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A radioactive substance contains 10000 nucle $i$ and itshalf-life period is 20 days. The number of nuclei present at the end of 10 days is
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2617 Upvotes
Verified Answer
The correct answer is:
7070
We know,
$$
\begin{array}{l}
\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \Rightarrow \frac{\mathrm{N}}{10000}=\left(\frac{1}{2}\right)^{\frac{10}{20}} \\
\Rightarrow \mathrm{N}=\frac{10000}{\sqrt{2}}=\frac{10000}{1.414}=7070
\end{array}
$$
$$
\begin{array}{l}
\frac{\mathrm{N}}{\mathrm{N}_{0}}=\left(\frac{1}{2}\right)^{\mathrm{t} / \mathrm{T}} \Rightarrow \frac{\mathrm{N}}{10000}=\left(\frac{1}{2}\right)^{\frac{10}{20}} \\
\Rightarrow \mathrm{N}=\frac{10000}{\sqrt{2}}=\frac{10000}{1.414}=7070
\end{array}
$$
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