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The ratio of the energies of the electron in the hydrogen atom in the first and second excited states is
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The correct answer is:
$9: 4$
(a) Energy in Hydrogen atom is given as:
$\begin{aligned}
& E=-13.6 \frac{z^2}{n^2} \\
& \therefore \quad E \propto \frac{1}{n^2} \\
& \frac{E_1}{E_2}=\frac{n_2^2}{n_1^2}=\left(\frac{3}{2}\right)^2 \\
& E_1: E_2=9: 4
\end{aligned}$
$\begin{aligned}
& E=-13.6 \frac{z^2}{n^2} \\
& \therefore \quad E \propto \frac{1}{n^2} \\
& \frac{E_1}{E_2}=\frac{n_2^2}{n_1^2}=\left(\frac{3}{2}\right)^2 \\
& E_1: E_2=9: 4
\end{aligned}$
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