Question

The half life of a radioactive sample is $T$. If the activities of the sample at time t${}_1$ and t${}_2$ ($t_1<t_2$) are $R_1$ and $R_2$ respectively, then the number of atoms disintegrated in time $t_2-t_1$ is proportional to

• A. $\frac{R_1{R}_2}{R_1+R_2}T$
• B. ($R_1$- $R_2$)T
• C. $\frac{R_1+R_2}{T}$
• D. ($R_1$+ $R_2$)T

Answer 1

Answer: (B)

Activity of a radioactive sample is given as $a=\lambda N$

Activity at time $t_1$ $R_1=\lambda N_1$

Activity at time $t_2$ $R_2=\lambda N_2$

Then, the number of atoms disintegrated in time $t_2-t_1$ is $N_1-N_2=\frac{R_1}{\lambda }-\frac{R_2}{\lambda }$

Half life time period is: $T_{\frac{1}{2}}=\frac{\ln 2}{\lambda }$

$N_1-N_2=\left(R_1-R_2\right)\frac{T}{\ln 2}$