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I have read that in a p-n junction, free electrons from n side diffuse to the p side and then recombine with the holes of p side. Here, it does not refer regarding energy bands. However, free electrons must be electrons in conduction band and hole must be in valence band. If free electrons recombine with the holes, they are degraded to the valence band from conduction band. For this, they must lose energy. So my question is "Do they really lose energy? If so, how?"

Wait! Before pondering about this, I have another question related to this question. It's due to this question, I came up with the above question. In p-n junction, if we have a look at the valence band, in case of p side, the valence band has more holes and less electrons whereas in case of n side, it has very less holes but more electrons. So don't electrons diffuse from valence band of n side to p side? My understanding says they diffuse as they are going to have lower energy when they are distributed evenly. So why shouldn't they? (Of course, before the barrier potential has been reached.)

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Yes charge carriers lose energy when they recombine. There are a few places the energy can go depending on the exact recombination process. The biggest two are emitting a photon, or transferring to heat in the semiconductor.

In the semiconductor band model, "electrons" are only found in the conduction band and "holes" are only in the valence band. Electrons in the n side of a pn nunction do diffuse to the p side and likewise holes do the opposite. Electrons and holes describe the contribution to current of the valence and conduction bands. To try to investigate diffusion of empty states in the valence band or filled states in the conduction band would be double counting.

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  • $\begingroup$ "or transferring to heat in the semiconductor" Just want to add that, in technical therms, this would be called "emitting a phonon". $\endgroup$
    – lnmaurer
    Jul 8, 2021 at 13:12

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