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=Tlog(1.3)`
B. `t=T/(log (1.3))`
C. `t=T/2 (log2)/(log 1.3)`
D. `t=T (log1.3)/(log 2)`

Answer

D

Explanation

At time `t, =(N_(B))/(N_(A))=0.3 :. N_(B)=0.3N_(A)`
If `N_(0)` is total number of nuclei at t=0, then
`N_(A)+N_(B)=N_(0)`
`N_(A)(1+0.3)=N_(0), N_(A)=(N_(0))/1.3`
Again, as `N_(A)=N_(0)e^(-lambdat)`
`(N_(0))/1.3=N_(0)e^(-lambdat)` or `e^(-lambdat)=1/1.3`
or `lambdat=In(1.3)`
`t=(ln(1.3))/lambda=(ln(1.3))/(ln(2)//T) =(ln(1.3))/(In(2))xxT`