Question

How many ${}^{\mathbf{131}}\mathit{I}$atoms are administered in a typical thyroid cancer treatment?

(a) $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}$;

(b) $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}$;

(c) $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}$;

(d) $\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}$.

Short answer

The right choice is (d) $3.71\times {10}^{15}$.

Step-by-step solution

Known

The relation between the activity R of a sample and the number of radioactive nuclei N in the sample is given by:

$R=\lambda N$ ..............(i)

The relation between the half-life $T_{1/2}$and the decay constant $\lambda$is given by

$\lambda =\frac{In2}{T_{1/2}}$ ............(ii)

Givens

The half-life of ${}^{131}I$is role="math" localid="1664207830185" $T_{1/2}=8.04days$and its activity in a typical thyroid cancer treatment is$R=3.70\times {10}^{9}Bq$.

Calculations

We substitute our value for $T_{1/2}$into equation (ii), so we get the decay constant of ${}^{131}I$:

$$\begin{gathered} \lambda =\frac{In2}{8.04d} \\ =0.086d^{-1} \\ =9.98\times {10}^{-7}{s}^{-1} \end{gathered}$$

Now, we plug our values for $\lambda$and R into equation (i) and evaluate for N, so we get the number of ${}^{131}I$atoms administered in a typical thyroid cancer treatment:

$$\begin{gathered} N=\frac{R}{\lambda } \\ =\frac{3.70\times {10}^{9}Bq}{9.98\times {10}^{-7}{s}^{-1}} \\ =3.71\times {10}^{15}atom \end{gathered}$$

Therefore, the right choice is (d) $3.71\times {10}^{15}$.