.

Solved Examples - Q-values

Example 17. Q of 1H+3H2H+2H^{1}H + {}^{3}H \to {}^{2}H + {}^{2}H? (NCERT 13.5i)

Solution: Q=[1.007825+3.0160492(2.014102)]×931.5=0.00433×931.54.03Q = [1.007825 + 3.016049 - 2(2.014102)] \times 931.5 = -0.00433 \times 931.5 \approx -4.03 MeV — endothermic.

Example 18. Q of 12C+12C20Ne+4He^{12}C + {}^{12}C \to {}^{20}Ne + {}^{4}He? (NCERT 13.5ii)

Solution: Q=[24.00000019.9924394.002603]×931.5=0.004958×931.5+4.62Q = [24.000000 - 19.992439 - 4.002603] \times 931.5 = 0.004958 \times 931.5 \approx +4.62 MeV — exothermic.

Example 19. Is 56Fe228Al^{56}Fe \to 2\,^{28}Al possible? (NCERT 13.6)

Solution: Q=[55.934942(27.98191)]×931.5=26.9Q = [55.93494 - 2(27.98191)] \times 931.5 = -26.9 MeV < 0 — not energetically possible.

Example 20. Q of the deuteron-deuteron → He-3 + n reaction, from masses m(2^2H) = 2.014102 u, m(3^3He) = 3.016029 u, mnm_n = 1.008665 u?

Solution: Q=[2(2.014102)3.0160291.008665]×931.5=0.003510×931.53.27Q = [2(2.014102) - 3.016029 - 1.008665] \times 931.5 = 0.003510 \times 931.5 \approx 3.27 MeV. ✔ (Matches NCERT Eq. 13.13b.)

Example 21. Complete and identify: 01n+92235U54140Xe+X+201n^{1}_{0}n + {}^{235}_{92}U \to {}^{140}_{54}Xe + X + 2\,^{1}_{0}n.

Solution: A = 236 - 140 - 2 = 94; Z = 92 - 54 = 38 → X=3894X = {}^{94}_{38}Sr.

Solved Examples - Fission & Fusion Budgets

Example 22. Energy from complete fission of 1 kg of 239^{239}Pu at 180 MeV each? (NCERT 13.7)

Solution: N=1000239×6.023×1023=2.52×1024N = \dfrac{1000}{239} \times 6.023 \times 10^{23} = 2.52 \times 10^{24}; E=4.53×1026E = 4.53 \times 10^{26} MeV 7.3×1013\approx 7.3 \times 10^{13} 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= 3.01 \times 10^{26}; E ≈ 1.58×10141.58 \times 10^{14} J; t=1.58×1012t = 1.58 \times 10^{12} s ≈ 5×1045 \times 10^4 years.

Example 24. Coulomb barrier of two touching deuterons (r = 2.0 fm)? (NCERT 13.9)

Solution: U=9×109(1.6×1019)24×1015=5.76×1014U = \dfrac{9 \times 10^9 (1.6 \times 10^{-19})^2}{4 \times 10^{-15}} = 5.76 \times 10^{-14} J ≈ 360 keV.

Example 25. Fissions per second in a 500 MW reactor (200 MeV each)?

Solution: 5×1083.2×10111.6×1019\dfrac{5 \times 10^8}{3.2 \times 10^{-11}} \approx 1.6 \times 10^{19} 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 (3.8×10263.8 \times 10^{26} W)?

Solution: per proton 6.68 MeV = 1.07×10121.07 \times 10^{-12} J → 3.6×10383.6 \times 10^{38} protons/s → ≈ 6×10116 \times 10^{11} kg/s.

Solved Examples - Decay Law

Example 28. T1/2T_{1/2} = 20 min. Fraction left after 1 hour?

Solution: 3 half-lives → 18\dfrac{1}{8}.

Example 29. Decay constant for T1/2T_{1/2} = 1386 s?

Solution: λ=0.6931386=5×104\lambda = \dfrac{0.693}{1386} = 5 \times 10^{-4} s⁻¹.

Example 30. Activity of 101810^{18} atoms with λ=104\lambda = 10^{-4} s⁻¹?

Solution: R=λN=1014R = \lambda N = 10^{14} Bq.

Example 31. Activity falls from 8000 to 500 Bq in 36 hours: half-life?

Solution: ratio 16 = 242^4 → 4 half-lives → T1/2T_{1/2} = 9 hours.

Example 32. Mean life of a nuclide with T1/2T_{1/2} = 693 years?

Solution: τ=T1/20.693=1000\tau = \dfrac{T_{1/2}}{0.693} = 1000 years.

Example 33. What fraction decays during one mean life?

Solution: remaining =e137%= e^{-1} \approx 37\%; decayed ≈ 63%.

Example 34. A sample is 1/64 of its original after 30 days. Find T1/2T_{1/2}.

Solution: 1/64=(1/2)61/64 = (1/2)^6 → 6 half-lives → T1/2T_{1/2} = 5 days.

Solved Examples - Mixed Mastery

Example 35. 92238^{238}_{92}U 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 90234^{234}_{90}Th; Q=[238.05079234.043634.002603]×931.5=0.004557×931.54.25Q = [238.05079 - 234.04363 - 4.002603] \times 931.5 = 0.004557 \times 931.5 \approx 4.25 MeV.

Example 36. Mass converted to energy per second by a 1000 MW source?

Solution: m˙=Pc2=1099×10161.1×108\dot{m} = \dfrac{P}{c^2} = \dfrac{10^9}{9 \times 10^{16}} \approx 1.1 \times 10^{-8} kg/s ≈ 1 gram per day.

Example 37. The Sun loses how much mass per second (3.8×10263.8 \times 10^{26} W)?

Solution: 3.8×10269×10164.2×109\dfrac{3.8 \times 10^{26}}{9 \times 10^{16}} \approx 4.2 \times 10^{9} 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=pdaughterp_\alpha = p_{daughter}; K=p22mK = \dfrac{p^2}{2m}Kd=Kα×mαmd5×42340.085K_d = K_\alpha \times \dfrac{m_\alpha}{m_d} \approx 5 \times \dfrac{4}{234} \approx 0.085 MeV — the heavy partner takes the momentum but little energy.

Example 40. How much 235^{235}U (200 MeV/fission) equals the chemical energy of 1000 kg of coal (10710^7 J/kg)?

Solution: need 101010^{10} J; per kg U ≈ 8.2×10138.2 \times 10^{13} J → m=10108.2×1013m = \dfrac{10^{10}}{8.2 \times 10^{13}} kg ≈ 0.12 g. A coal truck versus a dust grain.