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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 (1/2)^{t/T_{1/2}}$.For element $A$: $N_A = 10 (1/2)^{t/1}$.For

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$6.62$

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(A) The amount of a radioactive substance remaining after time $t$ is given by $N = N_0 (1/2)^{t/T_{1/2}}$.For element $A$: $N_A = 10 (1/2)^{t/1}$.For element $B$: $N_B = 1 (1/2)^{t/2}$.Given that the remaining quantities are equal,$N_A = N_B$.$10 (1/2)^t = (1/2)^{t/2}$.$10 = (1/2)^{t/2} / (1/2)^t = (1/2)^{t/2 - t} = (1/2)^{-t/2} = 2^{t/2}$.Taking $\log_{10}$ on both sides:$\log_{10} 10 = (t/2) \log_{10} 2$.$1 = (t/2) 	imes 0.3010$.$t = 2 / 0.3010 \approx 6.64 \, years$ (approx $6.62$ based on provided options).

$A$ and $B$ are two radioactive elements. Their half-lives are $1 \, year$ and $2 \, years$ respectively. Initially,$10 \, g$ of $A$ and $1 \, g$ of $B$ are taken. After how many years will their remaining quantities be equal?

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