A mixture consists of two radioactive materials $A_{1}$ and $A_{2}$ with half lives of $20\ s$ and $10\ s$ respectively. Initially the mixture has $40\ g$ of $A_{1}$ and $160\ g$ of $A_{2}$. The amount of the two in the mixture will become equal after

Question

Dr. Abubacker · Accepted Answer

(d) $N_{1} = \frac{N_{01}}{(2)^{t/20}},\ N_{2} = \frac{N_{02}}{(2)^{t/10}}$
$N_{1} = N_{2}$
$\frac{40}{(2)^{t/20}} = \frac{160}{(2)^{t/10}} \Rightarrow 2^{t/20} = 2^{\left( \frac{t}{10} - 2 \right)}$
$\Rightarrow \frac{t}{20} = \frac{t}{10} - 2 \Rightarrow \frac{t}{20} - \frac{t}{10} = - 2$
$\Rightarrow \frac{t}{20} = 2 \Rightarrow t = 40$

Q. A mixture consists of two radioactive materials $A_{1}$ and $A_{2}$ with half lives of $20\ s$ and $10\ s$ respectively. Initially the mixture has $40\ g$ of $A_{1}$ and $160\ g$ of $A_{2}$. The amount of the two in the mixture will become equal after

Explanation

Reviewed by Dr. Abubacker

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