A radioactive isotope X has a half-life of 20 | vedclass

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Question

(A) Initially, the rock contains only the radioactive isotope $X$. Let the initial amount be $N_0 = 8 \, g$.After time $t$, the ratio of $X$ to $Y$ is

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(A) Initially, the rock contains only the radioactive isotope $X$. Let the initial amount be $N_0 = 8 \, g$.After time $t$, the ratio of $X$ to $Y$ is $1:7$. This means that for every $1 \, g$ of $X$ remaining, $7 \, g$ of $Y$ has been formed.The total amount of substance remains constant, so the initial amount $N_0 = 1 \, g + 7 \, g = 8 \, g$.The amount of undecayed isotope $X$ remaining after time $t$ is $N = 1 \, g$.The law of radioactive decay states that $N = N_0 (1/2)^n$, where $n$ is the number of half-lives.Substituting the values: $1 = 8 	imes (1/2)^n$, which simplifies to $(1/2)^n = 1/8$.Since $1/8 = (1/2)^3$, we have $n = 3$.The age of the rock is $t = n 	imes T_{1/2}$, where $T_{1/2} = 20 \, years$.Therefore, $t = 3 	imes 20 = 60 \, years$.

$A$ radioactive isotope $X$ has a half-life of $20$ years and decays into a stable element $Y$. In a given rock sample, the ratio of the amounts of $X$ and $Y$ is found to be $1:7$. The estimated age of the rock is ......... years.

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