A radioactive nucleus A with a half life T, decays into nucleus B. At t=0, there is no nucleus B. At somewhat t, the ratio of the number of B to that of A is 0.3 . Then, t is given by

Options

A. `t=T/2(log 2)/(log 1.3)`
B. `t =T(log 1.3)/(log 2)`
C. `t=T log(1.3)`
D. `t=T/(log(1.3))`

Answer

B

Explanation

Let the total number of nuclei at t = 0 be `N_(0)`. Let the number of nuclei B and A at time t be `N_(B)` and `N_(A)` respectively.
Given th at the ratio of number of nuclei B to the nuclei A at time t is given by:
`N_(B)/N_(A) =0.3`
`rArr N_(B)=0.3 N_(A)`
`N_(A) + N_(B)=N_(0)`
`N_(A) = N_(0)/1.3`
By the law of radioactive decay.
`N_(A)=N_(0)e^(-lambdat)`
where `lambda` is the disintegration constant:
`N_(0)/1.3 = N_(0)e^(-lambdat)`
`1/1.3 =e^(-lambdat)`
Or, ln (1,3) `=lambda t` Or, `t=(ln(1.3))/lambda`
As we know the half-life of A is T.
The disintegration constant can be expressed as:
`lambda =(ln(2))/T`
`t =(ln(1.3))/((ln(2))/T) =(ln(1.3))/(ln(2))T`