The half-life of a radioactive isotope X is 1 | vedclass

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(C) The decay process is $X \rightarrow Y$. Initially,at $t = 0$,let the amount of $X$ be $N_0 = 8$ and $Y = 0$. At time $t$,the ratio of $X$ to $Y$ i

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$4.11 	imes 10^9 \, y$

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(C) The decay process is $X ightarrow Y$. Initially,at $t = 0$,let the amount of $X$ be $N_0 = 8$ and $Y = 0$. At time $t$,the ratio of $X$ to $Y$ is $1:7$,meaning $N_X = 1$ and $N_Y = 7$. The total initial amount $N_0 = N_X + N_Y = 1 + 7 = 8$. Using the radioactive decay law: $N = N_0 (1/2)^n$,where $n$ is the number of half-lives. $1 = 8 	imes (1/2)^n \Rightarrow (1/2)^3 = (1/2)^n \Rightarrow n = 3$. The age of the rock $t = n 	imes T_{1/2}$. Given $T_{1/2} = 1.37 	imes 10^9 \, y$. $t = 3 	imes 1.37 	imes 10^9 \, y = 4.11 	imes 10^9 \, y$.

The half-life of a radioactive isotope $X$ is $1.37 \times 10^9 \, y$. It decays to a stable isotope $Y$. In a rock sample from the moon,the ratio of $X$ to $Y$ is $1:7$. The age of the rock is:

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