Question

The half life period of ${}_{58}^{141}Ce$ is $13.11$ days. It is a $\beta$-particle emitter and the average energy of the $\beta$-particle emitted is $0.442MeV$. What is the total energy emitted per second in watts by $10mg$ of ${}_{58}^{141}Ce$?

Answer 1

Rate of disintegrations per sec $=\lambda \times$ Number of atoms
$=\frac{0.693}{13.11\times 24\times 60\times 60}\times \frac{6.023\times {10}^{23}}{141}\times 0.01$
Total $\beta$-particles emitted $=2.61\times {10}^{13}$
Total energy emitted $=2.61\times {10}^{13}\times 0.442=1.1536\times {10}^{13}MeV$
Energy in erg $=(1.1536\times {10}^{13})(1.6\times {10}^{-6})$
Energy in watt $=\frac{1.1536\times {10}^{13}\times 1.6\times {10}^{-6}}{{10}^7}=1.84$ watt