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(D) Let the amount of $A_1$ and $A_2$ become equal after $t \, s$.The amount of a radioactive substance remaining after time $t$ is given by $N = N_0

Vedclass · Accepted Answer

$40$

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(D) Let the amount of $A_1$ and $A_2$ become equal after $t \, s$.The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$,where $T_{1/2}$ is the half-life.For $A_1$: $N_1 = 40 \left( \frac{1}{2} \right)^{t/20}$.For $A_2$: $N_2 = 160 \left( \frac{1}{2} \right)^{t/10}$.Setting $N_1 = N_2$:$40 \left( \frac{1}{2} \right)^{t/20} = 160 \left( \frac{1}{2} \right)^{t/10}$.Dividing both sides by $40$:$\left( \frac{1}{2} \right)^{t/20} = 4 \left( \frac{1}{2} \right)^{t/10}$.Using the property $2^{-x} = \frac{1}{2^x}$:$\frac{1}{2^{t/20}} = 4 \cdot \frac{1}{2^{t/10}}$.Rearranging the terms:$\frac{2^{t/10}}{2^{t/20}} = 4$.$2^{(t/10 - t/20)} = 2^2$.Equating the exponents:$\frac{t}{10} - \frac{t}{20} = 2$.$\frac{2t - t}{20} = 2$.$\frac{t}{20} = 2 \Rightarrow t = 40 \, s$.

$A$ 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 how many seconds (in $, s$)?

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