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 $\frac{N}{N_0 - N} = \frac{1}{15}$.Adding $1$ to both sides: $\frac{N}{N_0 - N} + 1 = \frac{1}{15} + 1 \implies \frac{N + N_0 - N}{N_0 - N} = \frac{16}{15} \implies \frac{N_0}{N_0 - N} = \frac{16}{15}$.Taking the reciprocal: $\frac{N}{N_0} = \frac{1}{16}$.We know the radioactive decay law: $\frac{N}{N_0} = (\frac{1}{2})^n$,where $n$ is the number of half-lives.$\frac{1}{16} = (\frac{1}{2})^4$,so $n = 4$.The age of the rock $t = n 	imes t_{1/2} = 4 	imes 50 \, years = 200 \, years$.

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 ........ $years$.

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