Q88 / 120 · VITEEE 2008 · Physics

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

  • A
    $10 \mathrm{~min}$
  • B
    $20 \mathrm{~min}$
  • Correct answer
    C
    $30 \mathrm{~min}$
  • D
    $40 \mathrm{~min}$

Solution

We have, $\mathrm{N}_{\mathrm{t}}=\mathrm{N}_{0}\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}}$, When $\begin{array}{l} \mathrm{N}_{\mathrm{t}}=\text { number of atoms present after time t } \\ \mathrm{N}_{0}^{\mathrm{t}}=\text { initial number of atoms } \\ \mathrm{T}_{1 / 2}=\text { half life of the nuclide } \\ \therefore \frac{\mathrm{N}_{\mathrm{t}}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\frac{\mathrm{t}}{\mathrm{T}_{1 / 2}}} \quad \text { or } \frac{1}{16}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \\ \text { or }\left(\frac{1}{2}\right)^{4}=\left(\frac{1}{2}\right)^{\frac{2}{\mathrm{~T}_{1 / 2}}} \Rightarrow 4=\frac{2}{\mathrm{~T}_{1 / 2}} \\ \Rightarrow \mathrm{T}_{1 / 2}=\frac{2}{4} \mathrm{Hr}=\frac{1}{2} \mathrm{Hr} \quad \therefore \mathrm{T}_{1 / 2}=30 \mathrm{~min} \end{array}$

Hence, the correct answer is option: 'C'.

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