1. The Occupancy of States
and Distributions:
In statistical physics two particular
distributions occupy attention: the Bose-Einstein, and the Fermi-Dirac. For our
purposes we will spend much more time on the latter but it is useful to see how
both enter the picture.
In
the case of the Bose-Einstein formalism we are concerned with bosons which have
integral spin values (i.e. an orbital may be populated by any number of bosons
so the Pauli Exclusion principle doesn’t apply).
Consider
the distribution function for a system of non-interacting bosons in which the
system is in thermal and diffusive contact with a reservoir. Thus we may allow a circumstance such as
this:
Fig. 1: Illustrating systems in thermal and diffusive
contact.
Thus
heat can be exchanged between the systems A and B, as well as particles. In the formalism treatment we let e denote the energy of a
single orbital when occupied by one particle. Hence, when n particles occupy
the orbital then the energy must be n e.
To simplify, we treat one orbital as a system and ignore all
others. The grand sum taken over one
orbital would then be:
å¥
n = 0
å
e
exp ( Nm - e)/ t) = å¥ n = 0
ln e - n e/t
= å¥ n = 0
( l e - n e/t ) n
Now,
let x = l e - n e/t and sum the
series in closed form so the grand sum is now:
Ž = å¥ n = 0 (x) n = 1/ 1
– x = 1/ [ 1 - l e - n e/t ]
Provided: l e - n e/t < 1
The
ensemble average of the number of particles in the orbital is by definition of the average value:
<
n(e )> = å¥ n = 0
(nx) n / å¥ n = 0
(x) n =
x (d/dx) å¥ n = 0
(x) n
/ å¥ n = 0
(x) n =
x (d/dx) (1 – x) / (1 – x) -1
After
some calculus and messy algebra we find:
<
n(e )> = 1/ exp ( e - m)/ t
Which
is the Bose- Einstein distribution.
For
the Fermi-Dirac distribution we consider a simplified system represented by a
cubical box of volume V = L 3
The
number of electrons (Fermi particles) within occurs between energies e and e + de. The energies occur in a quantized way and are
depicted schematically below:
Fig 2. Box with Fermi
occupancy and energy levels.
The
diagram shows each level corresponding to two orbitals, one for spin up the
other for spin down. The Fermi energy:
e F = (ħ2 p
n F
/ L) 2 / 2m
Is
determined by the requirement that the system in the ground state hold N
electrons – with each orbital filled with one electron if the energy of the
orbital is less than:
e F.
Since for
a 3D box the quantized energy is:
ħ2 /2m [p
n2 / L]
where n2
= n x 2 + n y
2 + n z 2
Then the
number of states possessed by a 3D box is:
1/8 [4p n3 / 3]
For a
sphere, the number of orbitals in some radius is:
n= g [4p n3 / 3] (1/8) = g p n3 / 6
For
electrons, g = 2 so:
n= p n 3 / 3
Hence, if
the box holds N electrons the orbitals must be filled up to the quantum number
n F. So the total number of
electrons is:
N =
p n
F3 / 3 so n
F = (3N/p)
1/3
The Fermi
energy would then be:
e F = ħ2 ( p
/ L) 2 n F 2/ 2m =
ħ2 (
p / L) 2 (3N/p)
1/3/ 2m
Since V = L 3
we can write:
e F = ħ2 / 2m [3p2
N/ v] 2/3
This
yields good results for monovalent atoms because they have only one electron
per atom.
2. Treating the Nucleus as a Fermi Gas:
One
of the useful applications of the Fermi quantum statistics is to the atomic
nucleus, such as that for the deuteron.
The key to the application is assuming an independent particle model so
each nucleon moves in a smooth potential hole as opposed to being subject to
the actions of (A - 1) other nucleons.
In this sense, the model would exhibit similar properties to the
“electron gas” which we examined being confined to a 3D box in the previous
section. In this case, the nucleons will
be constrained to move in a spherical potential hole of radius:
R = r o
A 1/3
(I.e.
analogous to the case of electrons)
We
focus then on all momentum states being filled up to the Fermi momentum, i.e. p
F.
Since:
p F 2/ 2m = ħ2 /
2m [3p2
N/ v] 2/3
The
number of states up to p F per unit volume is:
N/ v
= 4p/
3 ( p F 3/
h3 )
Or the
volume in momentum space divided by the Planck constant.
Let’s now
apply this to the case of the deuteron, for which we have a neutron and proton
in each state and each can have spin up or spin down. Then we get a total of
four states so multiply the expression
by 4 to get:
4N = 4 [4p/ 3 ( c h3 ) (4p R 3/ 2 p h 3 ) 3]
Simplifying:
p F r o / h =
Ö (9 p/ 8) =
1.52
This
prescribes the maximum momentum present
in the nucleus in terms of the nuclear radius
r o .
p F = 1.52 (h/ r o)
Taking r o = 1.1 fm (e.g. 1.1 x 10 -15 m)
We learn
the maximum kinetic energies in the nucleus are as large as:
K F = p
F 2/ 2m = 39
MeV
If the binding energy of
the ‘last nucleon’, e.g. at the top of the Fermi distribution) is » 8 MeV then this result shows that the nucleus
looks to each nucleon (say to the neutron or proton in Deuterium) like a
potential hole of 39 MeV depth.
Suggested Problems:
1) Find the
Fermi sphere parameters: e F , v F and T F
for He 3 at absolute zero, viewed as a gas of non-interactive fermions. (The
density of the liquid is 81 kg/ m 3).
2)a) Show that (- ¶f / ¶ e) evaluated at the Fermi level (e = m) has value (4 kB T) -1.
Thus, the lower the temperature, then the steeper the slope of the Fermi-Dirac
function.
Hint: Use f(e) = 1/ {exp (m - e)/ t + 1}
b) Make a careful plot
of (- ¶f / ¶ e) vs. e/ kB for
the specific case: m/ kB = 5 x 10 4
K and
5 x 10 2 K .
3) Treating the Tritium
nucleus as a nucleonic electron gas, assume
an independent particle model (based on Fermi energy levels) so each nucleon moves in a smooth potential
hole.
Thereby
obtain the maximum momentum present in the nucleus in terms of the nuclear
radius r o .
p F = 1.52 (h/ r o)
(Take r o = 1.1 fm
)
Use this to estimate the
depth of the potential hole, i.e. the magnitude of the well in MeV for how the
nucleus looks to each nucleon (say to the neutrons or proton in Tritium) .
Give an idea of how errors enter your estimate
and explain why it can’t be taken too literally
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