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Consider a radioactive nuclide which follow decay rate given by $A(t)=A_0 2^{\left(t / t_0\right\}}$, where $A(t)$ is the fraction of radioactive material remaining after time $t$ from the initial $A_0$ at zero time. Let $A_1$ be the fraction of original activity which remains after 120 hours. Likewise $A_2$ is the fraction of original activity remaining after 200 hours. If $\frac{A_2}{A_2}=1.6$, then the half-life $\left(t_0\right)$ will be
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The correct answer is:
20 hours
Given, decay rate, $A=A_0 2^{-\left(\frac{t}{t_0}\right)}$
Let $\frac{t}{t_0}=n$
where, $t_0=$ half life.
As,
$\begin{aligned}
& \frac{A_1}{A_2}=16 \quad \Rightarrow \frac{A_0 \cdot 2^{-n_1}}{A_0 \cdot 2^{-n_2}}=16 \\
& \Rightarrow \quad 2^{i_2-n_1}=16=2^4 \quad \Rightarrow \quad n_2-n_1=4
\end{aligned}$
$\Rightarrow \quad \frac{t_2}{t_0}-\frac{t_1}{t_0}=4$
As, $t_1=120 \mathrm{~h}$ and $t_2=200 \mathrm{~h}$,
$\Rightarrow \quad \frac{200}{t_0}-\frac{120}{t_0}=4 \quad \Rightarrow \quad \frac{80}{t_0}=4$
$\Rightarrow \quad t_0=\frac{80}{4}=20 \mathrm{~h}$
Let $\frac{t}{t_0}=n$
where, $t_0=$ half life.
As,
$\begin{aligned}
& \frac{A_1}{A_2}=16 \quad \Rightarrow \frac{A_0 \cdot 2^{-n_1}}{A_0 \cdot 2^{-n_2}}=16 \\
& \Rightarrow \quad 2^{i_2-n_1}=16=2^4 \quad \Rightarrow \quad n_2-n_1=4
\end{aligned}$
$\Rightarrow \quad \frac{t_2}{t_0}-\frac{t_1}{t_0}=4$
As, $t_1=120 \mathrm{~h}$ and $t_2=200 \mathrm{~h}$,
$\Rightarrow \quad \frac{200}{t_0}-\frac{120}{t_0}=4 \quad \Rightarrow \quad \frac{80}{t_0}=4$
$\Rightarrow \quad t_0=\frac{80}{4}=20 \mathrm{~h}$
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