The half-life of a radioactive isotope X is 2 | vedclass

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(A) Let the initial number of atoms of $X$ be $N_0$.After time $t$,the number of atoms of $X$ remaining is $N$,and the number of atoms of $Y$ formed i

Vedclass · Accepted Answer

$60$

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(A) Let the initial number of atoms of $X$ be $N_0$.After time $t$,the number of atoms of $X$ remaining is $N$,and the number of atoms of $Y$ formed is $N_0 - N$.According to the problem,the ratio of $X$ to $Y$ is $\frac{N}{N_0 - N} = \frac{1}{7}$.This implies $7N = N_0 - N$,which simplifies to $8N = N_0$,or $\frac{N}{N_0} = \frac{1}{8}$.We know that $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.Thus,$(\frac{1}{2})^3 = (\frac{1}{2})^n$,which gives $n = 3$.The age of the rock $t$ is given by $t = n 	imes T_{1/2}$.Given $T_{1/2} = 20$ years,$t = 3 	imes 20 = 60$ years.Therefore,the age of the rock is $60$ years.

The half-life of a radioactive isotope $X$ is $20$ years. It decays to another element $Y$,which is stable. The two elements $X$ and $Y$ were found to be in the ratio $1:7$ in a sample of a given rock. The age of the rock is estimated to be:

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