Sometimes a radioactive nucleus decays into a nucleus which itsel… - Vidyadip

Question

Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An example is:
$^{38}\text{Sulphur}\xrightarrow[=2.48\text{h}]{\text{half-life}}\ \ ^{38}\text{Cl}\ \xrightarrow[=0.62\text{h}]{\text{half-life}}\ ^{38}\text{Ar}(\text{stable})$
Assume that we start with 1000 $^{38}S$ nuclei at time t = 0. The number of $^{38}Cl $ is of count zero at t = 0 and will again be zero at t = ∞. At what value of t, would the number of counts be a maximum?

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Answer

Suppose decay is given as shown below:
$^{38}\text{S}\xrightarrow[=2.48]{}\ \ ^{38}\text{Cl}\ \xrightarrow[0.62\text{h}]{}\ ^{38}\text{Ar}$
At any time t, let $^{38}S$ have $N_1(t)$ active nuclei and $^{38}Cl$ have $N_2 (t)$ active nuclei.
$\frac{\text{dN}_1}{\text{dt}}=-\lambda_1\text{N}_1=$ rate of formation of $Cl^{38}$.
Also, $\frac{\text{dN}_2}{\text{dt}}=-\lambda_\text{1}\text{N}_2=\lambda_1\text{N}_1$
But $\text{N}_1=\text{N}_0\text{e}^{-\frac{\lambda}{\text{t}}}$
$\frac{\text{dN}_2}{\text{dt}}=-\lambda_1\text{N}_0\text{e}^{-\frac{\lambda}{\text{t}}}-\lambda_2\text{N}_2\ .....(\text{i})$
Multiplying by $\text{e}^{\lambda2\text{t}}\text{ dt}$ and rearranging
$\text{e}^{\lambda_2\text{t}}\text{dN}_2+\lambda_2\text{N}_2\text{e}^{\lambda_2\text{t}}\text{dt}=\lambda_1\text{N}_0\text{e}^{(\lambda_2-\lambda_1)\text{t}}\text{dt}$
Integrating both sides
$\text{N}_2\text{e}^{\lambda_2\text{t}}=\frac{\text{N}_0\lambda_1}{\lambda_0-\lambda_1}\text{e}^{(\lambda_1-\lambda_1)\text{t}}+\text{C}$
Since, at $t = 0, N_2 = 0$, $\text{C}=-\frac{\text{N}_0\lambda_1}{\lambda_2-\lambda_1}$
$\therefore\ \text{N}_2\text{e}^{\lambda_2\text{t}}=\frac{\text{N}_0\lambda_1}{\lambda_2-\lambda_1}\big(\text{e}^{(\lambda_2-\lambda_1)\text{t}}-1\big)$
$\text{N}_2=\frac{\text{N}_0\lambda_1}{\lambda_2-\lambda_1}{(\text{e}^{\lambda_1\text{t}}-\text{e}^{\lambda_2\text{t}})}$
For maximum count, $\frac{\text{dN}_2}{\text{dt}}=0$
By using concepts of calculus and solving, we will get,
$\text{t}=\frac{\Big(\text{In}\frac{\lambda_1}{\lambda_2}\Big)}{(\lambda_1-\lambda_2)}$
$=\frac{\text{In}\Big(\frac{2.48}{0.62}\Big)}{(2.48-0.62)}$
$=\frac{\text{In}4}{1.86}=\frac{2.303\log4}{1.86}\ \ \bigg(\because\ \lambda=\frac{0.696}{\text{T}_{\frac{1}{2}}}\bigg)$
$=0.745\text{s}.$

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