A radioactive isotope X with a half-life of 1 | vedclass

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(C) Let $N_0$ be the initial amount of radioactive isotope $X$. After some time $t$,the amount of $X$ remaining is $N_X$ and the amount of $Y$ formed

Vedclass · Accepted Answer

$4.11 	imes 10^9$ years

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(C) Let $N_0$ be the initial amount of radioactive isotope $X$. After some time $t$,the amount of $X$ remaining is $N_X$ and the amount of $Y$ formed is $N_Y$. Given the ratio $N_X : N_Y = 1 : 7$,the total amount of substance is $N_X + N_Y = N_0$. Thus,$N_X = \frac{1}{1+7} N_0 = \frac{1}{8} N_0$. Using the radioactive decay law,$N_X = N_0 (\frac{1}{2})^n$,where $n$ is the number of half-lives. $\frac{1}{8} N_0 = N_0 (\frac{1}{2})^n \implies (\frac{1}{2})^3 = (\frac{1}{2})^n \implies n = 3$. The age of the rock $t = n 	imes T_{1/2} = 3 	imes 1.37 	imes 10^9$ years $= 4.11 	imes 10^9$ years.

$A$ radioactive isotope $X$ with a half-life of $1.37 \times 10^9$ years decays to $Y$,which is stable. $A$ sample of rock from the moon was found to contain both the elements $X$ and $Y$ in the ratio of $1 : 7$. The age of the rock is

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