Answer
380 hours, or 16 days.
Work Step by Step
The original activity is $R_0$ and the activity after 31.0 hours is $ (0.945)R_0$.
Use equation 30-5.
$$ (0.945)R_0=R_0e^{-\lambda t}$$
$$ ln 0.945= -\lambda t=-\frac{ln2}{T_{1/2}}t$$
Solve for the half-life.
$$ T_{1/2}= -\frac{ln2}{ ln 0.945}(31.0h)=380h$$
The half-life is approximately 16 days.
