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If $75 \%$ of a radioactive sample disintegrates in 16 days, the half-life of the radioactive sample is
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If $N_0$ be the initial amount of radioactive sample, then after $t=16$ day, remaining amount,
$\begin{aligned}
N & =N_0-75 \% \text { of } N_0 \\
& =N_0-\frac{75}{100} N_0=\frac{N_0}{4}
\end{aligned}$
Since, we know that,
$\begin{aligned}
& N=N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{4}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow & \frac{1}{4}=\left(\frac{1}{2}\right)^n \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^n \\
\Rightarrow & n=2 \Rightarrow \frac{t}{T_{1 / 2}}=2 \Rightarrow \frac{16}{T_{1 / 2}}=2 \\
\Rightarrow \quad & T_{1 / 2}=\frac{16}{2}=8 \text { days }
\end{aligned}$
$\begin{aligned}
N & =N_0-75 \% \text { of } N_0 \\
& =N_0-\frac{75}{100} N_0=\frac{N_0}{4}
\end{aligned}$
Since, we know that,
$\begin{aligned}
& N=N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{4}=N_0\left(\frac{1}{2}\right)^n \\
\Rightarrow & \frac{1}{4}=\left(\frac{1}{2}\right)^n \Rightarrow\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^n \\
\Rightarrow & n=2 \Rightarrow \frac{t}{T_{1 / 2}}=2 \Rightarrow \frac{16}{T_{1 / 2}}=2 \\
\Rightarrow \quad & T_{1 / 2}=\frac{16}{2}=8 \text { days }
\end{aligned}$
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