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after an electron moves from the n-type into the p-type silicon, it will combine with a hole and disappear. That makes an atom complete and more stable and it gives off a little burst of energy. But why does it emits this energy. I want the actual explanation.

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3 Answers 3

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You said it yourself : electron-hole recombination lowers the energy of the system ; by conservation of energy, energy must be emitted somehow. Think of the electron and the hole as quasiparticles of the system with a certain energy. When they recombine this energy is transferred to a photon.

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Generally speaking the electron and the atom, when separated, are in a state with greater energy respect to the situation in witch are bounded. They will tend to minimize their total energy thus expelling "something" containing this energy difference. In the case of direct gap semiconductors (as in LEDS) this "something" should contain a lot of energy and few momentum, due to the band structure of the semiconductors, and photons are the particle best suited for the job. In other cases (band structures), to reach the minimum of energy, we need energy as well as a lot of momentum, so exchanging energy through vibrations (phonons) is favorable, but in this case you wont see light.

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  • $\begingroup$ But why when separated, electrons are in a state with greater energy respect to the situation in which are bounded? $\endgroup$
    – Touhid
    Jan 20, 2016 at 10:50
  • $\begingroup$ Classically you would probably say: there is the em force witch is attractive and the particles getting closer decrease the potential energy, transforming it in kinetic energy (or in our case light). But this is really not a good example. In QM instead you would say: the electron is in a state which is not the state with minimum possible energy in the system, he wants to minimize his energy, so he jump in the lower state available $\endgroup$ Jan 20, 2016 at 11:01
  • $\begingroup$ Detailed explanation would be helpful. I actually want to understand the very basic. $\endgroup$
    – Touhid
    Jan 20, 2016 at 11:24
  • $\begingroup$ You've got the electron and the atom, the "states" are all the possible ways in witch we can found them (they are countless!). With QM you can calculate the probabilities of every possible transition among states. It turns out that if exists a state with lower energy in witch you can fall, there is a great probability that you will. You could ask, why is that? You could go deeper and deeper but at the end you would only find principles (least action, thermodynamics.) that are without explanation, but as far as we know they works. So it's like that because we've never seen something different. $\endgroup$ Jan 20, 2016 at 11:52
  • $\begingroup$ But what does this high energy state and lower energy state actually means? I think my main problem is not quite understanding this point. $\endgroup$
    – Touhid
    Jan 20, 2016 at 12:13
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Light is emitted from a p-n junction for the same reason it is emitted in any other atomic process: because of an oscillating charge distribution. During the finite time the electron spends crossing the gap, it is in a superposition of two states: the low-energy (low frequency) and high energy (high frequency) states. The charge density is calculated by adding the two states (and squaring the resultant amplitude). Because the states have different frequencies, they go in and out of phase at a rate corresponding to the difference of those frequencies. As a result, there is an oscillating charge density. The resulting emission of radiation can be calculated using Maxwell's equations.

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