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I am used to using the formula $ R_H (\frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}}) $ to calculate the transition wavenumbers when an electron in a Hydrogen atom emits a photon. Can you use the same formula to calculate the wavenumber when a electron absorbs a photon?

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  • $\begingroup$ Do you mean emit/absorb an electron? If so, your question is easily answered: take $$1/n_{initial) = 0 and n_{final}$$ to be whatever PQN the electron was absorbed into. $\endgroup$ Jul 27, 2020 at 11:32
  • $\begingroup$ @AravindSuresh no sorry I made a mistake, I mean when the electron absorbs/emits a photon, I suppose it doesn't matter if it's in Hydrogen apart from I've used the Rydberg constant for hydrogen. $\endgroup$
    – SSlats
    Jul 27, 2020 at 11:42

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Yes, the energy (and therefore the photon wavelength) required for the transition between two energy levels is independent of the "direction". One way to see this is observing absorption and emmision spectra of the hydrogen, where the position of the lines for both types is the same.

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  • $\begingroup$ Thank you, so does this mean you would have a negative wavenumber? Or do you rearrange to make it positive? $\endgroup$
    – SSlats
    Jul 27, 2020 at 20:23
  • $\begingroup$ You'd have to rearrange it, Maybe it's easier to visualize if you think of this equations in terms of energy rather than wavelengths: if you multiply both sides of the Rydberg equations by $hc$ you obtain $\Delta E = hcR_H (1/n_1^2-1/n_2^2)$. $\Delta E$ is the difference in energy between both levels, and its sign indicates if the transition requires or releases energy. $\endgroup$ Jul 27, 2020 at 21:37

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