固体物理学课件：热熔部分
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ωm
)3
∫ωm
0
kBT (eℏω / kBT
− 1) 2
ω 2dω
∫ CV
=
9
Nk
B
1 (
ωm
)3
ℏω m
kBT 0
(ξ )4 eξ
(eξ −1)2
dξ
ξ = ℏω / kBT θ D = ℏωm / kB
CV
=
9
NkB
(T
θD
)3
∫θ D T 0
ξ
(eξ
4eξ − 1) 2
dξ
=
3NkB
f
D
(θ D
−1
ρ
(ω
)dω
CV
=
(∂E ∂T
)V
=
∫ωm
0
k
B
(
ℏω kBT
)2
eℏω/ kBT (eℏω/ kBT −1)2
ρ(ω)dω
kBT >> ℏωi
∂Ei ∂T
=
kB
(
ℏωi
kBT
)
2
[
[1+ ℏω + ⋯⋯]
kBT
ℏω + 1 ( ℏω )2 + ⋯⋯]2
kBT 2 kBT
∂Ei ∂T
≈
kB
CV = 3∑N kB = 3NkB i=1
kBT << ℏωi
ω (ℏ ) e i 2 ℏωi / kBT
∂Ei ∂T
= kB
kBT (eℏωi / kBT
− 1) 2
∂Ei ∂T
≈
k
B
(
ℏωi
kBT
)
2
e
−ℏωi
/
kBT
2. Einstein model
( ℏω )2 eℏω / kBT
CV
= ∂E ∂T
= 3NkB
kBT (eℏω / kBT
Thermal conductivity at 300K, in W/cmK
B
C
N
O
F
Ne
... 2230 ...
...
...
75
0.27 1.29 ...
...
...
...
Na Mg
Al
Si
P
S
Cl Ar
158 400
...
428 645 ...
...
...
92
1.41 1.56
2.37 1.48 ...
q
dq = dqxdqydqz q
V
(2π
)3
dq
q
ω—ω+∆ω
∆n = ρ(ω)∆ω
A q= ω
CL
q—q+dq
q + dq = ω + dω
CL
V
(2π
)3
4πq2dq
=
V
2π
C2 3 L
ω
2dω
q—q+dq
2
×
V
2π
C2 3 T
ω
2
dω
ρ (ω )
=
3V
2π 2C
3
ω
2
1 C3
=
11
3
(
C
c in cal/gm K or Btu/lb F
0.215 0.0294 0.0923 0.092 0.0301 0.0305 0.0558 0.0321 0.0925 0.033 0.58 1.00 0.49 0.19 0.20
Molar C J/mol K
24.3 25.7 24.5 ... 25.6 26.4 24.9 24.8 25.2 28.3 111 75.2 36.9 ... ...
− 1) 2
CV
= 3NkB
f
E
(
ℏω
kBT
)
kBθE = ℏω
CV
=
3NkB
(θ E
T
)2
eθ E (eθE /T
/T
−
1)2
Diamond
Cp J/mol K 01 2 3 4 5 6
0 0.2 0.4 0.6 0.8 1.0 T/θE
3. Debye model
q 1 、2
ω = CLq
ω = CT q
...
Rb Sr
Y
Zr Nb Mo Tc
Ru
Rh
Pd
Ag Cd
In
Sn
Sb
Te
I
Xe
56 147 280 291 275 450
...
600 480 274 225 209 108 200 211 153 ...
64
0.58 ... 0.17 0.23 0.54 1.38 0.51 1.17 1.50 0.72 4.29 0.97 0.82 0.67 0.24 0.02 ...
