The half-life period of a radioactive element | vedclass

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(A) Let $N_0$ be the initial amount of radioactive element $A$ and $N$ be the amount remaining after time $t$.Given that $A$ decays into $B$, the amou

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$248$

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(A) Let $N_0$ be the initial amount of radioactive element $A$ and $N$ be the amount remaining after time $t$.Given that $A$ decays into $B$, the amount of $B$ present at time $t$ is $N_B = N_0 - N$.The ratio of $A$ to $B$ is given as $\frac{N}{N_B} = \frac{1}{15}$.Substituting $N_B = N_0 - N$, we get $\frac{N}{N_0 - N} = \frac{1}{15}$.Cross-multiplying gives $15N = N_0 - N$, which simplifies to $16N = N_0$, or $\frac{N}{N_0} = \frac{1}{16}$.The law of radioactive decay states that $\frac{N}{N_0} = (\frac{1}{2})^n$, where $n = \frac{t}{T_{1/2}}$ is the number of half-lives.Given $T_{1/2} = 62 	ext{ years}$, we have $\frac{1}{16} = (\frac{1}{2})^{\frac{t}{62}}$.Since $\frac{1}{16} = (\frac{1}{2})^4$, we equate the exponents: $4 = \frac{t}{62}$.Therefore, $t = 62 	imes 4 = 248 	ext{ years}$.

The half-life period of a radioactive element $A$ is $62 \text{ years}$. It decays into another stable element $B$. An archaeologist found a sample in which $A$ and $B$ are in $1 : 15$ ratio. The age of the sample is (in $\text{ years}$)

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