Question

How many 131I atoms are administered in a typical thyroid cancer treatment? (a) 4.2 * 1010; (b) 1.0 * 1012; (c) 2.5 * 1014; (d) 3.7 * 1015.

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:

$R\mathit{=}\lambda N$ (1)

the relation between the half-life and the decay constant is given by

$\lambda =\frac{\ln 2}{T_{1/2}}$ (2)

Givens

The half-life of ${}^{131}l$is$T_{1/2}=8.04days$ and its activity in a typical thyroid cancer treatment is $R=3.70x{10}^{9}Bq$.

Calculations

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

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

Now, we plug our values for and R into equation (1) and evaluate for N, so we get the number of atoms administered in a typical thyroid cancer treatment:

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

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