Free electron Fermi gas model: specific heat and Pauli paramagnetism Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton, (March 10, 2013)
Abstract As an example we consider a Na atom, which has an electron configuration of (1s)2(2s)2(2p)6(3s)1. The 3s electrons in the outermost shell becomes conduction electrons and moves freely through the whole system. The simplest model for the conduction electrons is a free electron Fermi gas model. In real metals, there are interactions between electrons. The motion of electrons is also influenced by a periodic potential caused by ions located on the lattice. Nevertheless, this model is appropriate for simple metals such as alkali metals and noble metals. When the Schrödinger equation is solved for one electron in a box, a set of energy levels are obtained which are quantized. When we have a large number of electrons, we fill in the energy levels starting at the bottom. Electrons are fermions, obeying the Fermi-Dirac statistics. So we have to take into account the Pauli’s exclusion principle. This law prohibits the occupation of the same state by more than two electrons. Sommerfeld’s involvement with the quantum electron theory of metals began in the spring of 1927. Pauli showed Sommerfeld the proofs of his paper on paramagnetism. Sommerfeld was very impressed by it. He realized that the specific heat dilemma of the Drude-Lorentz theory could be overcome by using the Fermi-Dirac statistics (Hoddeeson et al.).1 Here we discuss the specific heat and Pauli paramagnetism of free electron Fermi gas model. The Sommerfeld’s formula are derived using Mathematica. The temperature dependence of the chemical potential will be discussed for the 3D and 1D cases. We also show how to calculate numerically the physical quantities related to the specific heat and Pauli paramagnetism by using Mathematica, based on the physic constants given by NIST Web site (Planck’s constant ħ, Bohr magneton B, Boltzmann constant kB, and so on).2 This lecture note is based on many textbooks of the solid state physics including Refs. 3 – 10.
Content: 1. Schrödinger equation A. Energy level in 1D system B. Energy level in 3D system 2. Fermi-Dirac distribution function 3. Density of states A. 3D system B. 2D system C. 1D system 4. Sommerfeld’s formula 5. Temperature dependence of the chemical potential 6. Total energy and specific heat 7. Pauli paramagnetism 8. Physical quantities related to specific heat and Pauli paramagnetism
1 9. Conclusion
1. Schrödinger equation3-10 A. Energy level in 1D system We consider a free electron gas in 1D system. The Schrödinger equation is given by
p2 2 d 2 (x) H (x) (x) k (x) , (1) k 2m k 2m dx2 k k
where
d p , i dx
and k is the energy of the electron in the orbital. The orbital is defined as a solution of the wave equation for a system of only one electron:one-electron problem.
Using a periodic boundary condition: k (x L) k (x ) , we have
ikx k (x) ~ e , (2)
with
2 2 2 2 2 k k n , 2m 2m L
2 eikL 1 or k n , L where n = 0, ±1, ±2,…, and L is the size of the system.
B. Energy level in 3D system We consider the Schrödinger equation of an electron confined to a cube of edge L.
p2 2 H 2 . (3) k 2m k 2m k k k
It is convenient to introduce wavefunctions that satisfy periodic boundary conditions. Boundary condition (Born-von Karman boundary conditions).
k (x L, y, z) k (x, y, z) ,
k (x, y L, z) k (x, y, z) ,
2
k (x, y, z L) k (x, y, z) .
The wavefunctions are of the form of a traveling plane wave.
ikr k (r) e , (4) with
kx = (2/L) nx, (nx = 0, ±1, ±2, ±3,…..),
ky = (2/L) ny, (ny = 0, ±1, ±2, ±3,…..),
kz = (2/L) nz, (nz = 0, ±1, ±2, ±3,…..).
