A schoolroom polonium atomic battery

Problem

Polonium-210 ($^{210}\text{Po}$) has a half-life $\tau_{1/2} = 138\ \text{days}$ and undergoes $\alpha$ decay releasing an energy $E_\alpha = 5{,}3\ \text{MeV}$ per decay. You have a sample of mass $m = 1\ \text{mg}$ of $^{210}\text{Po}$. How much thermal power does the sample initially supply? Use $N_A = 6{,}022 \cdot 10^{23}$, $1\ \text{MeV} = 1{,}6 \cdot 10^{-13}\ \text{J}$.

Solution

Number of atoms. With molar mass $A = 210\ \text{g/mol}$ and $m = 1 \cdot 10^{-3}\ \text{g}$:

$N = \frac{m}{A}\,N_A = \frac{1 \cdot 10^{-3}}{210} \cdot 6{,}022 \cdot 10^{23} \approx 2{,}87 \cdot 10^{18}$

Decay constant. Converting the half-life to seconds ($138 \cdot 86400 \approx 1{,}192 \cdot 10^{7}\ \text{s}$):

$\lambda = \frac{\ln 2}{\tau_{1/2}} = \frac{0{,}693}{1{,}192 \cdot 10^{7}} \approx 5{,}81 \cdot 10^{-8}\ \text{s}^{-1}$

Initial activity.

$A = \lambda N = 5{,}81 \cdot 10^{-8} \cdot 2{,}87 \cdot 10^{18} \approx 1{,}67 \cdot 10^{11}\ \text{s}^{-1}$

Thermal power. Each decay deposits $E_\alpha$; the power is activity times energy:

$P = A \cdot E_\alpha = 1{,}67 \cdot 10^{11} \cdot 5{,}3 \cdot 1{,}6 \cdot 10^{-13}\ \text{W} \approx 0{,}14\ \text{W}$

$\ev{P \approx 0{,}14\ \text{W}}$

It is a modest but continuous power: this very principle (radioisotope sources) powers space probes far from the Sun.

Connections

Topics: Nuclear physics Concepts: Radioactive decay