Question

A radioactive sample $A$ decays to form $B$.$\left(A\to B\right)$. The half life of $A$ is $t_{\frac{1}{2}}=8\text{years}$. Only $A$ is present at $t=0$. Find the time at which if we are able to pick one atom out of the sample, then probability of getting $B$ is $15$ times of getting $A$.

• A. $32\text{Years}$
• B. $18\text{Years}$
• C. $12\text{Years}$
• D. $4\text{Years}$

Answer 1

The correct option is A $32\text{Years}$

$\begin{array}{ccccc}& & \text{A}& \to & \text{B}\\ \text{at}& t=0& N_0& & 0\\ \text{at}& t=t& N& & N_0-N\end{array}$

Probability of getting $A$,

$P_A=\frac{N}{N_0}$

Probability of getting $B$,

$P_B=\frac{N_0-N}{N_0}$

Given that, $P_B=15P_A$

$\Rightarrow \frac{N_0-N}{N_0}=15\frac{N}{N_0}$

$\Rightarrow N_0=16N$

$\Rightarrow \frac{N}{N_0}=\frac{1}{16}$

We know that,

$\frac{N}{N_0}=\frac{1}{2^{\frac{t}{t_{1/2}}}}$

$\Rightarrow \frac{1}{16}=\frac{1}{2^{\frac{t}{t_{1/2}}}}$

$\Rightarrow \frac{1}{2^4}=\frac{1}{2^{\frac{t}{t_{1/2}}}}$

$\Rightarrow \frac{t}{t_{1/2}}=4$

$\Rightarrow t=4\times 8=32\text{Years}$

Hence, option $\left(A\right)$ is correct.

Why this question? Probability of survival of nuclei at time t, $P=\frac{\text{Number of radioactive nuclei at time t}}{\text{Initial number of radioactive nuclei}}$