The components of the wavevector k are the quantum numbers, along with the quantum number ms of the spin direction. The energy eigenvalue is
2 2 (k) (k 2 k 2 k 2 ) k 2 . (5) 2m x y z 2m
Here
p (r) (r) k (r) . (6) k i k k k
So that the plane wave function k (r) is an eigenfunction of p with the eigenvalue k . The ground state of a system of N electrons, the occupied orbitals are represented as a point inside a sphere in k-space. Because we assume that the electrons are noninteracting, we can build up the N- electron ground state by placing electrons into the allowed one-electron levels we have just found. ((The Pauli’s exclusion principle)) The one-electron levels are specified by the wavevectors k and by the projection of the electron’s spin along an arbitrary axis, which can take either of the two values ±ħ/2. Therefore associated with each allowed wave vector k are two levels:
k, , k, .
In building up the N-electron ground state, we begin by placing two electrons in the one- electron level k = 0, which has the lowest possible one-electron energy = 0. We have
3 L3 4 V N 2 k 3 k 3 (2 )3 3 F 3 2 F where the sphere of radius kF containing the occupied one-electron levels is called the Fermi sphere, and the factor 2 is from spin degeneracy. The electron density n is defined by
N 1 n k 3 V 3 2 F
The Fermi wavenumber kF is given by
2 1/ 3 kF 3 n . (9)
The Fermi energy is given by
2 2 / 3 3 2n . (10) F 2m
The Fermi velocity is
k 1/ 3 v F 3 2n . (11) F m m
The Fermi temperature TF is defined by
F TF . kB
In this model, these quantities is dependent only on the electron density n.
We define the radius rs (the radius of a sphere that contains one electron) as
V 1 4 r 3 N n 3 s or
1/3 3 rs 4n
The dimensionless radius parameter is defined as
4 rs/a0 where a0 is the first Bohr radius, a0 = 0.52917721092 Å.
((Note)) The Fermi energy F can be estimated using the number of electrons per unit volume (the electron density) as
-15 2/3 2/3 F = 3.64645x10 n [eV] = 1.69253 n0 [eV],
-3 22 where n and n0 is in the units of (cm ) and n = n0×10 . The Fermi wave number kF is calculated as
7 1/3 -1 kF = 6.66511×10 n0 [cm ].
The Fermi velocity vF is calculated as
7 1/3 vF = 7.71603×10 n0 [cm/s].
((Example-1)) The Fermi energy of Au
Atomic molar mass M0 = 196.9666 g/mol Density = 19.30 g/cm3 There is one conduction electron per Au atom; p = 1.
N N M N N n p 0 p p A = 5.901 x 1022 /cm3. V V V M M 0 where N0 is the total number of atoms. Since n0 = 5.901, we have
2/3 F = 1.69253 n0 = 5.526 [eV].
7 1/3 8 -1 kF = 6.66511×10 n0 = 1.20442 x 10 [cm ].
7 1/3 8 vF = 7.71603×10 n0 = 1.39433 x 10 [cm/s].
4 TF = F/kB = 6.4136 x 10 K
((Example-2)) The Fermi energy of Cu
Cu has a fcc structure with the lattice constant a = 3.615 Å (conventional cell). There are 4 Cu atoms per unit volume for the conventional cell. Then the number electron density n is
5 4 n p = 8.50234 x 1022/cm3 a3 where p (= 1) is the number of conduction electron per Cu atom.
The Fermi energy is
2 2 / 3 3 2n = 1.12964 x 10-11 erg = 7.051 eV F 2m
The Fermi velocity is
k 1/ 3 v F 3 2n = 1.575 x 108 cm/s F m m
The Fermi wavenumber
2 1/ 3 8 kF 3 n =1.3604 x 10 /cm
The Fermi temperature is
F 4 TF = 8.1819 x 10 K. kB
The radius rs is
1/3 3 rs = 1.4108 Å. 4n or
r s = 2.66598. a0
((Table)) Ashcroft and Mermin
6
((Table)) Kittel
7
((Mathematica))
22 -3 Fermi energy F (eV) vs the number of electrons (n = n0×10 [cm ]).
EF eV 14 12
10
8
6
4
2
3 n0 1 cm 5 10 15 20 25
Fig.1 Fermi energy vs number density n (= n ×1022 [cm-3]). 0
2. Fermi-Dirac distribution function3-10
8 The Fermi-Dirac distribution gives the probability that a state at energy will be occupied in an ideal gas in thermal equilibrium
1 f ( ) , (12) e ( ) 1 where is the chemical potential and = 1/(kBT).
