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An electron in hydrogen atom makes a transition $n_1 \rightarrow n_2$ where $n_1$ are principle quantum numbers of the two states. Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of $n_1$ and $n_2$ are:
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The correct answer is:
$n_1=4$ and $n_2=2$
Time period of an electron in Bohr orbit is
$\begin{aligned}
& T=\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \\
\Rightarrow T & \propto n^3 \\
\Rightarrow \frac{T_1}{T_2} & =\frac{n_1^3}{n_2^3}
\end{aligned}$
As $T_1=8 T_2$ then
$\begin{aligned}
\Rightarrow & \left(\frac{8 T_2}{T_2}\right) & =\frac{n_1^3}{n_2^3} \\
\Rightarrow & \left(\frac{n_1}{n_2}\right)^3 & =\left(2^2\right)^3 \\
\Rightarrow & \frac{n_1}{n_2} & =2^2 \\
\Rightarrow & n_1 & =2 n_2
\end{aligned}$
$\begin{aligned}
& T=\frac{4 \varepsilon_0^2 h^3 n^3}{m e^4} \\
\Rightarrow T & \propto n^3 \\
\Rightarrow \frac{T_1}{T_2} & =\frac{n_1^3}{n_2^3}
\end{aligned}$
As $T_1=8 T_2$ then
$\begin{aligned}
\Rightarrow & \left(\frac{8 T_2}{T_2}\right) & =\frac{n_1^3}{n_2^3} \\
\Rightarrow & \left(\frac{n_1}{n_2}\right)^3 & =\left(2^2\right)^3 \\
\Rightarrow & \frac{n_1}{n_2} & =2^2 \\
\Rightarrow & n_1 & =2 n_2
\end{aligned}$
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