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?

Half-life of the radioactive isotope $= T$ years Original amount of the radioactive isotope $=N_{0}$

(a) After decay, the amount of the radioactive isotope $= N$

It is given that only $3.125 \%$ of $N_{0}$ remains after decay.

Hence, we can write:

$\frac{N}{N_{0}}=3.125 \%=\frac{3.125}{100}=\frac{1}{32}$

But $\frac{N}{N_{0}}=e^{-\lambda 1}$

Where, $\lambda=$ Decay constant

$ t =$ Time

$\therefore-\lambda t=\frac{1}{32}$

$-\lambda t=\ln l-\ln 32$

$-\lambda t=0-3.4657$

$t=\frac{3.4657}{\lambda}$

since $\lambda=\frac{0.693}{T}$

$\therefore t=\frac{\frac{3.466}{0.693}}{T} \approx 5 T$ Years

Hence, the isotope will take about $5 T$ years to reduce to $3.125 \%$ of its original value.

(b) After decay, the amount of the radioactive isotope $= N$ It is given that only $1 \%$ of $No$ remains after decay. Hence, we can write

$\frac{N}{N_{0}}=1 \%=\frac{1}{100}$

But $\frac{N}{N_{0}}=e^{-\lambda t}$

$\therefore e^{-\lambda t}=\frac{1}{100}$

$-\lambda t=\ln 1-\ln 100$

$-\lambda t=0-4.6052$

$t=\frac{4.6052}{\lambda}$

since, $\lambda=0.639 / T$

$\therefore t=\frac{4.6052}{\frac{0.693}{T}}=6.645 T$ years

Hence, the isotope will take about $6.645\, T$ years to reduce to $1 \%$ of its original value.