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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Aicharya · Accepted Answer

For hydrogen atom, $\bar{v}=109677\left[\frac{1}{{n_{\mathrm{f}}}^2}-\frac{1}{\mathrm{n}_{\mathrm{i}}^2}ight] \mathrm{cm}^{-1}$ Here, $\mathrm{n}_{\mathrm{i}}=2, \mathrm{n}_{\mathrm{f}}=1$ $\begin{aligned} 	herefore \quad \bar{v} & =109677\left[\frac{1}{1^2}-\frac{1}{2^2}ight] \mathrm{cm}^{-1} \ & =109677\left[\frac{1}{1}-\frac{1}{4}ight] \mathrm{cm}^{-1} \ & =109677\left[\frac{3}{4}ight] \mathrm{cm}^{-1} \ & =82257.8 \mathrm{~cm}^{-1} \end{aligned}$

Calculate the wavenumber of the photon emitted during transition from the orbit of $\mathrm{n}=2$ to…

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