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The half-life of a radiocative isotope is $30 \mathrm{~h}$. How long will it take to get reduced to $12.5 \%$ of its initial amount?
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Verified Answer
The correct answer is:
90 h
Half-life of radioactive isotope,
$T_{1 / 2}=30 \mathrm{~h}$
If $N_0$ be the initial amount of radioactive isotope, then remaining amount $(N)$ in time $t$ is given as
$N=125 \% \text { of } N_0=\frac{125}{100} \times N_0 \Rightarrow N=\frac{N_0}{8}$
We know that,
$\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow \quad\left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^n \Rightarrow n=3 \Rightarrow \frac{t}{T_{1 / 2}}=3 \\
\Rightarrow \quad t & =3 \times T_{1 / 2}=3 \times 30=90 \mathrm{~h}
\end{aligned}$
$T_{1 / 2}=30 \mathrm{~h}$
If $N_0$ be the initial amount of radioactive isotope, then remaining amount $(N)$ in time $t$ is given as
$N=125 \% \text { of } N_0=\frac{125}{100} \times N_0 \Rightarrow N=\frac{N_0}{8}$
We know that,
$\begin{aligned}
N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow \quad\left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^n \Rightarrow n=3 \Rightarrow \frac{t}{T_{1 / 2}}=3 \\
\Rightarrow \quad t & =3 \times T_{1 / 2}=3 \times 30=90 \mathrm{~h}
\end{aligned}$
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