8.3 Radioactive decay and half-life — Practise

1. The activity

Answer: 15 days is three half-lives, so $A = \dfrac{640}{2^3} = 80\,\text{Bq}$.

2. The decay constant

Answer: $T_{1/2} = 6.0 \times 3600 = 2.16 \times 10^{4}\,\text{s}$, so $\lambda = \dfrac{\ln 2}{T_{1/2}} = 3.2 \times 10^{-5}\,\text{s}^{-1}$.

3. The activity

Answer: $A = \lambda N = 1.2 \times 10^{-5} \times 4.0 \times 10^{15} = 4.8 \times 10^{10}\,\text{Bq}$.

4. The number of nuclei

Answer: $N = \dfrac{A}{\lambda} = \dfrac{2.0 \times 10^{6}}{5.0 \times 10^{-7}} = 4.0 \times 10^{12}$.

5. The activity

Answer: $A = A_0 e^{-\lambda t} = 8.0 \times 10^{8} \times e^{-0.3465} = 5.7 \times 10^{8}\,\text{Bq}$.

6. The time

Answer: One sixteenth is $\left(\tfrac{1}{2}\right)^4$, so four half-lives pass: $4 \times 8.0 = 32\,\text{days}$.

7. Fluorine-18, $^{18}_{9}\text{F}$, decays by beta-plus emission. Write the balanced nuclear equation.

• The parent nucleus is $^{18}_{9}\text{F}$.
• The daughter has 18 nucleons and $9 - 1 = 8$ protons, which is oxygen-18.
• A positron $^{0}_{+1}\text{e}$ is emitted.
• The equation is $^{18}_{9}\text{F} \rightarrow {}^{18}_{8}\text{O} + {}^{0}_{+1}\text{e} + \nu$, with a neutrino to conserve energy and momentum.

8. Americium-241, $^{241}_{95}\text{Am}$, decays by alpha emission. Write the balanced nuclear equation.

• The parent nucleus is $^{241}_{95}\text{Am}$.
• An alpha particle $^{4}_{2}\text{He}$ is emitted.
• The daughter has $241 - 4 = 237$ nucleons and $95 - 2 = 93$ protons, which is neptunium-237.
• The equation is $^{241}_{95}\text{Am} \rightarrow {}^{237}_{93}\text{Np} + {}^{4}_{2}\text{He}$.

9. The corrected count rate

Answer: The corrected count rate is $\dfrac{720 - 60}{60} = 11\,\text{counts per second}$, or 11 Bq.

10. The half-life

Answer: To two significant figures the half-life is 5.3 days.

11. The half-life

Answer: $110 \times 60 = 6600\,\text{s}$.

12. The half-life

Answer: The gradient is $-\lambda$, so $T_{1/2} = \dfrac{\ln 2}{1.05 \times 10^{-4}} = 6.6 \times 10^{3}\,\text{s}$.

13. Explain why the activity of a sample is predictable even though the decay of a single nucleus is random.

• The decay of one nucleus is random: there is a fixed probability each second that it decays, but no way to say when.
• The decay constant is that probability per second.
• A real sample contains a very large number of nuclei, and the average behaviour of a large number is steady.
• The activity is an average over all the nuclei, so it follows a smooth exponential curve even though each decay is unpredictable.

14. The method and the half-life

• Find how many times the activity halves. From 640 Bq to 40 Bq is four halvings: 320, 160, 80, then 40.
• Divide the total time by the number of halvings: 20 days divided by 4 gives a half-life of 5 days.

Answer: Count the halvings: 640 to 40 is four halvings. Then divide 20 days by 4 to get a half-life of 5.0 days.