The probability function of electrons occupying the donor state

[Dale12]

In Semiconductor physics and devices: basic principles
[By Donald A. Neamen], chapter 4, section 4.4, it says:
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One postulate used in the derivation of the Fermi-Dirac probability function was the Pauli exclusion principle, which states that only one particle is permitted in each quantum state. The Pauli exclusion principle also applies to the donor and acceptor states.

Suppose we have $$N_i$$ electrons and $$g_i$$ quantum states, where the subscript i indicates the ith energy level. There are $$g_i$$ ways of choosing where to put the first particle. Each donor level has two possible spin orientations for the donor eletron;thus each donor level has two quantum states. The insertion of an electron into one quantum state, however, precludes putting an electron into the second quantum state. By adding one electron, the vacancy requirement of the atom is satisfied, and the addition of a second electron in the donor level is not possible. The distribution function of donor electrons in the donor eneergy states is then slightly different than the Fermi-Dirac funtion.

The probability function of electrons occupying the donor state is

$$n_d=\frac{N_d}{1+\frac{1}{2}\exp{\frac{E_d-E_F}{kT}}}$$

where n_d is the density of electrons occupying the donor level and E_d is the energy of the donor level. The factor 1/2 in this equation is direct result of the spin factor just mentioned. The 1/2 factor is sometimes written as 1/g. Where g is called a degeneracy factor.
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My question are these:

1.how to understand sentences below? since there are two quantum states, why could only one electron be inserted?

Each donor level has two possible spin orientations for the donor eletron;thus each donor level has two quantum states. The insertion of an electron into one quantum state, however, precludes putting an electron into the second quantum state. By adding one electron, the vacancy requirement of the atom is satisfied, and the addition of a second electron in the donor level is not possible.

2.why the factor 1/2 is related to the spin factor? or why it is added before $$\exp{\frac{E_d-E_F}{kT}$$ but not before $$\frac{N_d}{1+\exp{\frac{E_d-E_F}{kT}}}$$ when considering the Fermi-Dirac distribution function.

Thanks for any help!
ps:
Semiconductor physics and devices: basic principles could be read in google books:
http://books.google.com/books?id=9oEifMuMAVsC&printsec=frontcover#v=onepage&q=&f=false

and my question is from Page 128 to 129.

 

[Anik Paul]

I am also struggling with these two sentences "The insertion of an electron into one
quantum state, however, precludes putting an electron into the second quantum state.
By adding one electron, the vacancy requirement of the atom is satisfied, and the
addition of a second electron in the donor level is not possible."

 

[Henryk]

The insertion of an electron into one quantum state, however, precludes putting an electron into the second quantum state

This is actually simple. What is a donor? . A donor is an atom that has one extra electron substituted for of the atoms of the crystal. Take an example of silicon (4-valent semiconductor). A donor would be a pentavalent element which would have 4 of its valence electrons participate in the bonding with neighbouring silicone atoms. The fifth electron of the donor is not contributing to the bonding and its energy is higher than that of the bonding electrons and it has to go to the next available energy band: the conduction band of the silicon. However, it is not quite free to go. It is still attracted to the positive charge of its atom thus creating a donor energy level. This is somewhat similar to a hydrogen atom energy band except the attractive potential between the donor ion and it electron is screened by the dielectric constant of the crystal and there are exclusion principle to be concerned with. The net result is that the donor energy state is quite close to the lowest energy of an electrons free to roam the lattice - the bottom of the conduction band.
Now, this state is double degenerate: we can have the electron in either of the two spin states. What we can't have is two electrons attracted to the same donor ion. Once an electron is attached to the donor ion, there is no force to attract another electron. In summary, there is a possibility of actually having a second electron in the donor state of the opposite direction but its energy would be much larger. Therefore, this the probability of such event is so small that it can be ignored.

Second point,

The probability function of electrons occupying the donor state is

Actually this is not probability. if N_d is the density of the donor sites, then n_d is the density of the occupied sites. Probability is never greater than 1 !.
Where does the factor come from?
It is actually easy to derive. From statistical physics, the probability of occupation of a level of energy $E$ is proportional to the factor $exp(\frac{-E}{kT})$.
We have three possible states: an electron bound to the donor site with spin up (energy E_d) , spin down (same energy) or the donor site empty and electron at the energy of the reference reservoir, that is Fermi energy, E_F. ( I.e. the electron chemical potential). Two electrons bound to a donor site have much higher energy and need not be considered. Therefore, the probability of a donor state being occupied is equal to the (probability of spin up occupied + spin down occupied)/( spin up + spin down + empty). Adding all the factors, we have $P(E_ d) = \frac{2 * exp(-E_d)}{2*exp(-E_d) + exp(-E_F)}$ = \frac 1{1+\frac 1 2 exp(\frac{E_d- E_F}{kT}##

 

[Henryk]

Sorry, pressed the wrong button

\frac 1{1+\frac 1 2 exp(\frac{E_d- E_F}{kT}##

was meant to display
$\frac 1 { 1+ \frac 1 2 exp(\frac {E_d - E_F}{kT}) }$
k is the Boltzmann constant.

 

[Anik Paul]

This makes sense now! Thank you so much for your valuable explanation!

Now, this state is double degenerate: we can have the electron in either of the two spin states. What we can't have is two electrons attracted to the same donor ion. Once an electron is attached to the donor ion, there is no force to attract another electron. In summary, there is a possibility of actually having a second electron in the donor state of the opposite direction but its energy would be much larger. Therefore, this the probability of such event is so small that it can be ignored.

Second point,

Actually this is not probability. if N_d is the density of the donor sites, then n_d is the density of the occupied sites. Probability is never greater than 1 !.
Where does the factor come from?
It is actually easy to derive. From statistical physics, the probability of occupation of a level of energy $E$ is proportional to the factor $exp(\frac{-E}{kT})$.
We have three possible states: an electron bound to the donor site with spin up (energy E_d) , spin down (same energy) or the donor site empty and electron at the energy of the reference reservoir, that is Fermi energy, E_F. ( I.e. the electron chemical potential). Two electrons bound to a donor site have much higher energy and need not be considered. Therefore, the probability of a donor state being occupied is equal to the (probability of spin up occupied + spin down occupied)/( spin up + spin down + empty). Adding all the factors, we have $P(E_ d) = \frac{2 * exp(-E_d)}{2*exp(-E_d) + exp(-E_F)}$ = \frac 1{1+\frac 1 2 exp(\frac{E_d- E_F}{kT}##