A radioisotope X with a half-life of 1.4 time | vedclass

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(C) Given the ratio of the amount of radioisotope $X$ to stable isotope $Y$ is $\frac{X}{Y} = \frac{1}{7}$.The total amount of the initial sample is $

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$4.2$

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(C) Given the ratio of the amount of radioisotope $X$ to stable isotope $Y$ is $\frac{X}{Y} = \frac{1}{7}$.The total amount of the initial sample is $X + Y$. The fraction of the remaining radioisotope $X$ is given by $\frac{X}{X+Y} = \frac{1}{1+7} = \frac{1}{8}$.We know that the remaining fraction after $n$ half-lives is $\left(\frac{1}{2}ight)^{n}$.Equating the two,$\left(\frac{1}{2}ight)^{n} = \frac{1}{8} = \left(\frac{1}{2}ight)^{3}$.Thus,the number of half-lives elapsed is $n = 3$.The age of the rock $t$ is given by $t = n 	imes T$,where $T$ is the half-life.$t = 3 	imes 1.4 	imes 10^{9} 	ext{ years} = 4.2 	imes 10^{9} 	ext{ years}$.

$A$ radioisotope $X$ with a half-life of $1.4 \times 10^{9} \text{ years}$ decays into $Y$,which is stable. $A$ sample of rock from a cave was found to contain $X$ and $Y$ in the ratio $1:7$. The age of the rock is ........ $\times 10^{9} \text{ years}$.

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