The problems are on the radioactive decay worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Radioactive decay is the random change of an unstable nucleus, and a sample decays exponentially with a fixed half-life. These worked answers correct the 20 problems of the Physics 11-12 worksheet on radioactive decay, from three kinds of radiation to a gram of radium: each solution shows the formula, the substitution and the final answer with its unit.
Every answer shows the formula, the substitution and the result with its unit. In nuclear equations, check the top row (nucleon number) and the bottom row (proton number) separately. In decay problems, turn the time into a number of half-lives before you do anything else.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB E.3-E.5
Problem 1 Three kinds of radiation
1. An alpha particle is a helium-4 nucleus (two protons, two neutrons); A falls by 4 and Z by 2.2. A neutron changes into a proton, with an electron and an antineutrino emitted; Z rises by 1 and A stays the same.
3. A and Z do not change; the nucleus only loses energy.
Problem 2 Radium alpha decay
A: 226 = A(radon) + 4, so A = 222. Z: 88 = 86 + 2, which balances. Radium-226 → radon-222 + alpha particle (helium-4).Problem 3 Cobalt-60 beta decay
In beta-minus decay A is unchanged, so A = 60, and Z rises by 1, so Z = 28. The daughter is nickel-60 (Z = 28).Problem 4 Three half-lives of cobalt-60
15.8 / 5.27 = 3.00 half-lives. Each half-life halves the activity: 8.0 → 4.0 → 2.0 → 1.0 (× 10⁶ Bq). Activity = 1.0 × 10⁶ Bq.Problem 5 Mass and energy units
1. 0.0305 × 931.5 = 28.4 MeV.2. 4.0 × 1.60 × 10⁻¹³ = 6.4 × 10⁻¹³ J.
Problem 6 Decay constant of caesium-137
T½ = 30.1 × 3.156 × 10⁷ = 9.50 × 10⁸ s. λ = 0.693 / T½ = 0.693 / 9.50 × 10⁸ = 7.30 × 10⁻¹⁰ s⁻¹.Problem 7 Activity of a radon sample
1. T½ = 3.82 × 86 400 = 330000 s, so λ = 0.693 / 330000 = 2.10 × 10⁻⁶ s⁻¹.2. A = λN = 2.10 × 10⁻⁶ × 2.0 × 10¹⁵ = 4.2 × 10⁹ Bq.
Problem 8 Phosphorus-32 tracer
Number of half-lives = 30 / 14.3 = 2.098. Fraction left = (½)2.098 = 0.234. 23.4 % remains.Problem 9 Reading a decay graph
1. The activity falls from 960 Bq to 480 Bq at 15 hours: T½ = 15 hours.2. 45 h is three half-lives: 960 → 480 → 240 → 120, so 120 Bq.
3. 240 Bq is two half-lives, 30 hours.
Problem 10 A PET tracer
A = A₀ (½)n, so n = log₂(2000 / 300) = 2.737 half-lives. t = 2.737 × 110 = 301 minutes (5.0 hours).Problem 11 Binding energy of the deuteron
Mass of parts = 1.007825 + 1.008665 = 2.016490 u. Δm = 2.016490 − 2.014102 = 0.002388 u. E = 0.002388 × 931.5 = 2.22 MeV; per nucleon 1.11 MeV.Problem 12 A milligram of mass
1. m = 1.0 × 10⁻⁶ kg; E = mc² = 1.0 × 10⁻⁶ × (3.00 × 10⁸)² = 9.0 × 10¹⁰ J.2. 9.0 × 10¹⁰ / 3.6 × 10⁶ = 25000 kWh.
3. 25000 / 5.0 = 5000 days = 13.7 years.
Problem 13 Energy of an alpha decay
Δm = 226.02541 − (222.01758 + 4.00260) = 226.02541 − 226.02018 = 0.00523 u. E = 0.00523 × 931.5 = 4.87 MeV, shared as kinetic energy of the alpha particle and the radon nucleus.Problem 14 Reading a decay series
1. Step 1: A falls by 4 and Z by 2, so alpha decay. Steps 2 and 3: A stays, Z rises by 1, so beta-minus decay. Step 4: A falls by 4, Z by 2, so alpha decay.2. Neutrons = A − Z = 238 − 92 = 146.
3. Both have Z = 92, so they are isotopes of uranium with different numbers of neutrons (146 and 142).
Problem 15 Spot the errors
1. Wrong: after two half-lives one quarter of the nuclei remain. 2. Wrong: the mass number is unchanged, because a neutron is replaced by a proton; the proton number rises by one. 3. Wrong: binding energy is the energy that must be supplied to pull the nucleus apart into separate nucleons. One quarter remains; A unchanged; energy supplied.Problem 16 Reading a binding energy chart
1. Iron-56, because it has the largest binding energy per nucleon (8.79 MeV), so the most energy is needed per nucleon to take it apart.2. Total = 56 × 8.79 = 492.2 MeV. Δm = 492.2 / 931.5 = 0.528 u.
3. Binding energy of helium-4 = 4 × 7.07 = 28.28 MeV; of two hydrogen-2 = 2 × 2 × 1.11 = 4.44 MeV. Energy released = 28.28 − 4.44 = 23.8 MeV.
Problem 17 Fission of uranium-235
1. A: 235 + 1 = 140 + 94 + x gives x = 2; Z: 92 = 54 + 38 balances. Two neutrons.2. Products: 140 × 8.29 + 94 × 8.59 = 1160.6 + 807.46 = 1968.1 MeV. Reactant: 235 × 7.59 = 1783.6 MeV. Energy released = 1968.1 − 1783.6 = 184 MeV.
3. N = (1000 / 235) × 6.02 × 10²³ = 2.56 × 10²⁴ nuclei. E = 2.56 × 10²⁴ × 184 × 1.60 × 10⁻¹³ = 7.54 × 10¹³ J.
Problem 18 Fusion fuel per kilogram
1. Binding energy of helium-4 = 4 × 7.07 = 28.28 MeV. Of H-2 = 2 × 1.11 = 2.22 MeV and of H-3 = 3 × 2.83 = 8.49 MeV, total 10.71 MeV. Energy released = 28.28 − 10.71 = 17.6 MeV.2. Mass of pair = 5.00 × 1.661 × 10⁻²⁷ = 8.30 × 10⁻²⁷ kg. Energy = 17.6 × 1.60 × 10⁻¹³ / 8.30 × 10⁻²⁷ = 3.39 × 10¹⁴ J per kg.
3. Ratio = 3.39 × 10¹⁴ / 8 × 10¹³ = 4.2, so fusion fuel gives about 4.2 times more energy per kilogram.
Problem 19 A straight line from a decay
1. A = A₀ e−λt, so ln A = ln A₀ − λt, which has the form y = c + mx with gradient −λ.2. ln 80.0 = 4.382; ln 31.8 = 3.459. Gradient = (3.459 − 4.382) / 8 = −0.115 h⁻¹.
3. λ = 0.115 h⁻¹; T½ = 0.693 / 0.115 = 6.0 hours.
Problem 20 A gram of radium
1. N = (1.00 / 226) × 6.02 × 10²³ = 2.66 × 10²¹.2. λ = 0.693 / (1600 × 3.156 × 10⁷) = 1.37 × 10⁻¹¹ s⁻¹. A = λN = 1.37 × 10⁻¹¹ × 2.66 × 10²¹ = 3.66 × 10¹⁰ Bq.
3. 400 / 1600 = 0.25 half-lives, so A = 3.66 × 10¹⁰ × 2−0.25 = 3.07 × 10¹⁰ Bq.
