LECTURE 12.
Prof.R.Parthasarathy
Thomas-Fermi Statistical Model
The method of determining the equivalent potential V (r) due to Thomas
and Fermi (1927-28) assumes that V (r) varies slowly in an electron wavelength so that many electrons can be localized within a volume over which
V (r) changes by a small fraction of itself. Then,the electrons can be treated
by statistical mechanics obeying Fermi-Dirac statistics. At normal temperatures, the thermal energy kT is very small compared with V (r) everywhere
except at the edge of the atom, where the chances of finding the electron is
small. FD statistics requires that the electron states fill in order of increasing
energy.
3
L
dkx dky dkz
The number of electron states in a cube of edge length L is 2π
times 2 (possible spin states). Then, the number of states for which the momentum p~ = h̄~k is < p~0 is
p
L 3 Z h̄0 Z π Z 2π 2
p0 3 L3
2
k dksinθdθdφ =
.
2π
0
0
0
3π 2 h̄3
(1)
If all the states are occupied, then the number of electrons per unit volume is
p2
p30
and the kinetic energy does not exceed 2m0 . The maximum kinetic energy
3π 2 h̄3
at any distance r from the nucleus is −V (r) and so p20 = −2mV (r). Then,
the number of electrons depends on r as
p30
,
3π 2 h̄3
3
(−2mV (r)) 2
.
=
3π 2 h̄3
n(r) =
(2)
The charge density is en(r) and so the Poisson equation for the electrostatic is
1 2
∇ V (r) = −4πen(r),
e
1 1 d 2 dV (r) r
= −4πen(r).
e r2 dr
dr
1
(3)
Equations 2 and 3 are the two simultaneous equations for n(r) and V (r). The
idea is to solve them! We recall that for neutral atoms of atomic number Z,
2
we realize that when r → 0, the leading term in V (r) should be − Zer , the
potential due to the nucleus. This is written as
Limit(r → 0)(rV (r)) = −Ze2 .
(4)
As r → ∞, there must be no net charge inside the sphere of radius r. So,
Limit(r → ∞)V (r) = 0. These are the boundary conditions necessary.
Substituting (2) in (3) for n(r), we have
3
1 d 2 dV (r) 4e2 2
r
=
−
−
2mV
(r)
.
r2 dr
dr
3πh̄3
(5)
2
In (5), let us put V (r)
= −Zer χ and note dVdr(r) = −Ze2 {− r12 χ + 1r dχ
}.
dr
d
2 dV (r)
2 d2 χ
Then it follows that: dr r dr = −Ze r dr2 . Let us set r = bx where b is
a constant to be fixed shortly. Then, (5) becomes
√
√
d2 χ
8e2 m 2mZe2 b b 3
√ χ2 .
=
(6)
dx2
x
3πh̄3
The constant b is fixed now as
√
b b =
3πh̄3
√
,
8me2 2mZe2
2
h̄2
1 3π 3
b =
1 .
2 4
me2 Z 3
(7)
The above result for b can be further simplified upon using the Bohr radius
h̄2
a0 since a0 = me
2 as
2
1 3π 3 a0
b =
1 ,
2 4
Z3
1
=
1 0.885a0 .
Z3
(8)
This shows that b has the dimension of length and so in r = bx, x is dimensionless. Now let us look at the boundary conditions for χ. As r → 0, we
2
2
have (rV (r)) = −Ze2 and as V (r) = − Zer χ, it reads as: limit(x → 0) χ = 1.
Similarly, we have limit(x → ∞) χ = 0. Thus, equation 6 upon using (7),
becomes,
3
d2 χ(x)
1
2,
√
χ(x)
=
with
dx2
x
χ(x) = 1, at x = 0;
χ(x) = 0, at x = ∞,
(9)
which is known as the Thomas-Fermi equation. In here χ(x) > 0. We know
that the there is a maximum for the kinetic energy and from (2), it follows
2
that when χ(x) < 0, n(r) must vanish. From (3) and (5), we have ddxχ2 = 0
when χ < 0. So, the Thomas-Fermi equations become:
3
d2 χ(x)
1
= √ χ(x) 2 , f or χ > 0,
2
dx
x
2
d χ(x)
= 0, f or χ < 0,
dx2
(10)
with the same boundary conditions as in (9). The second equation in (10) is
easy to solve and the solution is χ(x) = A(x − x0 ) with A negative. So χ(x)
has at most one zero in the interval (0, ∞) such that χ will be positive in
the interval (0, x0 ) and negative in the interval (x0 , ∞). Where is x0 ? The
number of electrons given in (2) should satisfy the normalization
4π
Z ∞
n(r)r2 dr = Z.
