Answer
See explanation.
Work Step by Step
The decay rate is cut by a factor of 2 after each half-life.
The activity obeys the same decay equation as N(t).
a. Calculate the activity after 24 days, which is 3 half-lives.
$$(325 Bq)2^{-24d/8.0d}=40.6 Bq$$
b. The activity is proportional to the number of radioactive nuclei. The percentage in the thyroid is $\frac{17.0}{40.6}=42\%$.
c. After emitting a $\beta^-$, the atomic number goes up by 1 and the mass stays the same. The nucleus $_{54}^{131}Xe$ is the result.
