Question

A certain radioactive element disintegrates with a decay constant of $7.9\times {10}^{-10}/sec$. At a given instant of time, if the activity of the sample is equal to $55.3\times {10}^{11}$ disintegration/ sec, then number of nuclei at that instant of time

• A. $7.0\times {10}^{21}$
• B. $4.27\times {10}^{13}$
• C. $4.27\times {10}^3$
• D. $6\times {10}^{23}$

Answer 1

The correct option is A $7.0\times {10}^{21}$

Given - Decay constant $\lambda =7.9\times {10}^{-10}/s;R=55.3\times {10}^{11}$disintegration/s

We have , $R=-\frac{dN}{dt}$ ...............eq1

and according to radioactive decay law ,

$\frac{dN}{dt}=-\lambda N$ ...............eq2

where $N=$ number of nuclei at any instant t ,

From eq1 and eq2 , we get

$R=\lambda N$

or $N=\frac{R}{\lambda }=\frac{55.3\times {10}^{11}}{7.9\times {10}^{-10}}=7.0\times {10}^{21}$