Two radioactive substances X and Y originally | vedclass

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(C) Let the half-life of $X$ be $T_{x} = t$ and the half-life of $Y$ be $T_{y} = 2t$.After three half-lives of $Y$,the time elapsed is $t_{total} = 3

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$\frac{8}{1}$

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(C) Let the half-life of $X$ be $T_{x} = t$ and the half-life of $Y$ be $T_{y} = 2t$.After three half-lives of $Y$,the time elapsed is $t_{total} = 3 	imes T_{y} = 3 	imes 2t = 6t$.Using the radioactive decay law $N(t) = N_{0} \left(\frac{1}{2}ight)^{t/T}$,the number of nuclei remaining for $X$ and $Y$ are:$N_{1}' = N_{1} \left(\frac{1}{2}ight)^{6t/t} = N_{1} \left(\frac{1}{2}ight)^{6} = \frac{N_{1}}{64}$$N_{2}' = N_{2} \left(\frac{1}{2}ight)^{6t/2t} = N_{2} \left(\frac{1}{2}ight)^{3} = \frac{N_{2}}{8}$Given that $N_{1}' = N_{2}'$,we have:$\frac{N_{1}}{64} = \frac{N_{2}}{8}$$\frac{N_{1}}{N_{2}} = \frac{64}{8} = \frac{8}{1}$

Two radioactive substances $X$ and $Y$ originally have $N_{1}$ and $N_{2}$ nuclei respectively. The half-life of $X$ is half of the half-life of $Y$. After three half-lives of $Y$,the number of nuclei of both are equal. The ratio $\frac{N_{1}}{N_{2}}$ will be equal to:

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