The half-life of a radioactive isotope X is 5 | vedclass

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(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.The amount of stable element $Y$ formed is $N_0 - N$.Gi

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) Let $N$ be the amount of radioactive isotope $X$ remaining and $N_0$ be the initial amount.The amount of stable element $Y$ formed is $N_0 - N$.Given the ratio $\frac{N}{N_0 - N} = \frac{1}{15}$.This implies $\frac{N_0 - N}{N} = 15$,so $\frac{N_0}{N} - 1 = 15$,which gives $\frac{N_0}{N} = 16$.We know the radioactive decay law: $N = N_0 (\frac{1}{2})^{t/T_{1/2}}$,where $T_{1/2} = 50$ years.Thus,$\frac{N_0}{N} = 2^{t/T_{1/2}} = 16$.Since $16 = 2^4$,we have $\frac{t}{T_{1/2}} = 4$.Therefore,$t = 4 	imes 50 = 200$ years.

The half-life of a radioactive isotope $X$ is $50$ years. It decays to another element $Y$ which is stable. The two elements $X$ and $Y$ were found to be in the ratio of $1 : 15$ in a sample of a given rock. The age of the rock was estimated to be:

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