(6.95)N = 2 ´ á á àk

®

£kF

â k®

H2 ΠL3 V = 2 V à0kF 4 Π k

2 â k

8 Π3=

V

Π2à

0

kF

k2

â k =V

Π2à

0

ΕF

k2

â k

â Εâ Ε

We know that

(6.96)

â Ε

â k=

â

â k

Ñ2

k2

2 m

=Ñ

2

m

k

So

(6.97)N =V

Π2à

0

ΕF

k2

â k

â Εâ Ε =

V

Π2à

0

ΕF k2 â Ε

Ñ2

mk

=m V

Π3 Ñ2à

0

ΕF

k â Ε

Because Ε = Ñ2 k2 2 m, k = 2 m Ε Ñ

(6.98)N =m V

Π2 Ñ2à

0

ΕF

k â Ε =m V

Π2 Ñ2à

0

ΕF

2 m Ε Ñ â Ε = à0

ΕF H2 mL32 V2 Π2 Ñ3

Ε â Ε = à0

ΕF V

2 Π2

2 m

Ñ2

32Ε â Ε = à

0

ΕF

DHΕL â ΕSo

(6.99)DHΕL = V2 Π2

2 m

Ñ2

32Ε

Another way to get DHΕL is(6.100)DHΕL = â N

â Ε

Because

(6.101)N =V k

3

3 Π2=

V H2 m ΕL323 Π2 Ñ3

(6.102)DHΕL = â Nâ Ε

=3

2

V H2 mL 323 Π2 Ñ3

Ε =V

2 Π2

2 m

Ñ2

32Ε

Here, we can also prove that DHΕL = 3 N 2 Ε,(6.103)DHΕL = â N

â Ε=

3

2

V H2 mL323 Π2 Ñ3

Ε =3

2 Ε

V H2 m ΕL323 Π2 Ñ3

=3 N

2 Ε

Total energy

(6.104)UHTL = à0

¥

Ε f HΕL DHΕL â ΕHeat capacity

(6.105)CV =¶U

¶TV = à

0

¥

Εâ f HΕL

â TDHΕL â Ε

First we prove that CV = Ù0

¥HΕ - ΜL â f HΕLâT

DHΕL â ΕIt is easy to prove that

(6.106)à0

¥ â f HΕLâ T

DHΕL â Ε = ââ T

à0

¥

f HΕL DHΕL â Ε = â Nâ T

= 0

(because the number of electrons are fixed, so â N

âT= 0)

Therefore,

44 Phys463.nb

(6.107)

CV =¶U

¶Tv = à

0

¥

Εâ f HΕL

â TDHΕL â Ε =

à0

¥

Εâ f HΕL

â TDHΕL â Ε - Μ ´ H0L = à

0

¥

Εâ f HΕL

â TDHΕL â Ε - Μ ´ à

0

¥ â f HΕLâ T

DHΕL â Ε = à0

¥HΕ - ΜL â f HΕLâ T

DHΕL â ΕWe know that

(6.108)

â f HΕLâ T

=â

â T

1

expJ Ε- ΜkB T

N + 1= B 1

expJ Ε- ΜkB T

N + 1F2 exp Ε - Μ

kB T

Ε - Μ

kB T2

(6.109)

â f HΕLâ Ε

=â

â Ε

1

expJ Ε- ΜkB T

N + 1= -B 1

expJ Ε- ΜkB T

N + 1F2 exp Ε - Μ

kB T

1

kB T

So,

(6.110)

â f HΕLâ T

= -â f HΕL

â Ε

Ε - Μ

T

As a result,

(6.111)CV = -à0

¥HΕ - ΜL â f HΕLâ Ε

Ε - Μ

T

DHΕL â Ε = -à0

¥ HΕ - ΜL 2T

DHΕL â f HΕLâ Ε

â Ε

At T

PlotB:SechB -1+Ε

2 tF

2

4 t

. t ® .005,SechB -1+Ε

2 tF

2

4 t

. t ® .01,

SechB -1+Ε2 t

F2

4 t

. t ® .1,SechB -1+Ε

2 tF

2

4 t

. t ® 1>, 8Ε, 0, 2, LabelStyle ® Medium,

PlotLegends ® :"T=1

100

Μ

kB

", "T=1

10

Μ

kB

", "T=Μ

kB

">, PlotPoints ® 1000F

0.0 0.5 1.0 1.5 2.0ΕΜ

10

20

30

40

50

60

-Μâ f HΕL

âΕ

T=1

100

Μ

kB

T=1

10

Μ

kB

T=Μ

kB

Notice that

(6.113)kB T lnI f -1 - 1M = Ε - Μ

(6.114)

CV =DHΕF L

Tà

0

1HΕ - ΜL 2 â f =DHΕF L

Tà

0

1AkB T lnI f -1 - 1ME2 â f = DHΕF L kB2 T à0

1AlnI f -1 - 1ME2 â f = DHΕF L kB2 T Π2

3

=Π2

3

DHΕF L kB2 TSo CV is a linear function of T at T > A T3, so the electron

contribution dominates the heat capacity.

46 Phys463.nb