44 phys463 - university of michigan · 2015. 2. 24. · n = 2·Æ Æ à (6.95) k fi £kf âk fi...
TRANSCRIPT
-
(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