1.

In a radioactive material the activity at time `t_1` is `R_1` and at a later time `t_2`, it is `R_1`. If the decay constant of the material is `lamda`, thenA. `R_1 = R_2 e^(-lamda(t_1 - t_2))`B. `R_1 = R_2 e^(lamda(t_1 - t_2))`C. `R_1 = R_2(t_2//t_1)`D. `R_1 = R_2`

Answer» Correct Answer - A
(a) The decay rate `R` of a radioactive material is the number of decays per second
From radioactive decay law,
`-(dN)/(dt) prop N` or `- (dN)/(dt) = lamda N`
Thus, `R = - (dN)/(dt)` or `R prop N`
Or `R = lamda N` or `R lamda N_0 e^(-lamda t)`...(i)
Where `R_0 = lamda N_0` is the activity of the radioactive material at time `t = 0`
At time `t_1, R_1 = R_0 e^(-lamda t_1)` ...(ii)
At `t_2, R_2 = R_0 e^(-lamda t_2)` ...(iii)
Dividing Eq. (ii) by (iii), we have
`(R_1)/(R_2) = (e^(-lamda t_1))/(e^(-lamda t_2)) = e^(-lamda(t_1 - t_2))`
or `R_1 = R_2 e^(-lamda(t_1 -t_2))`.


Discussion

No Comment Found

Related InterviewSolutions