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In a LED the electrons in the n-type move to the p-type where they are then 'combined' with the holes in the valence shell of the p-type which produces photons. My question is that when this happens are the electrons gone or have disappeared?

Also do the electrons 'run out' in this process because they are being used up to create photons?

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  • $\begingroup$ @SamapanBhadury Thank you for the explanation. I have one final question: as you stated above the electrons don't vanish but get 'replaced' with the hole which releases a photon. Does this take place with all impurities added to the semi-conductor like boron or specific types of elements? $\endgroup$ – Faheem Azeemi Oct 3 at 19:04
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By 'combined' with the holes in the valence band (not shell), you mean the recombination process. To answer your question, you must understand the concept of 'hole'. You probably already know 'hole' is not a separate particle rather it is a lack of electron. In overly simplified words, you can think a positively charged ion has a hole in it as it requires a electron to neutralize itself. If a electron is near by, the ion attracts it, captures it and becomes neutral and in the process emits a photon. In terms of 'hole' we say, electron recombines with the 'hole' and emits a photon. But note that the captured or recombined electron is not vanished. It sits in the electron shell of the atom. Thus it does not vanish, rather goes to a bound state within the atom.

From the above explanation, I guess the question of something 'running out' of electron is no longer a problem. The electron does not vanish. It just stays within the atom. The emitted photon is just the excess energy carried by the electron. Due to quantum nature of the electron's orbits, the emitted photons have discrete frequencies. If you have further confusion, let me know in the comments, I will be glad to answer it.

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  • $\begingroup$ Thank you for the explanation. I have one final question: as you stated above the electrons don't vanish but get 'replaced' with the hole which releases a photon. Does this take place with all impurities added to the semi-conductor like boron or specific types of elements? $\endgroup$ – Faheem Azeemi Oct 3 at 18:49
  • $\begingroup$ The impurities i.e. the doping agents, they usually differ from the electron configuration of the main semi-conducting material (Si or Ge etc.) Silicon has 4 valance electrons. Thus the doping agents are chosen to have 3 or 5 valance electrons. If you have exactly same number of valance electrons, then there is no creation of holes and thus no recombination. $\endgroup$ – Samapan Bhadury Oct 3 at 19:16
  • $\begingroup$ My final question is when the electrons combine with the holes in the valence band, is that captured electron able to be released back (into the conduction band) since if this did not happen the electrons would 'run' out, because they are stuck in the valence band, and the battery only provides electrons when one electron gets attracted so that there isn't a net charge? $\endgroup$ – Faheem Azeemi Oct 3 at 19:56
  • $\begingroup$ Surely, they can again acquire sufficient energy and rise back up to conduction band. But I would advise you to be careful with the phrase 'run out'. It's not like the semi-conductor turns to a electron-less material. It is just that the electrons in the valance band do not contribute to current conduction. I didn't get the last part of your question (the battery part). $\endgroup$ – Samapan Bhadury Oct 3 at 20:41
  • $\begingroup$ If you think I have answered all your questions regarding the present context, can you accept the answer? $\endgroup$ – Samapan Bhadury Oct 14 at 21:42
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Electrons absorb and emit photons IN response to local energy levels. They do not covert. They do not disappear. It's in every introduction to the atom, on a basic level. They do not die in the process of absorption or emission of photons.

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