Answer
a) ${\bf 236} \, \rm{\mu g}$
b) ${\bf 140.3} \, \rm{\mu g}$
c) ${\bf 0.775} \, \rm{\mu g}$
Work Step by Step
We know that the radioactive decay is given by:
$$
N = N_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
where:
- $ N $ is the remaining quantity after time $ t $,
- $ N_0 $ is the initial quantity ($250\;\rm \mu g $ in this case),
- $ t $ is the elapsed time,
- $ t_{1/2} $ is the half-life of the isotope (12 days for $^{131}\text{Ba}$, see appendix C).
Now, we can find the remaining mass for each specified time period.
$$\color{blue}{\bf [a]}$$
After $t=$1 Day:
$$
N = 250 \left( \frac{1}{2} \right)^{1 / 12}
\approx \color{red}{\bf 236} \, \rm{\mu g}
$$
$$\color{blue}{\bf [b]}$$
After $t=$10 Day:
$$
N = 250 \left( \frac{1}{2} \right)^{10 / 12}
\approx \color{red}{\bf 140.3} \, \rm{\mu g}
$$
$$\color{blue}{\bf [c]}$$
After $t=$100 Day:
$$
N = 250 \left( \frac{1}{2} \right)^{100 / 12}
\approx \color{red}{\bf 0.775} \, \rm{\mu g}
$$
