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`, then

Options

A. `R_1=R_2(t_2/t_1)`
B. `R_1=R_2e^((-lamda)(t_1-t_2))`
C. `R_1=R_2`
D. `R_1=R_2e^(lamda(t_1-t_2))`

Answer

B

Explanation

For the radioactive material , `R_1 = R_0e^(-lamdat_1) " at time " t_1`
and `R_2 = R_0e^(-lamdat_2) " at time " t_2`
where `R_0 = lamda N_0 `= activity at time t=0
`therefore R_1/R_2 = (e^(-lamdat_1))/(e^(-lamdat_2)) = e^(-lamda(t_1 - t_2))`
`therefore R_1 = R_2 e^(-lamda(t_1 - t_2))`