Question

Determine the amount of $\begin{array}{c}210\\ 84\end{array}Po$ (in micrograms) necessary to provide a source of alpha particles of $5$ millicurie strength. The half life of polonium is 138 days.

Answer 1

We are given that rate of decay, $R=5\text{mCi}=5(3.7\times {10}^7\text{dis/s)}=18.5\times {10}^7\text{dis/s}$
$T_{1/2}=138\text{days}=138(24\times 60\times 60)=1.192\times {10}^7\text{s}$.
Thus, $\lambda =\frac{ln2}{T_{1/2}}=\frac{0.693}{1.192\times {10}^{7}s}=0.581\times {10}^{-7}{s}^{-1}$
As $R=\lambda N,N=\frac{R}{\lambda }=\frac{18.5\times {10}^7\text{dis/s}}{0.581\times {10}^{-7}{s}^{-1}}=31.84\times {10}^{14}\text{atoms}$
Since $6.023\times {10}^{23}$ atoms of $\begin{array}{c}210\\ 84\end{array}Po$ have a mass of 210 g, mass of $31.84\times {10}^{14}$ atoms of $\begin{array}{c}210\\ 84\end{array}Po$
$=\left(\frac{210}{6.023\times {10}^{23}}\right)(31.84\times {10}^{14})g=1.11\times {10}^{-6}g$

Why this Question? Caution: The original unit for measuring the amount of radioactivity was the curie (Ci)–first defined to correspond to one gram of radium-226 and more recently defined as 1 curie $=3.7\times {10}^{10}$ radioactive decays per second.