Question
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.
$$t = T\log \left( {1.3} \right)$$
B.
$$t = \frac{T}{{\log \left( {1.3} \right)}}$$
C.
$$t = T\frac{{\log 2}}{{\log 1.3}}$$
D.
$$t = \frac{{\log 1.3}}{{\log 2}}T$$
Answer :
$$t = \frac{{\log 1.3}}{{\log 2}}T$$
Solution :
Let initially there are total $${N_0}$$ number of nuclei
At time $$t$$ $$\frac{{{N_B}}}{{{N_A}}} = 0.3\left( {{\text{given}}} \right)$$
$$\eqalign{
& \Rightarrow {N_B} = 0.3\,{N_A} \cr
& {N_0} = {N_A} + {N_B} = {N_A} + 0.3\,{N_A} \cr
& \therefore {N_A} = \frac{{{N_0}}}{{1.3}} \cr} $$
As we know $${N_t} = {N_0}{e^{ - \lambda t}}$$
$$\eqalign{
& {\text{or,}}\,\,\frac{{{N_0}}}{{1.3}} = {N_0}{e^{ - \lambda t}} \cr
& \frac{1}{{1.3}} = {e^{ - \lambda t}} \cr
& \Rightarrow \ln \left( {1.3} \right) = \lambda t \cr
& {\text{or,}}\,\,t = \frac{{\ln \left( {1.3} \right)}}{\lambda } \cr
& \Rightarrow t = \frac{{\ln \left( {1.3} \right)}}{{\frac{{\ln \left( 2 \right)}}{T}}} = \frac{{\ln \left( {1.3} \right)}}{{\ln \left( 2 \right)}}T \cr} $$