18.38

19.2 The calculation of μ(T)

• To obtain these curves, we must determine μ(T) . The calculation is considerably more complicated than it was for T = 0. We have

• the number of particles at T = 0 is the same as at T ≠ 0,

• At high temperatures the fermion gas approximates the classical ideal gas. In the classical limit,

• The spin degeneracy factor is 2 for fermions. • For T >> TF ,μ/kT takes on a large negative value and exp(-μ/kT) >>1. • As an example, consider a kilomole of 3He gas atoms (which are fermions)

at STP. The Fermi temperature is 0.069 K, so that T /TF = 3900. Using eqns (9.16) and (9.17), we find thatμ/kT = -12.7 and exp(-μ/kT) = 3.3x105. The average occupancy of single particle states is indeed very small, as in the case of an ideal dilute gas obeying the M-B distribution.

19.2) consider a kilomole of 3He gas atoms under STP conditions

(a) What is the Fermi temperature of the gas?

KTeTemperaturFermi

KK

v

N

mk

hT

F

F

07.0

07.0069.0

4.22504.1

10023.6

1038.11098.42

10626.6

504.12

32

26

2327

234

322

(b) Calculate

kTandkT exp

5

26

23

234

2327

23

2

1028.3

7.12

?exp

7.12

10023.6

4.22

10626.6

2731038.11098.422ln

22ln

?

kTe

kT

kT

kT

N

V

h

mkT

kT

thatknowwekT

(c) Find the average occupancy of a single particle state that has energy of

f

kT23

7

7

6

7.1223

108.6

108.6

10469.1

11

11

1

?:

7.12:23:

fOccupancyAverage

f

e

ef

thatknowWe

fOccupancyAverage

kTPotentialChemical

kTEnergy

kTkT

KT

19.3) For a system of noninteracting electrons, show that the probability of finding an electron in a state with energy ∆ above the chemical potential μ is the same as the probability of finding an electron absent from the with energy ∆ below μ at any given temperature T

19.4 Properties of a Fermion gas

The internal energy of a gas of N fermions

Integration by parts (I)

• In calculus, integration by parts is a rule that transforms the integral of products of functions into other, hopefully simpler, integrals. The rule arises from the product rule of differentiation.

• If u = f(x), v = g(x), and the differentials du = f '(x) dx and dv = g'(x) dx; then in its simplest form the product rule is

duvuvdvu

)(')()()('))'()(( xgxfxgxfxgxf

Integration by parts (II)• In the traditional calculus curriculum, this rule is often stated

using indefinite integrals in the form

• As a simple example, consider

Since ln x simplifies to 1/x when differentiated, we make this part of ƒ; since 1/x2 simplifies to −1/x when integrated, we make this part of g. The formula now yields

dxxgxfxgxfdxxgxf )()(')()()(')(

dxx

x 2

ln

dxxxx

xdx

x

x)/1)(/1(

lnln2

• At T = 0, U = (3/5)NεF , this energy is large because all the electrons must occupy the lowest energy states up to the Fermi level.

• The average energy of a free electron in silver at T = 0 is

• The mean kinetic energy of an electron, even at absolute zero, is two orders of magnitude greater than the mean kinetic energy of an ordinary gas molecule at room temperature.

Heat capacity

• The electronic heat capacity Ce can be found by taking the derivative of Equation (19.18):

• For temperatures that are small compared with the Fermi temperature, we can neglect the second term in the expansion compared with the first and obtain

• Thus the electronic specific heat capacity is 2.2 x 10-2 R. This small value explains why metals have a specific heat capacity of about 3R, the same as for other solids.

• It was originally believed that their free electrons should contribute an additional (3/2) R associated with their three translational degrees of freedom. Our last calculation shows that the contribution is negligible.

•The energy of the electrons changes only slightly with temperature (dU/dT is small) because only those electrons near the Fermi level can increase their energies as the temperature is raised, and there are precious few of them.

• At very low temperatures the picture is different. From the Debye theory, Cv is proportional to T3 and so the heat capacity of a metal takes the form Cv = AT + BT3,

where the first term is the electronic contribution and the second is associated with the crystal lattice.

• At sufficiently low temperatures, the AT term can dominate,

as the sketch of Figure 19.9 indicates.

Figure 19.9 Sketch of the heat capacity of a metal as a function of temperature showing the electronic and lattice contributions.

S = 0 at T = 0, as it must be. The Helmholtz function F = U -TS is

The fermion gas pressure is found from

• For silver we found that N/V = 5.9 x1028 m-3 and TF = 65,000K .

Thus P = 2/5 *5.9*1028 *(1.38*10-23) (6.5*104) = 2.1*1010 Pa = 2.1*105 atm.

• Given this tremendous pressure, we can appreciate the role of the surface potential barrier in keeping the electrons from evaporating from the metal.

19.5 Applications to White Dwarf Stars

• The temperature inside the core of a typical star is at the order of 107

K.

• The atoms are completely ionized at such a high T, which creates a hugh electron gas

• The loss of gravitational energy balances with an increase in the kinetic energy of the electrons and ions, which prevent the collapse of star!

Example: The pressure of the electron gas in Sirius B can be calculated with the formula

Using the following numbers Mass M = 2.09 × 1030

kg

Radius R= 5.57 × 106 m

Volume V= 7.23 × 1020 m3

• Assuming that nuclear fusion has ceased after all the core hydrogen has been converted to helium!

• The number nucleons =

• Since the ratio of nucleons and electrons is 2:1

there are electrons

Therefore, T(=107 K) is much smaller then TF .

i.e. is a valid assumption !

Thus: P can be calculated as

A white dwarf is stable when its total energy is minimum

For

Since can be expressed as

Where

For gravitational energy of a solid

With In summary

To find the minimum U with respect to R

19.7 a) Calculate Fermi energy for Aluminum assuming three electrons per Aluminum atom.

eV

V

NelectronsforDensity

molekilo

atoms

kilomoleKg

mKg

V

N

F 8.111.26.58

108.13

1011.92

1063.6

108.1

3#

1099.51002.627

1069.2

32

29

31

234

29

28263

3

19.7b) Show that the aluminum at T=100 K, μ differs from εF by less than 0.01%. (The density of aluminum is 2.69 x 103 kg m-3 and its atomic weight is 27.)

%01.0

1038.41

1062.88.11

1000

121

121

50

2

15

2

0

22

0

thanless

eVKeVK

T

T

F

19.7c) Calculate the electronic contribution to the specific heat capacity of aluminum at room temperature and compare it to 3R.

Using the following equation

FFe

kTNk

T

TNkC

22

2

19.13. Consider the collapse of the sun into a white dwarf. For the sun, M= 2 x 1030 kg, R = 7 x 108 m, V= 1.4 x 1027 m3.

(a) Calculate the Fermi energy of the Sun’s electrons.

eV

vm

vm

V

N

m

h

V

N

nucleousofelectronsof

electronsofNo

F

eF

eF

eF

6.20

10248.633.0

104.1

1023.7

26.1

205.133.0

8

3

2

104.1

102205.1

2

1#

10205.11066.1

102.

5

32

27

27

322

27

57

5727

30

(b) What is the Fermi temperature?

(c) What is the average speed of the electrons in the fermion gas (see problem 19-4). Compare your answer with the speed of light.

keVk

eV

kT F

F

5

151039.2

1062.8

6.20