(i) lim F . T 0
(ii) f() = 1/2 at = .
( ) (iii) For - >>kBT, f() is approximated by f ( ) e . This limit is called the Boltzmann or Maxwell distribution.
(iv) For kBT<<F, the derivative -df()/d corresponds to a Dirac delta function having a sharp positive peak at = .
f ( ) e ( ) [e ( ) 1]2
f ( ) [ ]d [ f ( )] 1 0 0
F E 1.0
0.8
0.6
0.4
0.2
0.0 E 0.90 0.95 1.00 1.05 1.10
9 Fig.2 Fermi-Dirac distribution function f() at various T (= 0.002 – 0.02). kB = 1. (T = 0) = F = 1.
120
100
80
60
40
20
0 0.90 0.95 1.00 1.05 1.10
Fig.3 Derivative of Fermi-Dirac distribution function -df()/d at various T (= 0.002 – 0.02). kB = 1. (T = 0) = F = 1.
3. Density of states3-10
A. 3D system
10
Fig. Density of states in the 3D k-space. There is one state per (2/L)3.
There is one state per volume of k-space (2/L)3. We consider the number of one- electron levels in the energy range from to +d; D()d
L3 D( )d 2 4k 2dk , (13) 2 3
2 1/ 2 where D() is called a density of states. Since k (2m / ) , we have
2 1/ 2 dk (2m / ) d /(2 ) .
Then we get the density of states
3/ 2 V 2m D( ) 2 2 . (14) 2
((Definition))
11 So far we assume that there are N free electrons in the volume V. We consider the case when each atom in metals has nv conduction electrons. If there are N0 atoms in the volume V, the number of the total free electrons N is expressed by
N nv N0
Here we define DA ( F ) [1/(eV atom)] which is the density of states per unit energy per atom.
D( F ) D( F ) DA ( F ) nv , (15) N N0
where D( F ) / N corresponds to the density of states per unit energy per electron. Noting that
F 3 / 2 F 3 / 2 V 2m 2 V 2m 3 / 2 N D( )d d . (16) 2 2 2 2 F 0 2 0 3 2 we have
3 DA ( F ) nv . (17) 2 F
For Al, we have F = 11.6 eV and nv = 3. Then DA(F) = 0.39/(eV atom).
Here we make a plot of f()D () as a function of using Mathematica.
3 / 2 V 2m f ( )D( ) 2 2 ( ) a ( ) , (18) 2 e 1 e 1 where
3/ 2 V 2m a 2 2 2
12 D E f E
1.0
0.8
0.6
0.4
0.2
E 0.2 0.4 0.6 0.8 1.0 1.2 Fig.4 D()f() at various T (= 0.001 – 0.05). kB = 1. (T = 0) = F = 1. The constant a of D() (= a ) is assumed to be equal to 1. The dashed line denotes the curve of D().
B. 2D system
13 k y
dk
k
kx 2p L
2 2 Fig. Density of states for the 2D k-space. There is one state per area of the L reciprocal lattice plane.
For the 2D system, we have
L2 D( )d 2 2kdk . (19) 2 2
2 The factor 2 comes from the spin weight. Since d ( / 2m)2kdk , we have the density of states for the 2D system as
mL2 D( ) 2 , (20) which is independent of .
C. 1D system
14 Ek
EF
k O2p L Fig. Energy k vs k for the one dimensional case. The discrete states are described |k,> and |k,> with k = (2/L)n. (nx = 0, ±1, ±2,...). All the states below the Fermi energy F are occupied at T = 0 K.
For the 1D system we have
1/ 2 L 2L 2m 1 1/ 2 D( )d 2 2dk 2 d (21) 2 2
The factor 2 before dk arises from the two states of k and -k. Thus the density of states for the 1D system is
1/ 2 L 2m 1/ 2 D( ) 2 . (22) which decreases with increasing .
