Physics - Modern Physics Question with Solution | TestHub

From radioactive decay law.

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

Thus,$R=-\frac{dN}{dt}$

Or$R= \lambda N \mathrm{o}\mathrm{r} R= \lambda N_0{e}^{-\lambda t} \dots \left(\mathrm{i}\right)$

Where$R_0 =\lambda N_0$is the activity of the radioactive materialat time$t=0.$

At time$t_1, R_1=R_0{e}^{-\lambda t} \dots \left(\mathrm{i}\mathrm{i}\right)$

At time$t_2, R_2=R_0{e}^{-\lambda t} \dots \left(\mathrm{i}\mathrm{i}\mathrm{i}\right)$

Dividing Eq. (ii) by (iii), we have

$\frac{R_1}{R_2}=\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda (t_1-t_2)}$

or$R_1=R_2{e}^{-\lambda (t_1-t_2)}$