91Ó°ÊÓ

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

$R\mathit{=}\mathrm{λ}N$ (1)

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

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

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

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

$\mathrm{λ}=\frac{In2}{8.04d}=0.086d^{-1}=9.98×{10}^{-7}{s}^{-1}$

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:

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

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