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 -

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

Options

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

Answer

D

Explanation

AT time t
`(N_(B))/(N_(A))=.3impliesN_(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^(-lamdat),(N_(0))/1.3=N_(0)e^(-lamdat)`
`1/1.3=e^(-lamdat)=ln(1.3)=lamdat` or `t=(ln (1.3))/(lamda)`
`t=(l (1.3))/((ln(2))/T)=(ln(1.3))/(ln(2))T`