An element X with a half-life of 1.4 times 10 | vedclass

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(A) Let $N_X$ be the amount of element $X$ and $N_Y$ be the amount of element $Y$ present in the rock at time $t$.Given the ratio $N_X : N_Y = 1 : 7$,

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$4.2 	imes 10^9$

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(A) Let $N_X$ be the amount of element $X$ and $N_Y$ be the amount of element $Y$ present in the rock at time $t$.Given the ratio $N_X : N_Y = 1 : 7$,we have $N_Y = 7N_X$.The total initial amount of the element $X$ (at $t=0$) was $N_0 = N_X + N_Y = N_X + 7N_X = 8N_X$.Using the radioactive decay law,$N_X = N_0 \left( \frac{1}{2} ight)^{n}$,where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.Substituting the values,$N_X = 8N_X \left( \frac{1}{2} ight)^{n}$.This simplifies to $\frac{1}{8} = \left( \frac{1}{2} ight)^{n}$,which means $\left( \frac{1}{2} ight)^3 = \left( \frac{1}{2} ight)^{n}$.Thus,$n = 3$.Since $n = \frac{t}{T_{1/2}}$,we have $t = 3 	imes T_{1/2}$.Given $T_{1/2} = 1.4 	imes 10^9$ years,the age of the rock is $t = 3 	imes 1.4 	imes 10^9 = 4.2 	imes 10^9$ years.

An element $X$ with a half-life of $1.4 \times 10^9$ years decays to form another stable element $Y$. $A$ sample is taken from a rock that contains both $X$ and $Y$ in the ratio $1:7$. If at the time of formation of the rock,$Y$ was not present in the sample,then the age of the rock in years is

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