Decay constants of two radio-active samples A and B are 15 x and 3x respectively . They have equal number of initial nuclei . The ratio of the number of nuclei left in A and B after a time `(1)/(6x)` is

Options

A. e
B. `e^(2)`
C. `e^(-1)`
D. `e^(-2)`

Answer

D

Explanation

According to radioactive decay , `N = N_(0) e^(-lambda t)`
where
`N_(0)`= Initial number of nuclei at t = 0
N = Number of nuclei left undecayed after time t
`lambda` = Decay constant
For A , `N_(A) = N_(0) A^(e^(-(15x) (1//6x)))`
= `N_(0) A^(e^(-5//2))`
For B , `N_(B) = N_(0) B^(e^(- (3x) (1//6 x)))`
`= N_(0) B^(e^(-1//2))`
As per question ,
`N_(0) A = N_(0) B`
`therefore (N_(A))/(N_(B)) = (e^(-5//2))/(e^(-1//2)) = e^(-2)`