Question

A solution contains a mixture of isotopes of $X^{A_1}(t_{1/2}=14days)$ and $X^{A_2}(t_{1/2}=25days)$. Total activity is 1 curie at t = 0. The activity reduces by $50\%$ in 20 days.

If the ratio of their initial number of nuclei is (x-1) . Find the value of x?

Answer 1

Let initially, the number of nuclei of isotopes $X^{A1}$ and $X^{A2}$ be x and 1-x respectiively.
Calculate the decay constants from the half life periods.
$k_1=\frac{0.693}{t_{1/2}}=\frac{0.693}{14}=0.0495$
$k_2=\frac{0.693}{t_{1/2}}=\frac{0.693}{25}=0.02772$Initially the activity is 1 curie and after 20 days, the activity is 0.5 curie.
The number of nuclei at time t is $A=A_0{e}^{-kt}$
For $X^{A1}$
$A=x{e}^{-0.0495\times 20}=0.372x$......(1)
For $X^{A2}$
$A=(1-x)e^{-0.0277\times 20}=0.575(1-x)$......(2)
From (1) and (2)
$0.5=0.372x+0.575-0.575x$
$x=0.370$ and (1-x) = 0.634
Thus, the ratio of the initial number of nuclei is:
$\frac{0.366\times 14}{0.634\times 25}=0.325$