$A$ radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to $(a)$ $3.125\%$ $(b)$ $1\%$ of its original value?

(N/A) The activity $A$ of a radioactive sample follows the law $A = A_0 e^{-\lambda t}$,where $A_0$ is the initial activity and $\lambda = \frac{\ln 2}{T} \approx \frac{0.693}{T}$.
$(a)$ Given $\frac{A}{A_0} = 3.125\% = \frac{3.125}{100} = \frac{1}{32}$.
Since $\frac{1}{32} = (\frac{1}{2})^5$,we have $e^{-\lambda t} = (\frac{1}{2})^5$.
Taking the natural logarithm on both sides: $-\lambda t = 5 \ln(\frac{1}{2}) = -5 \ln 2$.
Substituting $\lambda = \frac{\ln 2}{T}$,we get $-\frac{\ln 2}{T} t = -5 \ln 2$.
Thus,$t = 5T$ years.
$(b)$ Given $\frac{A}{A_0} = 1\% = \frac{1}{100}$.
$e^{-\lambda t} = \frac{1}{100} \implies -\lambda t = \ln(10^{-2}) = -2 \ln 10$.
$t = \frac{2 \ln 10}{\lambda} = \frac{2 \times 2.303}{0.693/T} = \frac{4.606}{0.693} T \approx 6.646T$ years.