Answer
a) ${\bf 58.05}\;\rm h$
b) ${\bf 188.5}\;\rm h$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The activity of a radioactive sample decays exponentially, so
$$
R = R_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
where:
- $ R = 95 \, \text{mCi} $: Activity after 16 hours.
- $ R_0 = 115 \, \text{mCi} $: Initial activity.
- $ t = 16 \, \text{hours} $.
- $ t_{1/2} $: Half-life of the tracer (unknown).
Solve for $ t_{1/2} $:
$$ \frac{R}{R_0} = \left( \frac{1}{2} \right)^{t / t_{1/2}} $$
Take the Natural Logarithm of Both Sides:
$$\ln\left[ \frac{R}{R_0} \right]=\ln\left[ \frac{1}{2} \right]^{t / t_{1/2}}$$
$$\ln\left[ \frac{R}{R_0} \right]=\dfrac{t}{t_{_{1/2}}}\ln\left[ \frac{1}{2} \right]$$
Thus,
$$t_{_{1/2}}=t\;\dfrac{ \ln\left[ \frac{1}{2} \right]} {\ln\left[ \frac{R}{R_0} \right]}$$
Substitute the known values:
$$t_{_{1/2}}=(16)\;\dfrac{ \ln\left[ \dfrac{1}{2} \right]} {\ln\left[ \dfrac{95}{115} \right]}$$
$$t_{_{1/2}}=\color{red}{\bf 58.05}\;\rm h$$
Thus, the half-life of the tracer is approximately 58.0 hours.
$$\color{blue}{\bf [b]}$$
Now we need to find the time $ t $ it takes for the activity to decay from $ 0.95 R_0 $ (95 mCi) down to the lowest usable level, 10 mCi.
$$
R = R_0 \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
where:
- $ R = 10 \, \text{mCi} $
- $ R_0 = 95 \, \text{mCi} $
- $ t_{1/2} = 58.05 \, \text{hours} $
Solve for $ t $:
$$
\frac{R}{R_0} = \left( \frac{1}{2} \right)^{t / t_{1/2}}
$$
Take the Natural Logarithm of Both Sides:
$$\ln\left[ \frac{R}{R_0} \right]=\ln\left[\left( \frac{1}{2} \right)^{t / t_{1/2}}\right]$$
$$\ln\left[ \frac{R}{R_0} \right]=\dfrac{t}{t_{_{1/2}}}\ln\left[ \frac{1}{2} \right]$$
$$t=t_{_{1/2}}\;\dfrac{\ln\left[ \frac{R}{R_0} \right]}{ \ln\left[ \frac{1}{2} \right]} $$
Substitute the known:
$$t=(58.05 )\;\dfrac{\ln\left[ \dfrac{10}{95} \right]}{ \ln\left[ \dfrac{1}{2} \right]} $$
$$t=\color{red}{\bf 188.5}\;\rm h$$
