Question

In a radioactive material the activity at time $ t_1 $ is $ R_1 $ and at a later time $ t_2 $ it is $ R_2 $ . If the decay constant of the material is $ \lambda $ , then

• A. $ R_{1} = R_{2 }e^{-\lambda\left(t_{1}-t_{2}\right)} $
• B. $ R_{1} = R_{2 }e^{\lambda\left(t_{1}-t_{2}\right)} $
• C. $ R_1 = R_2 (t_2/t_1) $
• D. $ R_1 = R_2 $

Answer 1

Answer: (A)

The decay rate R of a radioactive material is the number of decays per second.
From radioactive decay law.
$ - \frac{dN}{dt} \propto N $
or $ - \frac{dN}{dt} = \lambda N $
Thus, $ R = - \frac{dN}{dt} $
or $ R = \lambda N $
or $ R = \lambda N_0 e^{- \lambda t} ...(i) $
where $ R_0 = \lambda N_0 $ is the activity of the radioactive material at time t = 0.
At time $ t_1, R_1 = R_0 e^{- \lambda t_1} ...(ii) $
At time $ t_2, R_2 = R_0 e^{- \lambda t_2} ...(iii) $
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)} $