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On page 581 of Ashcroft and Mermin, Solid State Physics:

Acceptor Level In contrast to a donor level, an acceptor level, when viewed as an electronic level, can be singly or doubly occupied, but not empty. This is easily seen from the hole point of view. An acceptor impurity can be regarded as a fixed, negatively charged attractive center superimposed on an unaltered host atom. This additional charge $-e$ can weakly bind one hole (corresponding to one electron being in the acceptor level). The binding energy of the hole is $\epsilon_a - \epsilon_v$, and when the hole is "ionized" an additional electron moves into the acceptor level. However, the con­figuration in which no electrons are in the acceptor level corresponds to two holes being localized in the presence of the acceptor impurity, which has a very high energy due to the mutual Coulomb repulsion of the holes.

Bearing this in mind, we can calculate the mean number of electrons at an acceptor level from (28.30) by noting that the state with no electrons is now prohibited, while the two-electron state has an energy that is $\epsilon_a$ higher than the two one-electron states.

Therefore $$ \langle n\rangle = \frac{ 2e^{\beta\mu} + 2e^{-\beta(\epsilon_a-2\mu)} }{ 2e^{\beta\mu} + e^{-\beta(\epsilon_a-2\mu)} } \,. $$

In the quote, $\epsilon_a$ is the energy of the acceptor level, $\epsilon_v$ is the energy of the top of the valence band, and $\langle n\rangle$ is the average occupancy of the acceptor level. Equation (28.30) is $$ \langle n\rangle = \frac{ \sum N_je^{-\beta(E_j- \mu N_j)} }{ \sum e^{-\beta(E_j- \mu N_j)} } $$ where $E_j$ and $N_j$ are the energy and number of electrons in state $j$.

It seems like the two electron state corresponds to a situation where the acceptor ion has a full outer shell (hole filled), whereas the one electron state is when the hole is not filled. I'm not sure what the zero electron state refers to. Why can we not say then that the situation where the acceptor hole is filled is the single electron state?1

Also, if the acceptor level cannot be empty, at absolute zero is the Fermi level between $\epsilon_v$ and $\epsilon_a$? If so, would this not mean the acceptor level is empty?


1. Like the description on page 580 "The hole is bound when the level is empty."

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  • $\begingroup$ Have you found the answer? I'm looking for it but can't see it. $\endgroup$ Commented Apr 8, 2021 at 20:50
  • $\begingroup$ @KarimChahine Unfortunately not. I don't really understand their physical motivation for calculating the average occupancy of the acceptor level in this manner. On page 580, they say "The hole is bound when the level is empty" as well, so I'm assuming perhaps they mean something else by "acceptor level" here than on pg. 580 (and in general, since I think all levels apart from the valence should be empty at zero temperature). Maybe "acceptor level" is a akin to an electron orbital (1 electron filled is the 1 hole state and 2 electron filled is the zero holes state). $\endgroup$
    – eugenhu
    Commented Apr 9, 2021 at 13:11
  • $\begingroup$ It seems like you can equivalently do the $\langle n\rangle$ calculation by considering the acceptor level as capable of holding zero or one electron. The zero electron state corresponding to the one hole state, and the one electron state as the zero/filled hole state. $\endgroup$
    – eugenhu
    Commented Apr 9, 2021 at 13:12
  • $\begingroup$ My reasoning for this is as follows. A semiconductor has N, say silicon, atoms. Impurities, say boron atoms, are added to the material. A single impurity is modeled as a -1 ion with a bound +ve hole orbiting around it, superimposed on the pure N atom silicon crystal with N standing wave electrons. If the bound hole is ionised, gaining an energy of $\epsilon_a - \epsilon_v$, and becomes effectively a free particle, then presumably one of the N electrons loses an energy $\epsilon_a - \epsilon_v$ and drops into a different state, which we identify as the acceptor level. $\endgroup$
    – eugenhu
    Commented Apr 9, 2021 at 13:13
  • $\begingroup$ I seem the get the same answer doing it this way, but whether this physical interpretation of the mathematics of the band theory as a whole is accurate, I'm not sure and haven't put that much thought about it. $\endgroup$
    – eugenhu
    Commented Apr 9, 2021 at 13:13

1 Answer 1

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One should remember that each energy level has two spin states. You are right in that "two electron state corresponds to a situation where the acceptor ion has a full outer shell (hole filled), whereas the one electron state is when the hole is not filled". That one electron is assumed to be occupying the corresponding bound energy state. Every such energy state can accommodate two electrons with different spins.

Having two holes attached to the acceptor impurity is energetically unfavorable, similarly to how it is unfavorable to have two electrons on the same donor site: the negative ion is shielded by the first hole (which is assumed to be already there) and the second hole is not going to be attracted to the impurity strongly enough to form a bound state (this is a bit too handwavy argument to my taste, but I am quite sure this is what Ashcroft and Mermin mean).

The equation for the mean number of particles $\langle n\rangle$ is obtained from the assumption that there are two one-particle states and a single two-particle state.

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    $\begingroup$ Why does the electron occupy the bound energy state in the one electron state? And what do you think the answer is for OP's last question? $\endgroup$ Commented Apr 29, 2021 at 15:24
  • $\begingroup$ Think of acceptors inside a semiconductor as kind of halogens atoms in vacuum. And think of donor atoms inside a semiconductor as an alkali metal in vacuum. In other words acceptors behave similarly to fluorine and donors behave similarly to sodium. Fluorine does not like to be positively charged and sodium does not like to be negatively charged. Conventionally, we call a neutral state of a donor/acceptor to have one electron, positively charged state - zero electrons and negatively charged state - two electrons $\endgroup$
    – Pavlo. B.
    Commented Apr 29, 2021 at 18:59
  • $\begingroup$ As fluorine doesn't like to be positively charged, acceptors do not like either. Similarly, as sodium does not like to be negatively charged, donors do not like either. It is a long story why sodium and fluorine behave this way, but it all comes from interplay between filled energy shells and charge shielding. $\endgroup$
    – Pavlo. B.
    Commented Apr 29, 2021 at 19:00
  • $\begingroup$ Fermi distribution is a distribution for non-interacting electrons and it doesn't work as intuitively or cleanly for interacting systems, where placing electron in one level influences positions of all other levels. It is easier to do arguments from the more basic energy considerations $\endgroup$
    – Pavlo. B.
    Commented Apr 29, 2021 at 19:05

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