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

Answer

B

Explanation

After time t, number of nuclei of A , `N_A=N_0e^(-lambdat)`
`therefore N_B=N_0 -N_A =N_0(1-e^(-lambdat))`
`because N_B/N_A =0.3 therefore (1-e^(-lambdat))/(e^(-lambdat))=0.3`
or, `lambdat` =In 1.3 or , `("In2"/T)t` =In 1.3
or , `t=T"In1.3"/"In2"=T"log1.3"/"log2" `