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(A) The activity $R$ of a radioactive substance is given by $R = \lambda N$,where $N$ is the number of radioactive nuclei present at time $t$. Since t

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$1: 4$

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(A) The activity $R$ of a radioactive substance is given by $R = \lambda N$,where $N$ is the number of radioactive nuclei present at time $t$. Since the initial masses are the same,the initial number of nuclei $N_0$ is the same for both substances.The activity at time $t$ is given by $R = R_0 \left( \frac{1}{2} \right)^{t / T_{1/2}}$,where $T_{1/2}$ is the half-life.For the first substance: $T_{1/2, 1} = 1 \ yr$,$t = 4 \ yr$.$R_1 = R_0 \left( \frac{1}{2} \right)^{4/1} = R_0 \left( \frac{1}{2} \right)^4$.For the second substance: $T_{1/2, 2} = 2 \ yr$,$t = 4 \ yr$.$R_2 = R_0 \left( \frac{1}{2} \right)^{4/2} = R_0 \left( \frac{1}{2} \right)^2$.The ratio of their activities is:$\frac{R_1}{R_2} = \frac{R_0 (1/2)^4}{R_0 (1/2)^2} = \left( \frac{1}{2} \right)^{4-2} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Thus,the ratio is $1: 4$.

The masses of two radioactive substances are same and their half-lives are $1 \ yr$ and $2 \ yr$ respectively. The ratio of their activities after $4 \ yr$ will be

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