Question

A sample contains two radioactive nuclei $x$ and $y$ with half-lives $2$ hour and $1$ hour respectively. The nucleus $x-decays$ to $y$ and $y-decays$ into a stable nucleus $z$. At $t=0$ the activities of the components in the same were equal. Find the ratio of the number of the active nuclei of $y$ at $t=1$ hours to the number at $t=0$.

Answer 1

We know, $x\stackrel{{\lambda }_x}{\to }y\stackrel{{\lambda }_y}{\to }z$

$t=0\left[r_x{]}_0\right[r_y{]}_0$
$t=4\left[r_x{]}_t\right[r_y{]}_t$

$[r_x{]}_0\Rightarrow$ activity of $x$

$[r_y{]}_0\Rightarrow$ activity of $y$

Given:

At t=0, $[r_x{]}_0=[r_y{]}_0$
${\lambda }_x=\frac{0.693}{2}\Rightarrow$ Half-life for nuclei $x$
${\lambda }_x=\frac{0.693}{1}\Rightarrow$ Half-life for nuclei $y$
$[N_x{]}_0\Rightarrow$ No. of active nuclei of $x$
$[N_y{]}_0\Rightarrow$ No. of active nuclei of $y$

$\therefore {\lambda }_x[N_x{]}_0={\lambda }_y[N_y{]}_0$

or $\frac{[N_x{]}_0}{[N_y{]}_0}=\frac{{\lambda }_y}{{\lambda }_x}$

Now maximum yield of $y(N_y{)}_t$ at $t=4hr$ is :

$[N_y{]}_t=\frac{[N_x{]}_0.{\lambda }_x}{{\lambda }_y-{\lambda }_x}[e^{-{\lambda }_x.t}-e^{-{\lambda }_y.t}]+[N_y{]}_0{e}^{-{\lambda }_y.t}$

$=\frac{{\lambda }_y[N_y{]}_0.{\lambda }_x}{{\lambda }_x[{\lambda }_y-{\lambda }_x]}[e^{-{\lambda }_x.t}-e^{-{\lambda }_y.t}]+[N_y{]}_0{e}^{-{\lambda }_y.t}$

$=\frac{{\lambda }_y[N_y{]}_0}{[{\lambda }_y-{\lambda }_x]}[e^{-{\lambda }_x.t}-e^{-{\lambda }_y.t}]+[N_y{]}_0{e}^{-{\lambda }_y.t}$

$\therefore \frac{[N_y{]}_t}{[N_y{]}_0}=[\frac{{\lambda }_y}{{\lambda }_y-{\lambda }_x}.e^{-{\lambda }_x.t}-\frac{{\lambda }_y}{{\lambda }_y-{\lambda }_x}{e}^{-{\lambda }_y.t}+e^{-{\lambda }_y.t}]$

$=[\frac{0.693}{2\times [\frac{0.693}{1}-\frac{0.693}{2}]}\times e^{\left(\frac{-0.693\times 4}{2}\right)}+\frac{0.693}{2\times [\frac{0.693}{1}-\frac{0.693}{2}]}\times e^{\left(\frac{-0.693\times 4}{1}\right)}+e^{\left(\frac{-0.693\times 4}{1}\right)}]$

$=0.25-0.06+0.06$
$\frac{[N_y{]}_t}{[N_y{]}_0}=0.25$

Hence, the ratio of the number of the active nuclei of $y$ at $t=4hours$ to the number at $t=0$ is 0.25.

Hence answer is $4\times 0.25=1$