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Calculate the wavenumber of the photon emitted during transition from the orbit of $\mathrm{n}=2$ to $\mathrm{n}=1$ in hydrogen atom. $\left[R_H=109677 \mathrm{~cm}^{-1}\right]$
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The correct answer is:
$82257.8 \mathrm{~cm}^{-1}$
For hydrogen atom,
$\bar{v}=109677\left[\frac{1}{{n_{\mathrm{f}}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right] \mathrm{cm}^{-1}$
Here, $\mathrm{n}_{\mathrm{i}}=2, \mathrm{n}_{\mathrm{f}}=1$
$\begin{aligned}
\therefore \quad \bar{v} & =109677\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{1}{1}-\frac{1}{4}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{3}{4}\right] \mathrm{cm}^{-1} \\
& =82257.8 \mathrm{~cm}^{-1}
\end{aligned}$
$\bar{v}=109677\left[\frac{1}{{n_{\mathrm{f}}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}\right] \mathrm{cm}^{-1}$
Here, $\mathrm{n}_{\mathrm{i}}=2, \mathrm{n}_{\mathrm{f}}=1$
$\begin{aligned}
\therefore \quad \bar{v} & =109677\left[\frac{1}{1^2}-\frac{1}{2^2}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{1}{1}-\frac{1}{4}\right] \mathrm{cm}^{-1} \\
& =109677\left[\frac{3}{4}\right] \mathrm{cm}^{-1} \\
& =82257.8 \mathrm{~cm}^{-1}
\end{aligned}$
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