Q143 / 200 · NEET 2012 (Screening) · Physics

A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20 \mathrm{~s}$ and $10 \mathrm{~s}$ respectively. Initially the mixture has $40 \mathrm{~g}$ of $A_1$ and $160 \mathrm{~g}$ of $A_2$. The amount of the two in the mixture will become equal after

  • A
    $60 \mathrm{~s}$
  • B
    $80 \mathrm{~s}$
  • C
    $20 \mathrm{~s}$
  • Correct answer
    D
    $40 \mathrm{~s}$

Solution

For $40 \mathrm{~g}$ amount $40 \mathrm{~g} \underset{\text { half -life }}{\stackrel{20 \mathrm{~s}}{\longrightarrow}} 20 \mathrm{~g} \stackrel{20 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}$ For $160 \mathrm{~g}$ amount $\begin{array}{l} 160 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 80 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 40 \mathrm{~g} \\ 40 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 20 \mathrm{~g} \stackrel{10 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g} \end{array}$ So, after $40 \mathrm{~s} A_1$ and $A_2$ remains same.

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

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