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A nucleus has a half-life of 30 minutes. At $3 \mathrm{PM}$ its decay rate was measured as 120,000 counts/sec. What will be the decay rate at $5 \mathrm{PM} ?$
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The correct answer is:
7,500 counts/sec
Given $\mathrm{T}=30$ minutes. $\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=120 \mathrm{~K} \frac{\text { counts }}{\mathrm{sec}}$
After each half life, activity is reduced to half therefore after $\mathrm{n}$ half lives activity reduces to $\left(\frac{1}{2}\right)^{\mathrm{n}}$.
Also $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$
$\frac{\mathrm{dN}}{\mathrm{dt}}$ at $5 \mathrm{P} . \mathrm{M}$. will be equal to activity remaining after four half lives.
i.e. $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{16} \mathrm{th}\right)$ of the initial activity
$\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\text {at } 5 \mathrm{PM}}=\left(\frac{1}{16}\right.$ th $)$ of the initial activity
$$
\begin{array}{l}
\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{5 \mathrm{P}_{. \mathrm{M}}}=\left(\frac{1}{16}\right)\left(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)_{3 \mathrm{PM}}
\end{array}
$$
After each half life, activity is reduced to half therefore after $\mathrm{n}$ half lives activity reduces to $\left(\frac{1}{2}\right)^{\mathrm{n}}$.
Also $\frac{\mathrm{dN}}{\mathrm{dt}} \propto \mathrm{N}$
$\frac{\mathrm{dN}}{\mathrm{dt}}$ at $5 \mathrm{P} . \mathrm{M}$. will be equal to activity remaining after four half lives.
i.e. $\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{16} \mathrm{th}\right)$ of the initial activity
$\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{\text {at } 5 \mathrm{PM}}=\left(\frac{1}{16}\right.$ th $)$ of the initial activity
$$
\begin{array}{l}
\left(\frac{\mathrm{dN}}{\mathrm{dt}}\right)_{5 \mathrm{P}_{. \mathrm{M}}}=\left(\frac{1}{16}\right)\left(\frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}\right)_{3 \mathrm{PM}}
\end{array}
$$
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