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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$.For substance $A$: $N_

Vedclass · Accepted Answer

$6.62$

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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} ight)^{t/T_{1/2}}$.For substance $A$: $N_A = 10 \left( \frac{1}{2} ight)^{t/1}$.For substance $B$: $N_B = 1 \left( \frac{1}{2} ight)^{t/2}$.Given that the remaining quantities are equal, $N_A = N_B$, so $10 \left( \frac{1}{2} ight)^t = 1 \left( \frac{1}{2} ight)^{t/2}$.Dividing both sides by $(1/2)^{t/2}$, we get $10 = (1/2)^{t/2 - t} = (1/2)^{-t/2} = 2^{t/2}$.Taking the logarithm on both sides: $\log_{10} 10 = \frac{t}{2} \log_{10} 2$.$1 = \frac{t}{2} 	imes 0.3010$.$t = \frac{2}{0.3010} \approx 6.64 \, years$. The closest option is $6.62 \, years$.

$A$ and $B$ are two radioactive substances whose half-lives are $1$ and $2$ years respectively. Initially, $10 \, g$ of $A$ and $1 \, g$ of $B$ are taken. The time (approximate) after which they will have the same quantity remaining is ........... $years$.

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