How do you know if electron is deexciting or exciting itself? I got a question which goes like this:


  According to Bohr's theory, the electronic energy of H-atom in the n th Bohr's orbit is given by $$E_n=\frac{-21.7*10^{-12}}{n^2}J$$ Calculate the longest wavelength of electron from the third Bohr's orbit of the He+ ion.


For longest wavelength transition, we need lowest energy transition. Therefore the electron can de-excite to $n=2$ or excite to $n=4$. I do not need the answer to the question, but I need to know how to know if the electron is exciting or de-exciting. 

From what I know, I think it would be exciting to $n=4$ since the energy is lesser than de-excitation energy.
The 'longest wavelength of electron' does not make any sense. I assume that the question asks for the longest wavelength transition when a photon is absorbed or emitted (as fluorescence). One way to answer this is to look up the Balmer formula for hydrogenic like atoms and choose two principle quantum numbers for the transition you want. Alternatively you take the difference in energy between two levels using the formula given. As you start with n=3 the closest is either going to be n=4 (absorption) or n=2(emission). I leave it up to you to do the calculation but the answer must be positive as it corresponds to a photon's wavelength :) ( Note that the question gives the energy in Joules and you will want the wavelength in nanometres).