Physics - Modern Physics Question with Solution | TestHub

Initial number of nuclei of $A$ and $B$ is $N_0$.

Half-life of $A$, $T_A=30$ min.

Half-life of $B$, $T_B=60$ min.

Time $t=120$ min.

Number of half-lives for $A$, $n_A=\frac{120}{30}=4$.

Number of half-lives for $B$, $n_B=\frac{120}{60}=2$.

Number of nuclei of $A$ remaining, $N_A=\frac{N_0}{2^4}=\frac{N_0}{16}$.

Number of decayed nuclei of $A$, $D_A=N_0-\frac{N_0}{16}=\frac{15N_0}{16}$.

Number of nuclei of $B$ remaining, $N_B=\frac{N_0}{2^2}=\frac{N_0}{4}$.

Number of decayed nuclei of $B$, $D_B=N_0-\frac{N_0}{4}=\frac{3N_0}{4}$.

Ratio of decayed nuclei of $B$ to that of $A$ is $\frac{D_B}{D_A}=\frac{\frac{3N_0}{4}}{\frac{15N_0}{16}}=\frac{3}{4}\times \frac{16}{15}=\frac{4}{5}$.

Answer: $4:5$