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

Options

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

Answer

D

Explanation

At time t
`N_a/N_A=3 rArr N_B = 3N_A`
also let initially there are total `N_0` number of nuclei
`N_A+N_B=N_0`
`N_A=N_0/1.3`
Also as we know
`N_A=N_0 e^(-lambdat) , N_0/1.3=N_0e^(-lambdat)`
`1/1.3=e^(-lambdat) rArr ln(1.3) =lambdat` or `t=(ln(1.3))/lamda`
`t=(ln(1.3))/((ln(2))/T) =(ln(1.3))/(ln(2))T`