Question 1Multiple choice · 1 markA* challenge
A nuclide has a half-life of 4 hours. What fraction of a sample decays in 10 hours?
- A 0.18
- B 0.82
- C 0.90
- D 1.3
Show answer
Answer: B
Free E.3 Radioactive decay questions for IB Physics (SL & HL), part of E Nuclear & quantum physics. Each question has the answer and the marking points an examiner would look for.
Generate a full E.3 Radioactive Decay worksheet →All E Nuclear & quantum physics questionsA nuclide has a half-life of 4 hours. What fraction of a sample decays in 10 hours?
Answer: B
Which statements about β− decay are correct?
I. A neutrino is emitted.
II. The emitted electrons have a continuous range of energies.
III. A neutron changes into a proton.
Answer: C
A sample of iodine-131 has a half-life of 8.0 days and an initial activity of 4.0 × 106 Bq.
Calculate the decay constant.
Calculate the activity after 20 days.
State why the activity of a real sample may not follow the decay law exactly.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.0 × 10−6 s−1 λ = ln 2 / T½ ✓ 1.0 × 10−6 s−1 ✓ | 2 |
| (b) | 7.1 × 105 Bq A = A₀e−λt or 2.5 half-lives ✓ 7.1 × 105 Bq ✓ | 2 |
| (c) | random; background radioactive decay is random (and background radiation is present) ✓ | 1 |
Carbon-14 is a β− emitter with a half-life of 5730 years. Living material has a constant ratio of C-14 to C-12.
Write the equation for the decay of carbon-14 (proton number 6).
Calculate the decay constant of carbon-14 in s−1.
The activity per gram of carbon in an ancient wooden object is 35% of that in living wood. Determine the age of the object.
Explain why carbon dating is unreliable for objects more than about 50 000 years old.
State why the activity of a sample decreases even though the half-life is constant.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 146C → 147N + 0−1e + ν̄ 146C → 147N + 0−1e ✓ + electron antineutrino ✓ | 2 |
| (b) | 3.83 × 10−12 s−1 λ = ln 2T½ ✓ 3.83 × 10−12 s−1 ✓ | 2 |
| (c) | 8679 years A = A0e−λt ✓ t = ln(1/0.35)λ ✓ 8679 years ✓ | 3 |
| (d) | activity too low the activity is then very small ✓ comparable with the background count, so large uncertainty ✓ | 2 |
| (e) | fewer nuclei remain activity ∝ number of undecayed nuclei, which decreases ✓ | 1 |
A student measures the count rate from a radioactive source with a Geiger–Müller tube. Each reading is the number of counts in one minute. The background count is 20 counts per minute.
| t / min | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| counts per minute | 835 | 654 | 526 | 423 | 343 | 265 | 216 |
| ln(corrected count rate) | 6.70 | 6.45 | 6.23 | 6.00 | 5.78 | 5.50 | 5.28 |
Explain why the background count must be subtracted.
The uncertainty in a count N is √N. Calculate the percentage uncertainty in the count at t = 0.
Use the graph to determine the decay constant and hence the half-life.
Explain why the points scatter about the line even with careful measurement.
Suggest how the uncertainty in each reading could be reduced.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | only the source counts decay the background is not from the source; only the source’s count rate decays exponentially ✓ | 1 |
| (b) | 3.46% √835 = 28.9 ✓ 3.46% ✓ | 2 |
| (c) | λ ≈ 0.237 min−1; T ≈ 2.93 min line of best fit drawn ✓ λ = −gradient ✓ λ = 0.237 min−1 ✓ T1/2 = ln 2λ = 2.93 min (accept ±8%) ✓ | 4 |
| (d) | random decay radioactive decay is random, so counts fluctuate statistically ✓ | 1 |
| (e) | count for longer count for longer (or repeat and average) so that N is larger and √N/N smaller ✓ | 1 |
The mass of a helium-4 nucleus is 4.001 506 u. The mass of a proton is 1.007 276 u and the mass of a neutron is 1.008 665 u.
Calculate the binding energy per nucleon of helium-4, in MeV.
Explain, with reference to binding energy per nucleon, why the fusion of light nuclei releases energy.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 7.1 MeV mass of nucleons = 4.031 882 u ✓ mass defect = 0.030 376 u ✓ binding energy = 0.030 376 × 931.5 = 28.3 MeV ✓ per nucleon = 7.07 MeV ✓ | 4 |
| (b) | BE per nucleon increases the product has a higher binding energy per nucleon ✓ so the total binding energy increases and the difference is released ✓ | 2 |
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