Solid State Physics
1
Section 10-1
1
Topics
Structure of Solids
Free Electron Gas in Metals
Summary
2
Structure of Solids
Face-centered cubic (fcc)
Salts:
LiF, KF, KCl, KI, NaCl
Metals:
Ag, Al, Au, Ca, Cu, Ni, Pb
3
Structure of Solids
Na+
When two oppositely charged
ions are brought together they
experience an attractive
electrostatic force.
Cl-
But, when their wave functions begin
to overlap they experience the Coulomb
repulsion of their electrons as well as the
repulsion due to the exclusion principle
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Structure of Solids
Na+
The electric potential at a given
ion is the sum of the electric
potentials due to every other
ion in the solid
+
+
+
-
+
-
Cl-
N
V 
V
i
i 1
N

k
i 1
ei
ri
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Structure of Solids
Therefore, the net electrostatic
potential energy of an ion is
given by
U  eV   e  k
i
Na+
r
Cl-
ei
ri
For the 6 nearest neighbors of the chlorine ion
the electrostatic potential energy is
U  6
ke
r
2
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Structure of Solids
When the sum is extended over
the whole solid, the factor
of 6 is replaced by the factor a
called the Madelung constant
and the net electrostatic potential
energy is given by
2
U  a
Na+
r
Cl-
ke
r
For NaCl a = 1.7476 and the equilibrium
separation between the ions is r = 0.282 nm
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Structure of Solids
The potential energy associated
with the repulsive force can
be modeled using
U rep 
Na+
r
Cl-
A
r
n
So the total potential energy of an ion is given by
U  a
ke
r
2

A
r
n
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Structure of Solids
If r0 is the equilibrium spacing,
that is, the spacing for which
the net force between the ions
is zero, then we can write
Na+
r
Cl-
ke  r0 1  r0  
U  a
    
r  r
n  r  
2
n
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Structure of Solids
Example 10-1: Compute r0, given
that the density of NaCl is
r = 2.16 g/cm3
Na+
r
Cl-
Each ion is associated with a
cube of volume r03. The molar
mass of NaCl is m = 58.4 g, therefore,
r 
m
V

m
3
A 0
2N r
and
r0 
m
3
2N Ar
 0.282 nm
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Structure of Solids
The dissociation energy is the
energy required to break the
bonds that hold the solid
together. For NaCl it is
measured to be 770 kJ/mol.
Na+
r
Cl-
The dissociation energy per ion-pair in eV is
given by
[(7.7 x 105J/mol)/(1.602 x 10-19 J/eV)]/
6.022 x 1023 ion pairs/mol
= 7.98 eV per ion pair
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Structure of Solids
The cohesive energy is the
energy per atom (or atomic pair).
Na+
r
Cl-
For NaCl we need 7.98 eV to
remove a Na+ and a Cl- from
the solid. To go from Cl- to Cl
requires 3.62 eV and Na+ to Na
releases 5.14 eV. Therefore, the
cohesive energy is (7.98 + 3.62 – 5.14)/2 =
3.23 eV per atom.
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Structure of Solids
Simple cubic (sc)
Examples: CsCl, NH4Cl, Ba, Fe, K, Na
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Structure of Solids
Hexagonal close-packed (hcp)
Examples: Be, Cd, Mg, Zn
15
Free Electron Gas in Metals
Solid metals are bonded by the metallic bond
One or two of the valence electrons from
each atom are free to move throughout the
solid
All atoms share all the electrons. A metal is a
lattice of positive ions immersed in a gas of
electrons. The binding between the electrons
and the lattice is what holds the solid
together
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Free Electron Gas in Metals
We first consider the physics of the electron
gas. Since we are dealing with fermions, the
electron gas obeys Fermi-Dirac statistics.
Recall that the number of electrons with
energy in the range E to E+dE is given by
n ( E ) dE  f ( E ) g ( E ) dE
where, the density of states g(E) is given by
V 4 p dp
2
g ( E ) dE  W
h
3
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Free Electron Gas in Metals
The electron has two spin states, so W = 2.
If the electron’s speed << c, we can take its
2
energy to be
p
E 
2m
So the number of states in (E, E+dE) is given by
g ( E ) dE  V
  8m 
 2 
2 h 
3/2
E
1/ 2
dE
Extra Credit: Derive this expression
Due date: Friday, March 28
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Free Electron Gas in Metals
The total number of electrons N is given by
the integral
N V
  8m 
 2 
2 h 
3/2


0
E
e
1/ 2
dE
( E  E F ) / kT
1
At T=0, the Fermi-Dirac distribution is equal
to 1 if E < EF and 0 if E > EF.
At T=0, all energy levels are filled up to
the energy EF, called the Fermi energy.
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Summary
The ions in solids form regular lattices
A metal is a lattice of positive ions immersed
in a gas of electrons. All ions share all
electrons
The attraction between the electrons and the
lattice is called a metallic bond
At T = 0, all energy levels up to the Fermi
energy are filled
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