A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of $20 \mathrm{~s}$ and $10…
- $60 \mathrm{~s}$
- $80 \mathrm{~s}$
- $20 \mathrm{~s}$
- $40 \mathrm{~s}$
Solution
$40 \mathrm{~g} \underset{\text { half -life }}{\stackrel{20 \mathrm{~s}}{\longrightarrow}} 20 \mathrm{~g} \stackrel{20 \mathrm{~s}}{\longrightarrow} 10 \mathrm{~g}$
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.
Asked in: NEET 2012 (Screening)
Practice more Structure of Atoms and Nuclei questions on Aicharya