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(C) Let $N_0$ be the initial number of nuclei for both $X$ and $Y$.Given half-lives: $T_{1/2, X} = 1\, h$ and $T_{1/2, Y} = 2\, h$.After time $t = 2\,

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$1 : 1$

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(C) Let $N_0$ be the initial number of nuclei for both $X$ and $Y$.Given half-lives: $T_{1/2, X} = 1\, h$ and $T_{1/2, Y} = 2\, h$.After time $t = 2\, h$,the number of nuclei remaining is given by $N = N_0 (1/2)^{t/T_{1/2}}$.For $X$: $N_X = N_0 (1/2)^{2/1} = N_0 / 4$.For $Y$: $N_Y = N_0 (1/2)^{2/2} = N_0 / 2$.Activity $A = \lambda N$,where $\lambda = \ln(2) / T_{1/2}$.Ratio of activities: $\frac{A_X}{A_Y} = \frac{\lambda_X N_X}{\lambda_Y N_Y} = \frac{(\ln 2 / T_{1/2, X}) N_X}{(\ln 2 / T_{1/2, Y}) N_Y} = \frac{T_{1/2, Y}}{T_{1/2, X}} 	imes \frac{N_X}{N_Y}$.Substituting the values: $\frac{A_X}{A_Y} = \frac{2}{1} 	imes \frac{N_0/4}{N_0/2} = 2 	imes \frac{1}{2} = 1$.Thus,the ratio is $1 : 1$.

Two radioactive substances $X$ and $Y$ initially contain equal number of nuclei. $X$ has a half-life of $1\, hour$ and $Y$ has a half-life of $2\, hours$. After $2\, hours$,the ratio of the activity of $X$ to the activity of $Y$ is

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