$1\text{g}$ of a radioactive substance disintegrates at the rate of $3.7\times {10}^{10}$ disintegrations per second. The atomic mass of the substance is $226$. Calculate its mean life.

The correct option is B $7.2\times {10}^{10}\text{s}$
Given:
$\frac{dN}{dt}=3.7\times {10}^{10}{\text{disintegrations sec}}^{-1}$
$A=226\text{g}$ and $m=1\text{g}$

Number of nuclei in $1\text{gm}$

Avogadro's number, $N_A=6.023\times {10}^{23}\text{atoms}$

Now, total number of nuclei,

$N=\left(\frac{N_A}{A}\right)m=\frac{6.023\times {10}^{23}}{226}\times 1=2.67\times {10}^{21}$

From Decay law,

$\frac{dN}{dt}=-\lambda N$

Considering only magnitude,

$\Rightarrow 3.7\times {10}^{10}=\lambda \times 2.67\times {10}^{21}$

$\Rightarrow \lambda =\frac{3.7\times {10}^{10}}{2.67\times {10}^{21}}=1.39\times {10}^{-11}{\text{s}}^{-1}$

Mean life, $\tau =\frac{1}{\lambda }$

$\tau =\frac{1}{1.39\times {10}^{-11}}=0.719\times {10}^{11}$

$=7.19\times {10}^{10}\text{s}\approx 7.2\times {10}^{10}\text{s}$.

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

Why this question? Hint: In order to solve this problem, first calculate the number of nuclei.