Question

After a certain lapse of time, the fraction of radioactive polonium undecayed is found to be $12.5\%$ of initial quantity. The duration of this time-lapse is, (if the half-life of polonium is $138$ days)

• A. $407$ days
• B. $410$ days
• C. $421$ days
• D. $414$ days

Answer 1

The correct option is D $414$ days

Given, $\frac{N}{N_0}=12.5\%;T_{\frac{1}{2}}=138\text{days}$

$\Rightarrow \frac{N}{N_0}=12.5\%=\frac{12.5}{100}=\frac{1}{8}$

Number atoms undeacyed after, $n{T}_{\frac{1}{2}}$ is,

$N=N_0\left(\frac{1}{2^n}\right)\Rightarrow \frac{N}{N_0}=\left(\frac{1}{2^n}\right)$

$\therefore \frac{1}{8}=\frac{1}{2^n}\Rightarrow n=3$

$\therefore t=n{T}_{\frac{1}{2}}=3\times 138=414\text{days}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.