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7/28/2019 5. Free e Fermi Gas
http://slidepdf.com/reader/full/5-free-e-fermi-gas 1/22
Metallic structure
Crystals can be classified by binding:
ionic, covalent, Van der Waals, metallic
In case of metallic bond,the mean free path ~ 108 inter-atomic distance
Alkali metals: Li, Na, K, Cs, Rb have bcc structure
Noble metals: Lu, Ag, Au have fcc structure
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Free e Fermi gas
Why condensed matter so transparent to conduction es?
(1) es are not scattered by cores arranged on periodiclattice because matter waves travel freely in a periodic lattice
(2) conduction e
s are scattered only infrequently by other conduction es Pauli exclusion principle
Free e Fermi gas = gas of free es subject to Pauli principle
This theory preceded quantum mechanics
+ + +
+ + +
Eg. Na metal
• Conduction es are 3s valence es
• Atomic core contains 10 es
• Cores occupy only 15% volume!
• Free es move throughout the bulk
volume
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•In an alkali metal the atomic cores occupy a relatively small
part (-15 percent) of the total volume of the crystal, hut in a
noble metal (Cu, Ag, Au) the atomic cores are relatively larger
and may be in contact with each other.
•The common crystal structure at room temperature is bcc for
the alkali metals and fcc for the noble metals. •A monovalent crystal which contains N atoms will have N
conduction electrons and N positive ion cores.eg Na+ •The Fermi energy eF is defined as the energy of the topmost
filled level in the ground state of the N electron system
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Energy levels in one dimension
L
e
Consider an e of mass m confined to L by
an infinite barrier
( ) of e is solution of the Schordinger eqnn x
H 0V 2, / 2 , / H p m p i d dx
22
2
2
nn n n
d H
m dx
Boundary conditions: (0) 0, ( ) 0n n L
2 1sin ;
2n n
n
x n L
cos ;nd n n A x
dx L L
sin ;n
n A x
L
22
2 sin
nd n n
A xdx L L
22
2n
n
m L
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22
2n
n
m L
n m s
e occupancy
1 1
1
2 1
13 1
1
4 0
0
If there are 6 e s,
If = top most filled level F n
If = total number of e s N
2 ; F n N
22
2
F F
n
m L
22
2 2 F
N
m L
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Free e gas in 3DFree-particle Schrodinger equation in 3D:
2 2 2 2
2 2 2 ( ) ( )2k k k r r m x y z
If e s are confined to cube of edge , the w.f. is a standing wave L
( ) sin( / )sin( / )sin( / )n x y z r A n x L n y L n z L
Periodic boundary conditions: ( , , ) ( , , )k k x L y z x y z
Wave fns. satisfying above conditions are that of travelling plane wave
( ) exp( )k r ik r 2 4
with 0, , ,... xk L L
& similarly for & y z k k
exp[ ( )] exp[ 2 ( ) / ] xik x L i n x L L exp( 2 / ) exp( 2 )i n x L i n
exp( ) xik x2 2
2 2
( ) ik r
k
r e x x
2
( )
ik r
xik e
2
( ) x k k r
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2
2 2 2( ) ( )
2x y z k k k k k k r r
m
2 2 2 2
2 2 2( ) ( )
2k k k r r
m x y z
2
2
2 ( ) ( )k x k r k r x
22
2k k
m 2& k
( ) ( )k k p r i r ( )k k r
p mv k k
vm
22
2 F F
k m
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3There exists one distinct triplet , , for (2 /L) of -space x y z k k k k
3In sphere of volume 4 / 3, total number of orbitals F k
3
3
4 / 32
(2 / )
F k N
L
3 3
23
F L k
3
23
F Vk
1/32
3 F N k V
2/32 2
3
2 F
N
m V
F
F
k v
m
1/32
3 N
m V
6 1~10 m
~1 to 15 eV
6~10 m/s
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Density of states
( )dN
d
D
2/32 2
3
2
F
N
m V
3/2
2 2
2( ) 2
V mD
3/2
2 2
2{No of orbitals of energy
3
V m N
3The expression for can be written as ln ln constant
2 N N
3
2
dN d
N
3( )
2
dN N
d D
3( )
2
F
F
N D
3
2
B F
N
k T
ff f
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Effect of temperature
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H t it f
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Heat capacity of e gas
Total electronic thermal energy B
F
NT E k T
T
= 2 el B
F
E T C Nk T T
At low ( ) we can use ( ) to get B F el T k T C D
0 0( ) ( ) ( )
F
E d f d D D
3We know that ( )
2 F
B F
N
k T D
2 21( )
3 el F BC k T D
21
2
el B
F
T C Nk
T
E i t l H t C it
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C = gT + AT 3
At temperatures very much lower the T F , the heat
capacity may be written as the sum of electronic and
lattice contributions
where g and A are constants characteristic of the material
Experimental Heat Capacity
El t i l d ti it
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Electrical conductivity
1
F e E v B
c{ 0 eE B
dv
F mdt
dk
dt
dk eE
dt / dk e Edt
( ) (0) / k t k eEt / k eE
Wh t ff t th ti f ?
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What affects the motion of es? Phonons
◦ Quantization of lattice vibrations
◦ Analogous to photon with wave-particle duality
◦ Sound waves in crystals propagate through phonons
◦ Thermal vibrations thermally excited phonons
◦
A phonon of wave vector k interacts with other particles if it had a momentum k
Impurities
Lattice imperfections
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e s collide with impurities, lattice imperfections and phonons
steady state in electric field can be achieved
e s drift due to application of electric field
incremental velocity
eE
vm
/ k eE
In constant , with e s of charge , per unit volume, E n q e
Electric current density j nq v2
Ohm's law
ne E
jm
Conductivity / j E
2
ne
m
2Electrical resistivity
m
ne
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Mean Free Path
How do we determine the collision time?
Electrons have a mean free path that they travel before a
collision occurs = v F
At low temperatures, most of the mobile electrons are right at the
Fermi surface, so v = v F (Fermi velocity) At these temperature, one can have mean free paths of the
order of ~ 1 cm for very pure crystals (even upto 10 cm for some
extremely pure metals!)
Eg. Cu v F = 1.57×108cm/s
(300 K) 3 × 10-6 cm
(4 K) 0.3 cm
More collisions at high temperatures lead to shorter collision
times and hence tends to increase with temperature.
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Experimental Resistivities
At room temperature (300 K), the electrical resistivity is
dominated by electron collisions with phonons
At low temperatures (~ 4 K), it is determined by collisions
with impurities
The rates of these collisions are dependent on one another
Matthiessen’s rule: = L + i
Phonon resistivity
(Related to
concentration of
phonons & T dependent)
Imperfection
resistivity (T
independent)
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References
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References
A. Beiser – “Concepts of Modern Physics”, 6 Ed., Tata
McGraw-Hill (New Delhi, 2003)
Charles Kittel – “Introduction to Solid State Physics”, 7
Ed., John Wiley and Sons (New York, 1996)
www.wikipedia.org