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A radioactive material has a half-life of 10 days. What fraction of the material would remain after 30 days?
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Verified Answer
The correct answer is:
0.125
$T_{1 / 2}=10$ days, $T=30$ days
$$
\begin{aligned}
& T=n \times T_{1 / 2} . \Rightarrow n=\frac{T}{T_{1 / 2}}=\frac{30}{10}=3 \\
& \frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^3=\frac{1}{8}=0.125 .
\end{aligned}
$$
$$
\begin{aligned}
& T=n \times T_{1 / 2} . \Rightarrow n=\frac{T}{T_{1 / 2}}=\frac{30}{10}=3 \\
& \frac{N}{N_0}=\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^3=\frac{1}{8}=0.125 .
\end{aligned}
$$
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