Two radioactive substance `A` and `B` have decay constants `5 lambda` and `lambda` respectively. At `t=0` they have the same number of nuclei. The ratio of number of nuclei of nuclei of `A` to those of `B` will be `(1/e)^(2)` after a time interval

Options

A. `(1)/(lambda)`
B. `(1)/(2 lambda)`
C. `(1)/(3 lambda)`
D. `(1)/(4 lambda)`

Answer

B

Explanation

According to radioactive decay, `N = N_(0) e^(-lambda t)` where, `N_(0)` = Number of radioactive nuclei present in the sample at t = 0, N = Number of radioactive nuclei left undecayed after time t, `lambda` = decay constant.
For A, `N_(A) = (N_(0))_(A)e^(-5 lambda t)" "...(i)`
For B, `N_(B) = (N_(0))_(B) e^(-lambda t)" "...(ii)`
As per question, `(N_(0))_(A) = (N_(0))_(B)` (Given)
`therefore` Dividing (i) by (ii), we get
`(N_(A))/(N_(B)) = (e^(-5 lambda t))/(e^(-lambda t)) = e^(-4 lambda t)`
`rArr" "((1)/(e))^(2) = e^(- 4 lambda t) or (1)/e^(2) = (1)/(e^(4 lambda t))`
`rArr" "e^(4 lambda t) = e^(2) or 4 lambda t = 2 or t = (2)/(4 lambda) = (1)/(2 lambda)`