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The radioactivity of a certain material drops to $\frac{1}{16}$ of the initial value in 2 hours. The half life of this radionuclide is
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The correct answer is:
$30 \mathrm{~min}$
We have, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$,
When
$\begin{array}{l}
\mathrm{N}_{\mathrm{t}}=\text { number of atoms present after time t } \\
\mathrm{N}_{0}^{\mathrm{t}}=\text { initial number of atoms } \\
\mathrm{T}_{1 / 2}=\text { half life of the nuclide } \\
\therefore \frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad \text { or } \frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \\
\text { or }\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \Rightarrow 4=\frac{2}{\mathrm{~T}_{1 / 2}} \\
\Rightarrow \mathrm{T}_{1 / 2}=\frac{2}{4} \mathrm{Hr}=\frac{1}{2} \mathrm{Hr} \quad \therefore \mathrm{T}_{1 / 2}=30 \mathrm{~min}
\end{array}$
When
$\begin{array}{l}
\mathrm{N}_{\mathrm{t}}=\text { number of atoms present after time t } \\
\mathrm{N}_{0}^{\mathrm{t}}=\text { initial number of atoms } \\
\mathrm{T}_{1 / 2}=\text { half life of the nuclide } \\
\therefore \frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad \text { or } \frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \\
\text { or }\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \Rightarrow 4=\frac{2}{\mathrm{~T}_{1 / 2}} \\
\Rightarrow \mathrm{T}_{1 / 2}=\frac{2}{4} \mathrm{Hr}=\frac{1}{2} \mathrm{Hr} \quad \therefore \mathrm{T}_{1 / 2}=30 \mathrm{~min}
\end{array}$
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