Physics - Modern Physics Question with Solution | TestHub

Activity is given by $A=\lambda N$

For sample $S_1$, $A_1={\lambda }_1{N}_1$

For sample $S_2$, $A_2={\lambda }_2{N}_2$

Given that $N_1=2N_2$ and $A_2=2A_1$

Substituting the values in the activity relation:

${\lambda }_2{N}_2=2({\lambda }_1{N}_1)$

${\lambda }_2{N}_2=2{\lambda }_1(2N_2)$

${\lambda }_2=4{\lambda }_1$

The half-life is inversely proportional to the decay constant, $T_{1/2}=\frac{\ln 2}{\lambda }$

The ratio of half-life of $S_1$ to $S_2$ is:

$\frac{(T_{1/2}{)}_1}{(T_{1/2}{)}_2}=\frac{{\lambda }_2}{{\lambda }_1}=4$

Answer: $4$