Question

Question: How many radioactive disintegrations occur per minute in a $\mathbf{0}\mathbf{.}\mathbf{0010}\mathbf{}\mathbf{g}$ sample of ${}^{\mathbf{209}}\mathbf{Po}$that has been freshly separated from its decay products? The half-life of ${}^{\mathbf{209}}\mathbf{Po}$ is 103 years.

Short answer

The number of radioactive disintegrations per minute$=3.7\times {10}^{10}{\min }^{-}$

Step-by-step solution

Half-life

The time taken by a substance to decay to half of its initial amount is known as the half-life. Since, radioactive decay processes are first-order reactions the half-life is related to the rate constant for the radioactive decay process by the following expression:

${\mathrm{t}}_{1/2}=\frac{0.693}{\mathrm{k}}$

Determining the rate constant for the radioactive decay process

The ratio constant for the radioactive decay process is:

$$\begin{gathered} \mathrm{k}=\frac{0.693}{{\mathrm{t}}_{1/2}} \\ \mathrm{k}=\frac{0.693}{103}{\mathrm{yr}}^{-1} \\ \mathrm{k}=0.00672{\mathrm{yr}}^{-1} \end{gathered}$$

Determining the decay rate

Amount of sample present initially$=0.0010\mathrm{g}$

Number of moles of sample present initially$=\frac{0.0010}{209}\mathrm{mol}$

Number of atoms of sample present$=\frac{0.0010}{209}\times 6.023\times {10}^{23}\mathrm{atoms}$

The decay rate gives the measure of the number of disintegrations per unit time.

$$\begin{gathered} \mathrm{A}=\mathrm{kN} \\ =\frac{0.00672}{1\mathrm{yr}}\times \frac{1\mathrm{yr}}{365\times 24\times 60\min }\times \frac{0.0010\mathrm{g}}{209\mathrm{g}/\mathrm{mol}}\times 6.023\times {10}^{23}\mathrm{atoms} \\ =3.68\times {10}^{10}\mathrm{disintegrations}\mathrm{per}\mathrm{minute} \end{gathered}$$