(11)
0
This expression using (2) and (7) reads as
Z x0 √
3
xχ(x) 2 dx = 1.
(12)
0
The upper limit is taken to be x0 since after this, n(r) is taken to be zero as
in here χ is negative. From (10), this becomes
1 =
Z x0
x
0
d2 χ
dx,
dx2
dχ
− χ}|x0 0 ,
dx
= x0 χ0 (x0 ) + 1,
= {x
3
(13)
using the boundary conditions. This condition requires the derivative χ0 to
vanish at the same point as does χ itself. This means that the point in
question must be at infinity. The Thomas-Fermi function χ(x) satisfying
(10) is independent of Z. It is an universal function, valid for large Z. The
differential equation (10) is numerically solved by V.Bush and S.H.Caldwell,
Phys.Rev. 38 (1931) 1898. We give the graph of the solutions below.
1×
χ(x) 0.5 ×
0,0
×
5
×
10
x
The Thomas-Fermi χ(x) function for any value of Z.
4
×
15
Applications
Atomic Radius
The notion of atomic radius now needs to be defined. Why? The electronic density given in (2) and expresses in terms of χ becomes zero only at
infinity (as the point x0 is at infinity) and so the atom is not occupying a
well defined region in space! This is unphysical. By atomic radius, we mean
the radius R(α) of the sphere centered at the origin and containing a given
fraction (1 − α)Z of the Z-atomic electrons. Then, it follows that
(1 − α)Z = 4π
Z R(α)
n(r)r2 dr.
(14)
0
From (2) and following the steps in (12), we find,
d2 χ
dx, R(α) = bX(α),
dx2
0
= Z{X(α)χ0 (X(α)) − χ(X(α)) + χ(0)}.
(1 − α)Z = Z
Z X(α)
x
(15)
Using the boundary condition χ(0) = 1, we have
α = χ(X(α)) − X(α)χ0 (X(α)).
(16)
This equation must be solved numerically. If we use the same value of α for
1
all atoms, X will be the same for them. So R(α) ∝ Z − 3 since R(α) = bX(α)
1
1
and b = Z − 3 ×0.885a0 . For α = Z1 , we have R( Z1 ) = R̄ = Z − 3 ×0.885a0 X( Z1 )
is the radius of the sphere containing all but one of the electrons. The figure
below gives the Z dependence of R̄.
5
4×
3×
R̄(10−8 cm)
2×
1×
0
×
25
×
50
Z
×
75
×
100
Z dependence of R̄
It is seen that R̄ is practically independent of Z. R̄ ' (2 − 3) × 10−8 cm.
Charge Form Factor
In the Thomas-Fermi model, all atoms are identical in shape and the scale
1
1
of the length is proportional to Z − 3 since r = bx and b = 0.885a0 × Z − 3 .
Thus the charge form factor will be an universal function for all heavy atoms.
The electron density in Thomas-Fermi model can be written as
n(r) =
χ 23
Z2
.
4π(0.885a0 )3 x
6
(17)
χ is the universal THomas-Fermi function. The form factor is
F (q) =
Z
n(r)ei~q·~r d~r,
4π Z ∞
=
n(r)sin(qr)rdr,
q 0
4πb2 Z ∞
=
n(x)sin(qbx)xdx,
q 0
Z ∞
3
1
4πb2
Z2
=
(χ(x)) 2 sin(qbx)x− 2 dx,
3
q 4π(0.885a0 ) 0
3
=
Z ∞
3
1
Z2
(χ(y)) 2 sin((q 0.885a0 )y)y − 2 dy,
q(0.885a0 ) 0
1
3
1
(18)
where y = Z − 3 x. The quantity G(Q) = Q1 0∞ [χ(y)] 2 y − 2 sin(Qy)dy is
tabulated in the literature. Here Q = 0.885a0 q. From the table, the charge
form factor can be read.
R
7