Chapter 13: Nuclei
Key Formulas:
• Mass defect: Δm = [Zmp + (A−Z)mn] − mnucleus
• Binding energy: B.E. = Δm × 931.5 MeV
• Radioactivity: N = N₀e−λt
• Half-life: t1/2 = 0.693/λ
Q1. The half-life of a radioactive substance is 30 days. What fraction of the original substance remains after 90 days?
Solution
N = N₀(1/2)t/t1/2
= N₀(1/2)90/30
= N₀(1/2)³
= N₀/8
Fraction remaining = 1/8 = 0.125
Q2. Find the binding energy per nucleon of 56Fe if the mass of 56Fe atom is 55.9349 u. (Mass of proton = 1.00783 u, mass of neutron = 1.00867 u)
Solution
Δm = [26(1.00783) + 30(1.00867)] − 55.9349
= [26.20358 + 30.26010] − 55.9349
= 56.46368 − 55.9349
= 0.52878 u
B.E. = 0.52878 × 931.5 = 492.6 MeV
B.E./nucleon = 492.6/56 = 8.80 MeV/nucleon
Q3. Explain the difference between nuclear fission and fusion. Give one example of each.
Solution
Fission: Heavy nucleus splits into lighter nuclei.
Example: 235U + n → 236U* → 141Ba + 92Kr + 3n
Fusion: Light nuclei combine to form a heavier nucleus.
Example: 22H → 4He + energy
Both release energy because the products have higher binding energy per nucleon.
Q4. A radioactive isotope has a half-life of 5 years. How much of a 100 g sample will remain after 20 years?
Solution
N = N₀(1/2)t/t1/2
= 100(1/2)20/5
= 100(1/2)&sup4;
= 100/16 = 6.25 g
Q5. The activity of a radioactive sample drops to 1/16th of its original value in 28 days. Find its half-life.
Solution
N/N₀ = (1/2)n where n = number of half-lives
1/16 = (1/2)n
(1/2)&sup4; = (1/2)n
n = 4 half-lives
t1/2 = 28/4 = 7 days
Q6. The mass of a hydrogen atom is 1.00783 u. Find the binding energy of the hydrogen nucleus.
Solution
Hydrogen nucleus = 1 proton
Mass of proton = 1.007276 u
Mass defect: Δm = 1.007276 − 1.00783
Actually, mass of hydrogen atom = mass of proton + mass of electron
Binding energy refers to nuclear binding energy.
For hydrogen (1 proton only, no binding):
B.E. ≈ 0 (hydrogen has no neutrons, so no nuclear binding energy in the conventional sense)
Q7. A radioactive sample has a half-life of 5 hours. Starting with 1000 atoms, find (a) the number of atoms remaining after 10 hours (b) the number of atoms that have decayed in the first 10 hours.
Solution
(a) N = N₀(1/2)t/t1/2
= 1000(1/2)10/5
= 1000(1/2)² = 1000/4 = 250 atoms
(b) Decayed atoms = 1000 − 250 = 750 atoms
Q8. Define the terms (i) mass defect (ii) binding energy (iii) binding energy per nucleon. How does binding energy per nucleon vary with mass number?
Solution
(i) Mass defect: The difference between the total mass of individual nucleons and the actual mass of the nucleus.
Δm = [Zmp + (A−Z)mn] − Mnucleus
(ii) Binding energy: The energy equivalent of mass defect.
B.E. = Δm × 931.5 MeV
(iii) B.E. per nucleon: B.E./A
Variation: B.E./nucleon increases with A, reaches maximum (~8.8 MeV) around Fe-56, then slowly decreases. Most stable nuclei are around A ≈ 56.
Q9. The decay constant of a radioactive substance is 0.05 s−1. Find (i) the half-life (ii) the time for the activity to drop to one-tenth of its original value.
Solution
(i) t1/2 = 0.693/λ = 0.693/0.05 = 13.86 s
(ii) N/N₀ = e−λt = 1/10
λt = ln 10 = 2.303
t = 2.303/0.05 = 46.06 s
Q10. A nuclear reaction releases 200 MeV of energy. If the mass of the reactants is 5 u, what is the mass of the products? (1 u = 931.5 MeV)
Solution
Energy released: E = Δm × 931.5 MeV
200 = Δm × 931.5
Δm = 200/931.5 = 0.2147 u
Δm = mreactants − mproducts
mproducts = 5 − 0.2147 = 4.785 u
Q11. Write the nuclear equations for (a) α-decay of U-238 (b) β-decay of C-14 (c) γ-emission.
Solution
(a) α-decay of U-238:
23892U → 23490Th + 42He
(b) β-decay of C-14:
146C → 147N + 0−1e + ν̄
(c) γ-emission (example):
6027Co* → 6028Ni + 0−1e + γ
(nucleus in excited state → ground state + γ)
Q12. The number of radioactive nuclei in a sample at t = 0 is N₀ and at t = t₀ is N₀/2e. Find the half-life of the sample.
Solution
N = N₀e−λt
N₀/2e = N₀e−λt₀
1/(2e) = e−λt₀
e−1 · e−1 = e−λt₀
e−2 = e−λt₀
λt₀ = 2
t1/2 = 0.693/λ = 0.693 × t₀/2 = 0.347t₀
Q13. Calculate the energy released in the fusion reaction: 22H → 4He + 1H, given masses: m(2H) = 2.014 u, m(3He) = 3.016 u, m(4He) = 4.0026 u, m(1H) = 1.0078 u.
Solution
Reaction: 22H → 4He (mass given is for 3He, but let's use 4He)
mreactants = 2 × 2.014 = 4.028 u
mproducts = 4.0026 u
Δm = 4.028 − 4.0026 = 0.0254 u
E = 0.0254 × 931.5 = 23.7 MeV
Q14. A radioactive substance has a half-life of 3 hours. What fraction of the original activity will remain after 9 hours?
Solution
Number of half-lives: n = 9/3 = 3
A = A₀(1/2)n = A₀(1/2)³
= A₀/8
Fraction remaining = 1/8 = 12.5%
Q15. The binding energies of He-4 and Fe-56 are 28.3 MeV and 492 MeV respectively. Find their binding energy per nucleon and state which is more stable.
Solution
He-4: B.E./nucleon = 28.3/4 = 7.075 MeV/nucleon
Fe-56: B.E./nucleon = 492/56 = 8.786 MeV/nucleon
Fe-56 is more stable (higher binding energy per nucleon means more tightly bound nucleus).