4. Sommerfeld’s formula
When we use a formula
L3 F(k) dkF(k) . (23) 3 k 2 the total particle number N and total energy E can be described by
2L3 N 2 f ( ) dkf ( ) dD( ) f ( ) , (24) k 3 k k 2
15 and
2L3 E 2 f ( ) dk f ( ) dD( )f ( ) . (25) k k 3 k k k 2
First we prove that
f ( ) g( )[ ]d g'( ) f ( )d
1 2 2 2 (2) 7 4 4 4 (4) g() kB T g () kB T g () 6 360 31 127 k 6T 6 6 g (6) () k 8T 8 8g (8) () 15120 B 604800 B 73 1414477 k 10T 10 10 g (10) () k 12T 12 12 g (12) () ... 3421440 B 653837184000 B
(26) using Mathematica.
((Mathematica))
Sommerfeld's formula
1 Clear "Global`" ;f x_ : ;h x D f x , x Simplify; x Exp 1 kB T g1 Series g x , x, ,14 Normal; G1 kB T g1 h x . x kB T y Expand; K n_ : G1 n y; G2 Sum K n , n,1,13
1 7 31 kB6 6 T6 g 6 g kB2 2 T2 g kB4 4 T4 g 4 6 360 15 120 8 8 8 8 10 10 10 10 12 12 12 12 127 kB T g 73 kB T g 1 414 477 kB T g 604 800 3421 440 653 837 184 000 So we get a final result
df 1 7 31 kB6 T6 6 g 6 g E E g kB2 T2 2 g kB4 T4 4 g 4 dE 6 360 15 120 8 8 8 8 10 10 10 10 12 12 12 12 127 kB T g 73 kB T g 1 414 477 kB T g 604800 3 421440 653 837 184 000 ______ Here we note that
16
f ( ) g( )[ ]d f ( )g( ) | g'( ) f ( )]d 0 0 0 . (27) g'( ) f ( )]d 0
We define
( ) g'( ) and g( = 0) =0, or
g( ) ( ')d ' . (28) 0
Then we have a final form (Sommerfeld’s formula).
1 7 f ( )( )d ( ')d ' k 2T 2 2'() k 4T 4 4 (3) () B B 0 0 6 360 31 127 k 6T 6 6 (5) () k 8T 8 8 (7) () 15120 B 604800 B 73 1414477 k 10T 10 10 (9) () k 12T 12 12 (11) () ... 3421440 B 653837184000 B
(29)
5. T dependence of the chemical potential We start with
N dD( ) f ( ) , where
3 / 2 V 2m D( ) 2 2 a , 2 and 3 / 2 V 2m a 2 2 2
17 1 2a 1 a N f ( )D( )d D( ')d ' k 2T 2 2D'() 3 / 2 k 2T 2 2 . B B 0 0 6 3 6 2
But we also have F (T 0) . Then we have
F 2a N D( )d 3 / 2 . F 0 3
Thus the chemical potential is given by
2a 2a 1 a 3 / 2 3 / 2 k 2T 2 2 , 3 F 3 6 B 2 or
3 / 2 2 1/ 2 2 kBT 1 . F 8 F F
2 kBT which is valid for the order of in the above expansion formula. The ContourPlot F of /F and kBT/F can be obtained by using the Mathematica .
18 2 Clear "Global`" ; eq1 y1 2 y2 x2 ; 8 f1 ContourPlot Evaluate eq1 , x,0,0.1 , y, 0.99, 1.00 , ContourStyle Red, Thick ; f2 Graphics Text Style "kBT EF", Black, 12 , 0.09, 0.99 , Text Style " EF", Black, 12 , 0.004, 0.999 , Text Style "3D system", Black, 15 , 0.07, 0.9995 ; Show f1, f2, PlotRange All 1.000 3D system m EF
0.998
0.996
0.994
0.992
0.990 kBT EF 0.00 0.02 0.04 0.06 0.08 0.10
Fig. ContourPlot of /F and kBT/F for small values of kBT/F
______The chemical potential is approximated by the forms,
2 kBT 2 F [1 ( ) ] (3D case). (30) 12 F
For the 1D case, similarly we have
2 kBT 2 F [1 ( ) ] (1D case). (31) 12 F
19 ______((Mathematica)) We now discuss the T dependence of by using the Mathematica. The higher order terms (proportional to T4, T6,..) are also taken into account. The results are as follows.
m 1.0
0.9
0.8
0.7
0.6
T 0.05 0.10 0.15 0.20
Fig.5 T dependence of chemical potential for the 3D system. kB = 1. F = (T = 0) = 1.
