At a certain instant a piece of radioactive material contains ${10^{12}}$ atoms. The half life of material is 40 days. Calculate the number of disintegrations in one second

Solution:

Half Life (t_1/2)= 40 days

Now,

Decay Constant =$\lambda  = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$

$ = \frac{{0.693}}{{40{\rm{ }}days}}$

$\begin{array}{l} = \frac{{0.693}}{{40 \times 24 \times 60 \times 60}}\\ = 2.005 \times {10^{ - 7}}\end{array}$

N₀ = 10¹² atoms

And,

The Disintegration rate in 1 second 

= Rate of Disintegration 

=$\frac{{ - dN}}{{dt}} = \lambda {N_0}$

$\begin{array}{l} = 2.005 \times {10^{ - 7}} \times {10^{ - 7}}\\ = 2.005 \times {10^5}\end{array}$

The number of disintegrations in one second is $2.005 \times {10^{ - 5}}$