Question

In an experiment on two radioactive isotopes of an element (which do not decay into each other), their molar ratio at a given instant is 3. The rapidly decaying isotope has larger mass and an initial activity of $1.0\mu Ci$. The half-lives of the two isotopes are 12 and 16 hr respectively.
Now,if their molar ratio after two days is $N_A/N_B$,then calculate the value of $4\left(N_A/N_B\right)$ will be:

Answer 1

We have, after two days, i.e., 48 hours

$\therefore N_A=N_0^1(\frac{1}{2}{)}^4=N_1^0/16$
$\therefore N_B=N_0^2(\frac{1}{2}{)}^3=N_1^0/8$

Mass ratio = $\frac{N_A}{N_B}=\frac{N_A^0}{N_B^0}.\frac{8}{16}=\frac{3\times 8}{16}=\frac{3}{2}$
$⟹4\left(N_A/N_B\right)=6$