20 m
1.35
1.30
1.25
1.20
1.15
1.10
1.05
T 0.05 0.10 0.15 0.20 0.25 0.30
Fig.6 T dependence of chemical potential for the 1D system. kB = 1. F = (T = 0) = 1.
6. Total energy and specific heat Using the Sommerfeld’s formula, the total energy U of the electrons is approximated by
(T ) 1 U f ()D()d D()d 2 (k T)2{D[(T)] (T)D'[(T)]} B . 0 0 6
The total number of electrons is also approximated by
(T ) 1 N f ( )D( )d D( )d 2 (k T )2 D'[(T )]. B 0 0 6
Since N / T 0 (N is independent of T), we have
21 1 '(T)D[(T)] 2k 2TD'[(T )] 0 , 3 B or
1 D'[(T )] '(T ) 2k 2T . 3 B D[(T)]
The specific heat Cel is defined by
dU 1 1 C 2k 2TD[(T )] { 2k 2TD'[(T)]} '(T )D((T ))}(T ) . el dT 3 B 3 B
The second term is equal to zero. So we have the final form of the specific heat
1 C 2k 2TD[(T)] . el 3 B
When (T ) F ,
1 C 2k 2D( )T . (32) el 3 B F
7. The heat capacity of electrons per mol atom From the above discussion, we have
D( F ) a F ,
2a N 3 / 2 , 3 F and
D( ) 3 F N 2 F
The heat capacity of N electrons is given by
1 C 2k 2 D( )T el 3 B F
So the heat capacity per electron is given by
22 C 1 D( ) el 2k 2 F T . N 3 B N
Using the expression of D( F ) / N 3/(2 F ) , we have
2 2 Cel 1 2 2 3 kB kB T T N 3 2 F 2 F
Suppose that each atom has nv conduction electrons. The total number of electrons is N;
N nv N0 .
So each atom has the heat capacity as
Cel Cel nv 1 2 2 nv kB D( F )T N N 0 N 3
1 2 2 D( F ) kB T 3 N 0 1 2k 2 D ( )T 3 B A F where
D( F ) DA ( F ) N0 or
D( F ) 3nv DA ( F ) nv N 2 F
The heat capacity per mol atom is
1 C (M ) 2k 2D ( )N T el 3 B A F A where NA is the Avogadro number.
1 2 N k 2 =2.35715 mJ eV/K2. 3 A B
is related to DA ( F ) as
23
1 2 N k 2D ( ) , 3 A B A F or
2 (mJ/mol K ) = 2.35715 DA ( F ) . (33)
We now give the physical interpretation for Eq.(32). When we heat the system from 0 K, not every electron gains an energy kBT, but only those electrons in orbitals within a energy range kBT of the Fermi level are excited thermally. These electrons gain an energy of kBT. Only a fraction of the order of kBT D(F) can be excited thermally. The total 2 electronic thermal kinetic energy E is of the order of (kBT) D(F). The specific heat Cel is 2 on the order of kB TD(F).
((Note))
For Pb, = 2.98, DA ( F ) =1.26/(eV at)
For Al = 1.35, DA ( F ) =0.57/(eV at)
For Cu = 0.695, DA ( F ) =0.29/(eV at)
Table The value of for typical metals (H.P. Myers, Introductory Solid State Physics (Taylor & Francis, London, 1990).
