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A radioactive substance has a half-life of $10^8$ years and an activity of $10^4 \mathrm{~Bq}$. The number of atoms of this substance present is
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Verified Answer
The correct answer is:
$4.5 \times 10^{19}$
Half-life of a radioactive substance,
$\begin{aligned}
T_{1 / 2} & =10^8 \text { years } \\
& =10^8 \times 365 \times 24 \times 60 \times 60=3.15 \times 10^{15} \mathrm{~s}
\end{aligned}$
Activity, $R=10^4 \mathrm{~Bq}$
Let $N$ be the number of atoms of substance, then we know that
$\begin{aligned}
R & =\lambda N \Rightarrow \lambda=R / \lambda \\
& =\frac{R}{\frac{0.693}{T_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{T_{1 / 2}}\right) \\
& =\frac{R \times T_{1 / 2}}{0.693}=\frac{10^4 \times 3.15 \times 10^{15}}{0.693} \\
& =4.54 \times 10^{19} \simeq 4.5 \times 10^{19} .
\end{aligned}$
$\begin{aligned}
T_{1 / 2} & =10^8 \text { years } \\
& =10^8 \times 365 \times 24 \times 60 \times 60=3.15 \times 10^{15} \mathrm{~s}
\end{aligned}$
Activity, $R=10^4 \mathrm{~Bq}$
Let $N$ be the number of atoms of substance, then we know that
$\begin{aligned}
R & =\lambda N \Rightarrow \lambda=R / \lambda \\
& =\frac{R}{\frac{0.693}{T_{1 / 2}}} \quad\left(\because \lambda=\frac{0.693}{T_{1 / 2}}\right) \\
& =\frac{R \times T_{1 / 2}}{0.693}=\frac{10^4 \times 3.15 \times 10^{15}}{0.693} \\
& =4.54 \times 10^{19} \simeq 4.5 \times 10^{19} .
\end{aligned}$
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