Two radioactive nuclei $\text{A}$ and $\text{B}$ initially contain equal number of radioactive atoms. Source $\text{A}$ has a half-life of $1$-hour and source $\text{B}$ has a half-life of $2$-hours. At the end of $2$-hours, the ratio of the rate of disintegration of $\text{A}$ to that of $\text{B}$ is-

The correct option is A $1:2$
Given, $(t_{1/2}{)}_A=1\text{hr};(t_{1/2}{)}_B=2\text{hr}$

Rate of disintegration (or) Rate of activity $=-\frac{dN}{dt}\propto N$

For element $\text{A}$;

No. atoms remain after $2\text{hr}$ is,

$N=N_0{\left(\frac{1}{2}\right)}^n$

$N_A=\frac{N_0}{4}[\because n=2]$

i.e.,

$N_0\stackrel{1hr}{\to }\frac{N_0}{2}\stackrel{1hr}{\to }\frac{N_0}{4}$

For element $\text{B}$;

No. atoms remain after $2\text{hr}$ is,

$N=N_0{\left(\frac{1}{2}\right)}^n$

$N_B=\frac{N_0}{2}[\because n=1]$

Now, ratio of activity after $2\text{hr}$ is,

$\therefore \frac{{(-\frac{dN}{dt})}_A}{{(-\frac{dN}{dt})}_B}=\frac{N_A}{N_B}=\frac{\frac{N_o}{4}}{\frac{N_o}{2}}=\frac{1}{2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.