The ratio of the number of active nuclei in t | vedclass

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Question

(C) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$. Given the initial ratio of nuclei $N_1/N_2 = 2/3$.

Vedclass · Accepted Answer

$1:4$

Vedclass · Answer

(C) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{\ln 2}{T_{1/2}} N$. Given the initial ratio of nuclei $N_1/N_2 = 2/3$. The number of nuclei remaining after time $t$ is $N(t) = N_0 (1/2)^{t/T_{1/2}}$. The activity at time $t$ is $A(t) = \lambda N(t) = \frac{\ln 2}{T_{1/2}} N_0 (1/2)^{t/T_{1/2}}$. For the two samples at $t = 12 \ hours$: $A_1 = \frac{\ln 2}{2} N_1 (1/2)^{12/2} = \frac{\ln 2}{2} N_1 (1/2)^6 = \frac{\ln 2}{2} N_1 \cdot \frac{1}{64}$. $A_2 = \frac{\ln 2}{3} N_2 (1/2)^{12/3} = \frac{\ln 2}{3} N_2 (1/2)^4 = \frac{\ln 2}{3} N_2 \cdot \frac{1}{16}$. The ratio of activities is $\frac{A_1}{A_2} = \frac{N_1}{N_2} \cdot \frac{T_{1/2, 2}}{T_{1/2, 1}} \cdot \frac{(1/2)^{12/T_{1/2, 1}}}{(1/2)^{12/T_{1/2, 2}}}$. Substituting the values: $\frac{A_1}{A_2} = \frac{2}{3} \cdot \frac{3}{2} \cdot \frac{(1/2)^6}{(1/2)^4} = 1 \cdot (1/2)^2 = 1/4$. Wait,re-evaluating: $\frac{A_1}{A_2} = \frac{N_1}{N_2} \cdot \frac{T_2}{T_1} \cdot 2^{(t/T_2 - t/T_1)} = \frac{2}{3} \cdot \frac{3}{2} \cdot 2^{(12/3 - 12/2)} = 1 \cdot 2^{(4-6)} = 2^{-2} = 1/4$.

The ratio of the number of active nuclei in two different samples is $2:3$. Their half-lives are $2 \ hours$ and $3 \ hours$ respectively. After $12 \ hours$,the ratio of their activities will be:

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