The half-life of a radioactive isotope X is 5 | vedclass

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(B) Let $N_0$ be the initial amount of radioactive isotope $X$ and $N$ be the amount remaining after time $t$.Given the ratio of $X$ to $Y$ is $1:15$,

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$200$

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(B) Let $N_0$ be the initial amount of radioactive isotope $X$ and $N$ be the amount remaining after time $t$.Given the ratio of $X$ to $Y$ is $1:15$, the total amount is $N_0 = N + N_Y = N + 15N = 16N$.Therefore, the fraction of the remaining isotope is $\frac{N}{N_0} = \frac{1}{16}$.Using the radioactive decay formula $\frac{N}{N_0} = \left(\frac{1}{2}ight)^n$, where $n$ is the number of half-lives:$\frac{1}{16} = \left(\frac{1}{2}ight)^n \implies \left(\frac{1}{2}ight)^4 = \left(\frac{1}{2}ight)^n \implies n = 4$.The age of the rock $t$ is given by $t = n 	imes T_{1/2}$.Given $T_{1/2} = 50$ years, $t = 4 	imes 50 = 200$ years.

The half-life of a radioactive isotope $X$ is $50$ years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio of $1 : 15$ in a sample of a given rock. The age of the rock was estimated to be..........$years$.

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