3.8.1.4
Radioisotope power systems convert decay heat to electricity
Radioactive Decay & Half-Life — AQA A-Level Physics
- Radioactive decayThe spontaneous and random disintegration of an unstable nucleus, emitting radiation (alpha, beta, or gamma) to become more stable. releases energyThe capacity to do work. Measured in joules (J). as heat. Radioisotope powerThe rate of energy transfer. Measured in watts (W). systems convert this heat to electrical powerThe rate of energy transfer. Measured in watts (W)..
- Used to powerThe rate of energy transfer. Measured in watts (W). space probes and satellites in locations too far from the Sun for solar panels.
- Plutonium-238 is the typical fuel: 1 g generates about 500 mW.
- The power output decays exponentially, following the same law as activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second.: $P = P_{0} e^{-\lambda t}$.
Worked Example
A space probe carries 4.0 kg of Pu-238 (half-lifeThe time taken for half the number of radioactive nuclei in a sample to decay, or for the activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second. to halve. 87.7 years, 5.5 MeV per alpha decay, 32% conversion efficiencyThe ratio of useful energyThe capacity to do work. Measured in joules (J). output to total energyThe capacity to do work. Measured in joules (J). input, expressed as a fraction or percentage.). It needs at least 0.4 kW. Estimate the operating time.
Show Solution
1
Find the initial number of nuclei
$$N_0 = \frac{4000 \times 6.02 \times 10^{23}}{238} = 1.012 \times 10^{25}$$
2
Find the initial activityThe number of nuclear decays per unit time. Measured in becquerels (Bq), where 1 Bq = 1 decay per second.
$$A_0 = \frac{N_0 \ln 2}{t_{1/2}} = \frac{1.012 \times 10^{25} \times \ln 2}{87.7 \times 365 \times 24 \times 3600} = 2.54 \times 10^{15} \text{ Bq}$$
3
Find the initial power output
$$P_0 = A_0 \times E = 2.54 \times 10^{15} \times 5.5 \times 10^6 \times 1.6 \times 10^{-19} = 2231 \text{ W}$$
Electrical: $P_0 = 2231 \times 0.32 = 714$ W
4
Find the time until power drops to 400 W
$$t = -\frac{t_{1/2}}{\ln 2} \ln\left(\frac{P}{P_0}\right) = -\frac{87.7}{\ln 2} \ln\left(\frac{400}{714}\right) = 73.3 \text{ years}$$
Answer
The source supplies power for approximately 73 years.