Two radioactive materials have decay constant `5lambda&lambda`. If initially they have same no. of nuclei. Find time when ratio of nuclei become `((1)/(e))^(2)` :

Options

A. `(1)/(lamda)`
B. `4 lamda`
C. `2 lamda`
D. `(1)/(2 lamda)`

Answer

D

Explanation

(d) Number of nuclei remained after time `t` can be written as
`N = N_0 e^(-lamda t)`
where `N_0` initial number of nuclei of both the substances.
`N_1 = N_0 e^(-5 lamda t)` …(i)
and `N_2 = N_0 e^(-lamda t)` …(ii)
Dividing Eq. (i) by Eq. (ii), we obtain
`(N_1)/(N_2) = e^((-5 lamda + lamda)t) = e^(-4 lamdat) = (1)/(e^(4 lamda t))`
But, we have given `(N_1)/(N_2) = ((1)/(e))^2 = (1)/(e^2)`
Hence, `(1)/(e^2) = (1)/(e^(4 lamda t))`
Comparing the powers, we get `2 = 4 lamda t` ltbr or `t = (2)/(4 lamda) = (1)/(2 lamda)`.