8.3 Radioactive decay and half-life — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Keep the time and the half-life in the same unit.

Warm-up: remember

A quantity halves every 4.0 minutes. What fraction of it remains after 12 minutes?
Write 3.0 hours in seconds.
A sample contains $5.0 \times 10^{20}$ undecayed nuclei and has a decay constant of $2.0 \times 10^{-6}\,\text{s}^{-1}$ . Find its activity.

Answers

12 minutes is three half-lives, so the fraction is $\left(\tfrac{1}{2}\right)^3 = \dfrac{1}{8}$ .
$3.0 \times 3600 = 1.08 \times 10^{4}\,\text{s}$ .
$A = \lambda N = 2.0 \times 10^{-6} \times 5.0 \times 10^{20} = 1.0 \times 10^{15}\,\text{Bq}$ .

If question 1 goes wrong, count the halvings. If question 2 goes wrong, one hour is 3600 s. If question 3 goes wrong, the activity is the decay constant multiplied by the number of nuclei.

Practise with help

1. A source has an activity of 640 Bq. Its half-life is 5.0 days. Find its activity after 15 days.

Hint

Find how many half-lives fit into 15 days.

Next step

15 days is three half-lives. Divide 640 by $2^3$ .

Full working

A = \frac{A_0}{2^3} = \frac{640}{8} = 80\,\text{Bq}

The activity after 15 days is 80 Bq.

2. A radioisotope has a half-life of 6.0 hours. Find its decay constant in $\text{s}^{-1}$ .

Hint

Change the half-life to seconds first, then use the link with $\ln 2$ .

Next step

Write $T_{1/2} = 6.0 \times 3600 = 2.16 \times 10^{4}\,\text{s}$ , then $\lambda = \dfrac{\ln 2}{T_{1/2}}$ .

Full working

\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{2.16 \times 10^{4}} = 3.2 \times 10^{-5}\,\text{s}^{-1}

The decay constant is $3.2 \times 10^{-5}\,\text{s}^{-1}$ .

3. A sample contains $4.0 \times 10^{15}$ nuclei of a radioisotope whose decay constant is $1.2 \times 10^{-5}\,\text{s}^{-1}$ . Find its activity.

Hint

The activity is the decay constant multiplied by the number of undecayed nuclei.

Next step

Write $A = \lambda N$ with the two values given.

Full working

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

The activity is $4.8 \times 10^{10}\,\text{Bq}$ .

4. A source has an activity of $2.0 \times 10^{6}\,\text{Bq}$ and a decay constant of $5.0 \times 10^{-7}\,\text{s}^{-1}$ . Find the number of radioactive nuclei in the source.

Hint

Rearrange the activity equation to make the number of nuclei the subject.

Next step

Write $N = \dfrac{A}{\lambda}$ .

Full working

N = \frac{A}{\lambda} = \frac{2.0 \times 10^{6}}{5.0 \times 10^{-7}} = 4.0 \times 10^{12}

The source contains $4.0 \times 10^{12}$ radioactive nuclei.

5. A source has an initial activity of $8.0 \times 10^{8}\,\text{Bq}$ and a decay constant of $1.05 \times 10^{-4}\,\text{s}^{-1}$ . Find its activity after 3300 s.

Hint

Use the exponential form $A = A_0 e^{-\lambda t}$ .

Next step

Work out the exponent $-\lambda t$ first, then find $e^{-\lambda t}$ .

Full working

\begin{aligned} \lambda t &= 1.05 \times 10^{-4} \times 3300 = 0.3465 \\ A &= A_0 e^{-\lambda t} = 8.0 \times 10^{8} \times e^{-0.3465} \\ &= 8.0 \times 10^{8} \times 0.707 \\ &= 5.7 \times 10^{8}\,\text{Bq} \end{aligned}

The activity after 3300 s is about $5.7 \times 10^{8}\,\text{Bq}$ .

6. Americium-241, $^{241}_{95}\text{Am}$ , decays by emitting an alpha particle. Write the balanced nuclear equation.

Hint

An alpha particle is a helium nucleus. Subtract its nucleon number and its proton number from those of americium.

Next step

Find the new nucleon number, $241 - 4$ , and the new proton number, $95 - 2$ . Then name the element with that proton number.

Full working

^{241}_{95}\text{Am} \rightarrow {}^{237}_{93}\text{Np} + {}^{4}_{2}\text{He}

The nucleon numbers balance: $241 = 237 + 4$ . The proton numbers balance: $95 = 93 + 2$ .

7. Carbon-14, $^{14}_{6}\text{C}$ , decays by beta-minus emission. Write the balanced nuclear equation.

Hint

Beta-minus emission is an electron leaving the nucleus. The nucleon number does not change, and the proton number rises by one.

Next step

The new proton number is $6 + 1 = 7$ , which is nitrogen.

Full working

^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{0}_{-1}\text{e} + \bar{\nu}

The antineutrino $\bar{\nu}$ carries away energy and momentum. The nucleon number stays at 14, as it must.

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

Hint

Think about the number of nuclei in a sample and what a probability means over a large number of trials.

Next step

Contrast the fate of one nucleus with the average behaviour of many millions of nuclei.

Full working

The decay of one nucleus is random: there is a fixed probability each second that it decays, but no way to say when it will. A real sample contains many millions of nuclei. The number that decay in the next second is an average over all of them, and averages of large numbers are steady. This is why the activity follows a smooth exponential curve even though each individual decay is unpredictable.