Question

A sample of radioactive material $A$, having an activity of $10\text{mCi}$ $(1\text{Ci}=3.7\times {10}^{10}\text{decays/sec})$ has twice the number of nuclei as another sample of a different radioactive material $B$, which has an activity of $20\text{mCi}$. The correct choices for half-lives of $A$ and $B$ would be, respectively,

• A. $20$ days and $5$ days
• B. $10$ days and $20$ days
• C. $5$ days and $10$ days
• D. $20$ days and $10$ days

Answer 1

The correct option is A $20$ days and $5$ days

The activity is given by,
$A=\lambda N$

For substance $A$:-

$A_A={\lambda }_A{N}_A\Rightarrow 10=N_A{\lambda }_A$

As, $N_A=2N_B$

$\therefore 10=2N_B{\lambda }_A......\left(i\right)$



For B:-

$A_B=N_B{\lambda }_B$

$20=N_B{\lambda }_B......\left(ii\right)$

Dividing $\left(i\right)$ by $\left(ii\right)$, we get,

$\frac{10}{20}=\frac{2N_B{\lambda }_A}{N_B{\lambda }_B}$

$\therefore {\lambda }_B=4{\lambda }_A$

Since, $T\propto \frac{1}{\lambda }$

$\frac{T_A}{T_B}=\frac{{\lambda }_B}{{\lambda }_A}=\frac{4{\lambda }_A}{{\lambda }_A}$

$\therefore T_A=4T_B$

Considering all the options, the only option that satisfies the above condition is $T_B=5$ days and $T_A=20$ days

Hence, $\left(D\right)$ is the correct answer.