Answer
Number of ${^{131}_{53}I}$ atoms administrated $=3.7\times10^{15} atoms$
The answer is option $(d)$
Work Step by Step
A thyroid treatment typically administrates $3.7\: GBq$ of ${^{131}_{53}I}$.
The half-life of ${^{131}_{53}I}$ is $8.04 \:days$.
We know the activity of radioactive specimens is given by
$-\frac{dN}{dt} =\lambda N$
where $N$ is the number of radioactive specimens, $\lambda$ is the decay constant, and $t$ is the time. The negative sign implies that the number of radioactive specimens decreases with time.
Also, the half-time of a radioactive specimen is given by $T_{\frac{1}{2}}=\frac{ln2}{\lambda}$, which can be modified to get $\lambda$ as
$\lambda=\frac{ln2}{T_{\frac{1}{2}}}$
Substituting this relation for $\lambda$ in the activity equation, we get
$-\frac{dN}{dt} =\frac{ln2}{T_{\frac{1}{2}}} N$
since we are concerned with the number of radioactive specimens we can ignore the negative sign
$\frac{dN}{dt} =\frac{ln2}{T_{\frac{1}{2}}} N$
Rearranging for N gives
$N=\frac{T_{\frac{1}{2}}\times\frac{dN}{dt}}{ln2}$
Substituting values for the half-life and activity we will get the number of radioactive specimens used. However, before substituting we need to convert half-life in days and activity in GBq to seconds.
hald-life$=8.04 \: days=8.04*24*60*60=694656\: s$
activity$=3.7\: GBq=3.7\times10^{9} decays/s$
Therefore,
$N=\frac{694656\times 3.7\times10^{9}}{ln2}=3.7\times10^{15} atoms$