______((Mathematica))
24 Heat capacity for the 3D case. We use the Sommerfeld's formula for the calculation of the total energy and the total number
Clear "Global`" ; T 1 7 U x x kB T 2 2 D T , T ,1 kB T 4 4 D T , T ,3 ; 0 6 360 replace De & ; U1 U . replace; Ce D U1,T Expand 1 1 kB2 2 TDe T kB22 T T De T De T T T 3 3 1 7 1 kB2 2 T2 De T T kB4 4 T3 De T kB2 2 T2 T T De T 3 30 6 7 7 7 kB4 4 T3 T De 3 T kB4 4 T4 T De 3 T kB4 4 T4 T T De 4 T 90 90 360 The chemical potential m[T] can be estimated from the expression of N N1 U . De & ; N2 D N1, T Expand
1 1 kB2 2 TDe T De T T kB2 2 T2 T De T 3 6 7 7 kB4 4 T3 De 3 T kB4 4 T4 T De 4 T 90 360 For simplicity we use the approximation to the order of 0 (T 2) for the total energy and number. Note that D[N1,T]=0 since N is independent of T.
eq1N2 1 N2 2 0;eq2Solve eq1, T
kB2 2 TDe T T 3De T
Ce1Ce 1 Ce 2 Ce 3 ;Ce2Ce1 .eq2 1 Simplify 1 kB2 2 TDe T 3
8. The relation between effective mass and the value of The value of is expressed by
1 2 2 1 2 2 3N 1 2 2 3N kB D( F ) kB kB 2 2 m 3 3 2 F 3 kF
So the value of g is proportional to the mass m. It is common practice to express the ratio of the observed to the free electron values of the electronic heat capacity as a ratio of a thermal effective mass to the electron mass m, is defined by the relation,
m (observed) th . m ( free)
9. Pauli paramagnetism The magnetic moment of spin is given by
25 2 Sˆ ˆ B z ˆ z Bz (quantum mechanical operator).
Then the spin Hamiltonian (Zeeman energy) is described by
2 Sˆ ˆ ˆ B z ˆ H z B ( )B Bz B, (34) in the presence of a magnetic field, where the Bohr magneton µB is given by
e B . (e>0) 2mc with
-24 B 9.27400915(23) x 10 J/T (S.I. unit)
-21 B 9.27400915(23) x 10 erg/Oe (cgs unit)
erg/Oe = emu
Fig. The magnetic field is applied along the z axis. (a) Spin–up state z . The spin magnetic moment is antiparallel to the magnetic field. The Zeeman energy is +BH.
26 (b) Spin-down state z . The spin magnetic moment is parallel to the magnetic field. The Zeeman energy is -BH.
(i) The magnetic moment antiparallel to H. Note that the spin state is given by a up-state,
z .
The energy of electron is given by
k B H ,
2 2 with k ( / 2m)k . The density of state for the spin-up state (the down-state of the magnetic moment) is
3 L 2 V 2m 3 / 2 D ( )d 3 4k dk 2 ( 2 ) B H d , (2 ) 4 or
1 D ( ) D( H ) . (35) 2 B
The factor ½ comes from the fact that D ( ) is the density of states per spin. Then we have
1 N D( H ) f ( )d . (36) 2 B B H
(ii) The magnetic moment parallel to H. Note that the spin state is given by
z .
The energy of electron is given by
k B H ,
The density of state for the spin down-state (the up-sate of the magnetic moment) is
3 L 2 V 2m 3 / 2 D ( )d 3 4k dk 2 ( 2 ) B H d , (2 ) 4
27 or
1 D ( ) D( H ) . (37) 2 B
Then we have
1 N D( H ) f ( )d . (38) 2 B B H
Fig. Density of states for the Pauli paramagnetism of free electron. Left: (D+()
for the z , the direction of the spin magnetic moment is parallel to
28 that of magnetic field). Right: (D-() for z ; the direction of the spin magnetic moment is antiparallel to that of magnetic field).
The magnetic moment M is expressed by
M (N N ) B [ D( H ) f ( )d D( H ) f ( )d ], (39) B 2 B B BH BH or
M B D( )[ f ( H ) f ( H )]d 2 B B 0 (40) f ( ) 2H D( )( )d 2HD( ) B B F 0
Here we use the relation;
f ( ) ( ) ( ) F
(see Fig.3).
