The problems are on the nuclear decay worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Atoms have discrete energy levels, so they emit and absorb photons of fixed energy equal to the gap between levels. These worked answers correct the 20 problems of the AP Physics 2 worksheet on nuclear decay, from counting protons and neutrons to half-life from lab data: each solution shows the formula, the substitution and the final answer with its unit.
Each answer lists the relationship, the substitution and the result with its unit. For nuclear equations, check that the top numbers and the bottom numbers each add up on both sides before you accept an answer.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 15
Problem 1 Counting protons and neutrons
1. Protons = Z = 6, neutrons = 14 − 6 = 8.2. Protons = 92, neutrons = 235 − 92 = 143.
Problem 2 Hydrogen energy levels
1. −13.6 eV, −3.40 eV, −1.51 eV, −0.85 eV.2. The electron must reach 0 eV, so the energy needed is 3.40 eV.
Problem 3 A visible hydrogen line
1. ΔE = −0.85 − (−3.40) = 2.55 eV.2. λ = 1240 / 2.55 = 486 nm, a blue-green line.
Problem 4 Absorbing a photon
1. ΔE = −3.40 − (−13.6) = 10.2 eV.2. No. An absorbed photon must match a gap exactly, and 11.0 eV does not match 10.2 eV (n = 2) or 12.1 eV (n = 3).
Problem 5 Alpha decay equation
1. ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He.2. A = 222, Z = 86.
Problem 6 Beta decay equation
1. ⁶⁰₂₇Co → ⁶⁰₂₈Ni + ⁰₋₁e + ν̄ (antineutrino).2. Z = 28: a neutron changes into a proton.
Problem 7 Counting half-lives
1. 15 days is 3 half-lives: 64 → 32 → 16 → 8.0 g.2. 20 days is 4 half-lives: 4.0 g.
3. Fraction = (½)³ = 1/8.
Problem 8 Reading a decay graph
1. The rate halves from 800 to 400 in 6 hours, so T = 6.0 h.2. 30 h is 5 half-lives: 800 / 32 = 25 per minute.
3. Each half-life removes half of what is left, and decay is random, so a few nuclei always remain.
Problem 9 Mass to energy
1. E = mc² = 1.0 × 10⁻³ × (3.00 × 10⁸)² = 9.0 × 10¹³ J.2. 9.0 × 10¹³ / 4.2 × 10⁹ = 2.1 × 10⁴, about 21 000 tonnes of TNT.
Problem 10 Unit conversion for mass defect
1. E = 0.0050 × 931.5 = 4.7 MeV.2. 4.66 MeV × 1.60 × 10⁻¹³ = 7.5 × 10⁻¹³ J.
Problem 11 Identify the radiation
1. Paper stops almost nothing, so there is no significant alpha. Aluminium cuts the rate a lot, so beta is present. Lead removes the rest, so gamma is present too.2. Beta ≈ (1190 − 400) / 1200 ≈ 66 %; gamma ≈ 400 / 1200 ≈ 33 %.
Problem 12 Carbon dating
1. 12.5 % is (½)³, so the age is 3 × 5730 = 17 190 years.2. 30 % lies between 50 % (one half-life) and 25 % (two half-lives), so the age is between 5730 and 11 460 years.
3. t = T ln(0.30) / ln(0.5) = 5730 × 1.737 = 9.95 × 10³ years.
Problem 13 Why spectra are lines
1. Electrons in the gas atoms can only be at fixed energy levels, so transitions release photons with only certain energies, giving certain wavelengths.2. The cool gas absorbs photons whose energies match its level gaps, removing those wavelengths from the continuous light and leaving dark lines at the same wavelengths as the emission lines.
Problem 14 Lyman and Balmer lines
1. 10.2 eV and 1.89 eV.2. 1240 / 10.2 = 122 nm and 1240 / 1.89 = 656 nm.
3. The 122 nm line is ultraviolet; the 656 nm line is visible (red).
Problem 15 Binding energy of the deuteron
1. Δm = 1.007276 + 1.008665 − 2.013553 = 0.002388 u.2. E = 0.002388 × 931.5 = 2.22 MeV.
3. 2.22 / 2 = 1.11 MeV per nucleon.
Problem 16 Energy from radium decay
1. Δm = 226.025410 − (222.017578 + 4.002603) = 0.005229 u.2. E = 0.005229 × 931.5 = 4.87 MeV.
3. The alpha receives 222 / 226 of the energy: 4.87 × 222 / 226 = 4.78 MeV.
Problem 17 Activity of a medical tracer
1. T = 6.0 × 3600 = 2.16 × 10⁴ s.2. A = 0.693 / 2.16 × 10⁴ × 2.0 × 10¹⁰ = 6.4 × 10⁵ Bq.
3. 18 h is 3 half-lives, so A = 6.4 × 10⁵ / 8 = 8.0 × 10⁴ Bq.
Problem 18 Power of the Sun
1. Δm = 4 × 1.007825 − 4.002603 = 4.031300 − 4.002603 = 0.028697 u.2. E = 0.028697 × 931.5 = 26.7 MeV.
3. m = P / c² = 3.8 × 10²⁶ / 9.0 × 10¹⁶ = 4.2 × 10⁹ kg per second.
Problem 19 Lines from a hypothetical atom
1. Transitions: 4→3, 4→2, 4→1, 3→2, 3→1 and 2→1, giving 6 lines.2. The smallest gap is 4→3, 1.0 eV, so λ = 1240 / 1.0 = 1240 nm (infrared).
3. The largest gap is 4→1, 7.5 eV, so λ = 1240 / 7.5 = 165 nm.
Problem 20 Half-life from lab data
1. Half of 480 is 240, which occurs between 2 min (276) and 3 min (209): interpolating gives about 2.5 min.2. 120 is a quarter of 480, which means two half-lives in 5 min, so T = 2.5 min.
3. Background counts do not come from the source and do not decay, so they would distort the ratio of counts and give a wrong half-life.
