Question

The rate of decay of a radioactive sample is $3.02\times {10}^6$ dpm at time 10 min and $1.20\times {10}^6$ dpm at a time 20 min. Evaluate the average life of sample in min.

(Write the answer to the nearest integer)

Answer 1

The rate of disintegration r, is related to the decay constant $\lambda$ through the expression $r=\lambda N$
Here N represents the amount of radioactive species present at time t.
The above expression for two different times $t_1=10$ min and $t_2=$ 20 min becomes

$r_1=\lambda .N_1,r_2=\lambda .N_2$

The ratio of the rate of disintegration at 10 min to the rate of disintegration at 20 minutes becomes

$\frac{r_1}{r_2}=\frac{N_1}{N_2}=\frac{3.02\times {10}^6}{1.20\times {10}^6}=2.52$

The integrated form of rate law expression for the first order decay process is $t=\frac{2.303}{\lambda }log\frac{N_0}{N}$

At the end of 10 minutes, $10=\frac{2.303}{\lambda }\log \frac{N_0}{N_1}$ ... (ii)

At the end of 20 min, $20=\frac{2.303}{\lambda }\log \frac{N_0}{N_2}$ ... (iii)

Subtract equation (ii) from equation (iii)

$20-10=\frac{2.303}{\lambda }[\log \frac{N_0}{N_2}-\log \frac{N_0}{N_1}]$

$10=\frac{2.303}{\lambda }\left[\log \frac{N_1}{N_2}\right]=\frac{2.303}{\lambda }\log 2.52$

The decay constant $\lambda =0.092mi{n}^{-1}$

The average life of the sample is $T_{av}=\frac{1}{\lambda }=\frac{1}{0.092}=10.87$ min