The susceptibility (M/H) thus obtained is called the Pauli paramagnetism.
2 p B D( F ) . (41)
10. Pauli susceptibility per atom mol Each atom has the Pauli susceptibility as
n n Pauli v v N N Pauli
Pauli 2 D( F ) 2 B B DA ( F ) N0 N0 where each atom has nv conduction electrons;
N nv N0 and
29 D( F ) DA ( F ) N0
Then the Pauli susceptibility per atom mol is given by
M 2 P N AB DA ( F ) where NA is the Avogadro number.
11. Experimental values of Pauli susceptibility Experimentally we measure the susceptibility per mol, p (emu/mol)
D( ) M 2 F N 2 N D ( ) , (42) P B N A B A A F
2 -5 whereB NA = 3.23278×10 (emu eV/mol) and DA(F) [1/(eV atom)] is the density of states per unit energy per atom. Since
1 2 N k 2D ( ) , (43) 3 A B A F
2 we have the following relation between P (emu/mol) and (mJ/mol K ),
M 5 P 1.3714810 . (44) where
1 2 N k 2D ( ) 3 A B A F
((Exampl-1)) Rb atom has one conduction electron.
2 -5 = 2.41 mJ/mol K , P = (1.37x10 )×2.41 (emu/mol) 1 mol = 85.468 g -6 P =0.386×10 emu/g (calculation)
((Exampl-2)) K atom has one conduction electron.
2 -5 = 2.08 mJ/mol K , P = (1.37x10 )×2.08 (emu/mol) 1 mol = 39.098 g -6 P =0.72x10 emu/g (calculation)
((Exampl-3)) Na atom has one conduction electron.
30 2 -5 = 1.38 mJ/mol K , P = (1.37x10 )×1.38 (emu/mol) 1 mol = 29.98977 g -6 P =0.8224x10 emu/g (calculation)
The susceptibility of the conduction electron is given by
P L P P /3 2 P /3 , (45) where L is the Landau diamagnetic susceptibility due to the orbital motion of conduction electrons. Using the calculated Pauli susceptibility we can calculate the total susceptibility:
Rb: = 0.386×(2/3)×10-6 = 0.26×10-6 emu/g K: = 0.72×(2/3)x10-6 = 0.48×10-6 emu/g Na: = 0.822×(2/3)×10-6 = 0.55×10-6 emu/g
These values of are in good agreement with the experimental results.6
12. Physical quantities related to specific heat and Pauli paramagnetism Here we show how to evaluate the numerical calculations by using Mathematica. To this end, we need reliable physics constant. These constants are obtained from the NIST Web site: http://physics.nist.gov/cuu/Constants/index.html
-27 Planck’s constant, =1.05457168×10 erg s -16 Boltzmann constant kB = 1.3806505×10 erg/K -21 Bohr magneton B = 9.27400949×10 emu 23 Avogadro’s number NA = 6.0221415×10 (1/mol) Velocity of light c = 2.99792458×1010 cm/s electron mass m = 9.1093826×10-28 g electron charge e = 1.60217653×10-19 C e = 4.803242×10-10 esu (this is from the other source) 1 eV = 1.60217653×10-12 erg 1 emu = erg/Gauss 1mJ = 104 erg
Using the following program, one can easily calculate many kinds of physical quantities. Here we show only physical quantities which appears in the previous sections.
((Mathematica)) Physics constants
31 Use the physical constants to calculate the physical quantities (in the units of cgs)
Clear "Global`" ; rule1 B 9.27400949 1021 , kB 1.3806505 1016 , NA 6.0221415 1023 , c 2.99792458 1010 , — 1.05457168 1027 , m 9.1093826 1028 , e 4.803242 1010 , eV 1.60217653 1012 , mJ 104 ; Fermi energy —2 3 2 1022 n0 2 3 1 eV . rule1 2 m
1.69253 n02 3 Fermi wavenumber
3 2 1022 n0 1 3 . rule1 N
7 1 3 6.66511 10 n0 Fermi velocity
— 3 2 1022 n0 1 3 . rule1 m
7.71603 107 n01 3 heat capacity
1 2 NA kB2 eV mJ . rule1 3 2.35715 Pauli paramagnetism
B2 NA eV . rule1 ScientificForm
3.23278 105 Relation between Pauli paramagnetrism and heat capacity
3 B2 mJ . rule1 ScientificForm 2 kB2 1.37148 105 13. Conclusion The temperature dependence of the specific heat is discussed in terms of the free electron Fermi gas model. The specific heat of electrons is proportional to T. The Sommerfeld’s constant for Na is 1.38 mJ/(mol K2) and is close to the value [1.094 mJ/(mol K2)] predicted from the free electron Fermi gas model. The linearly T dependence of the electronic specific heat and the Pauli paramagnetism give a direct evidence that the conduction electrons form a free electron Fermi gas obeying the Fermi- Dirac statistics.
32 It is known that the heavy fermion compounds have enormous values, two or three orders of magnitude higher than usual, of the electronic specific heat. Since is proportional to the mass, heavy electrons with the mass of 1000 m (m is the mass of free electron) move over the system. This is due to the interaction between electrons. A moving electron causes an inertial reaction in the surrounding electron gas, thereby increasing the effective mass of the electron.
REFERENCES 1. L. Hoddeson, E. Braun, J. Teichmann, and S. Weart, Out of the Crystal Maze (Oxford University Press, New York, 1992). 2. NIST Web site: http://physics.nist.gov/cuu/Constants/index.html 3. A.H. Wilson, The Theory of Metals (Cambridge University Press, Cambridge, 1954). 4. A.A. Abrikosov, Introduction to the Theory of Normal Metals (Academic Press, New York, 1972). 5. N.W. Ashcroft and N.D. Mermin, Solid State Physics (Holt, Rinehart, and Wilson, New York, 1976). 6. C. Kittel, Introduction to Solid State Physics, seventh edition (John Wiley and Sons, New York, 1996). 7. C. Kittel and H. Kroemer, Thermal Physics, second edition (W.H. Freeman and Company, New York, 1980). 8 S.L. Altmann, Band Theory of Metals (Pergamon Press, Oxford, 1970). 9. H.P. Myers, Introductory Solid State Physics (Taylor & Francis, London, 1990). 10. H. Ibach and H. Lüth, Solid-State Physics An Introduction to Principles of Materials Science (Springer Verlag, Berlin, 2003). ______APPENDIX Analysis of the specific heat data of typical metals at low temperatures
H.P. Myer , Introductory Solid State Physics (Taylor & Francis, London, 1990). Problem 6-8
33 ((Solution))
The lattice contribution to the heat capacity per mol atom is
3 3 M T T CL 233.782N AkB 233.782R mJ/(mol K) where NA is the Avogadro number and R is the gas constant
R = 8,314.4621 mJ/(mol K).
The electron contribution to the heat capacity per mol atom is
1 C (M ) 2 N k 2D ( )T T mJ/(mol K) el 3 A B A F where
2 = 2.35715 DA ( F ) , mJ/(mol.K )
where the unit of DA ( F ) is
DA ( F ) [1/eV.at].
Experimentally the specific heat at low temperatures can be expressed by
C(mJ / mol.K) T AT 3 . (1)
In order to determine the values of and A, we make a plot of the specific heat data: C/T vs T2.
C(mJ / mol.K) AT 2 (1) T
34 C/T
Data 1 40
35
30
25
20 y = m1 + m2 * M0 Value Error 15 m1 1.9858 0.01913 m2 3.0071 0.0029607 10 Chisq 0.0051388 NA R 1 NA
5
0 02468101214
T^2 Fig. Plot of C/T vs T2 for the substance under study.
It is found that C/T is linearly proportional to T2. Using the least-squares fit of the data to Eq.(1), we obtain
2 1.986 ± 0.019 = 2.35715 DA ( F ) [mJ/(mol K )]
R A 3.007 ± 0.003 =233.782 3
The values of the Debye temperature is obtained as
86.5K .
These values of and indicate that the substance under study is K (potassium)
35