3 L
+
2 CT3
)
ω →0~∞
∫0∞ ρ(ω)dω
∫ωm
0
ρ (ω )dω
=
3N
。 ωm ωm
ωm
ωm
=
C[6π2
(
N
1
)]3
V
( ℏω ) e2 ℏω / kBT
∫ CV
=
3kBV
2π 2C 3
ωm
0
kBT (eℏω / kBT
−1)2
ω 2dω
( ℏω )2 eℏω / kBT
CV
=
9
R( 1
)
→
9 Nk B
(T
θD
)3
∫0∞
ξ3
(eξ −
1)
dξ
=
12π
5
4
NkB
(T
θD
)3
T3
Li
Be
344 1440
0.85 2.00
Debye temperature and thermal conductivity Low temperature limit of Debye temperature in Kelvin
T
)
fD
(θD T
)
=
3( T θD
)3
∫θD T 0
ξ4eξ (eξ −1)2
dξ
CV J/mol K 0 5 10 15 20 25
Yb
0 0.2 0.4 0.6 0.8 1.0 T/θD
θD
100 105 110 115 120 125
In
0 5 10 15 20 25
TK
CV
(T
/θ
D
1. Dulong—Petit
Cv
=
(∂E ∂T
)v
KBT
E = 3NkBT
Dulong—Petit law
Cv = 3NkB
Ge Si
Cp J/mol K 01 2 3 4 5 6
0 100 200 300 TK
(ni + 12)ℏωi
ω ω Ei
=
1ℏ 2
∑ n ℏ e−niℏωi / kBT ii
ni
+ ∑ e i
−niℏωi / kBT
ni
β= 1
kBT
Ei
=
1 2
ℏωi
−
∂
∂β
ln ∑ e−niβℏωi
ni
Ei
=
1 2
ℏω
i
+
ℏωi
eβℏωi −1
E
=
3∑N Ei
i=1
=3∑N
i =1
(1 2
ℏωi
+
e
ℏωi
− βℏωi
) 1
ρω
∫ωm
0
ρ (ω )dω
=
3N
E
=
∫ωm
0
ℏω
eℏω / kBT
: 3纵7采采采3纵8
Substance
Aluminum Bismuth Copper Brass Gold Lead Silver Tungsten Zinc Mercury Alcohol(ethyl) Water Ice (-10 C) Granite Glass
c in J/gm K
0.900 0.123 0.386 0.380 0.126 0.128 0.233 0.134 0.387 0.140 2.4 4.186 2.05 .790 .84
...
...
...
K
Ca
Sc
Ti
V
Cr Mn Fe
Co
Ni
Cu Zn Ga Ge
wenku.baidu.com
As
Se
Br Kr
91 230 360 420 380 630 410 470 445 450 343 327 320 374 282 90
...
72
1.02 ... 0.16 0.22 0.31 0.94 0.08 0.80 1.00 0.91 4.01 1.16 0.41 0.6 0.50 0.02 ...
)3
∫ωm
0
kBT (eℏω / kBT
− 1) 2
ω 2dω
∫ CV
=
9
Nk
B
1 (
ωm
)3
ℏω m
kBT 0
(ξ )4 eξ
(eξ −1)2
dξ
ξ = ℏω / kBT θ D = ℏωm / kB
CV
=
9
NkB
(T
θD
)3
∫θ D T 0
ξ
(eξ
4eξ − 1) 2
dξ
=
3NkB
f
D
(θ D
−1
ρ
(ω
)dω
CV
=
(∂E ∂T
)V
=
∫ωm
0
k
B
(
ℏω kBT
)2
eℏω/ kBT (eℏω/ kBT −1)2
ρ(ω)dω
kBT >> ℏωi
∂Ei ∂T
=
kB
(
ℏωi
kBT
)
2
[
[1+ ℏω + ⋯⋯]
kBT
ℏω + 1 ( ℏω )2 + ⋯⋯]2
kBT 2 kBT
∂Ei ∂T
≈
kB
CV = 3∑N kB = 3NkB i=1
kBT << ℏωi
ω (ℏ ) e i 2 ℏωi / kBT
∂Ei ∂T
= kB
kBT (eℏωi / kBT
− 1) 2
∂Ei ∂T
≈
k
B
(
ℏωi
kBT
)
2
e
−ℏωi
/
kBT
2. Einstein model
( ℏω )2 eℏω / kBT
CV
= ∂E ∂T
= 3NkB
kBT (eℏω / kBT
Thermal conductivity at 300K, in W/cmK
B
C
N
O
F
Ne
... 2230 ...
...
...
75
0.27 1.29 ...
...
...
...
Na Mg
Al
Si
P
S
Cl Ar
158 400
...
428 645 ...
...
...
92
1.41 1.56
2.37 1.48 ...
q
dq = dqxdqydqz q
V
(2π
)3
dq
q
ω—ω+∆ω
∆n = ρ(ω)∆ω
A q= ω
CL
q—q+dq
q + dq = ω + dω
CL
V
(2π
)3
4πq2dq
=
V
2π
C2 3 L
ω
2dω
q—q+dq
2
×
V
2π
C2 3 T
ω
2
dω
ρ (ω )
=
3V
2π 2C
3
ω
2
1 C3
=
11
3
(
C
c in cal/gm K or Btu/lb F
0.215 0.0294 0.0923 0.092 0.0301 0.0305 0.0558 0.0321 0.0925 0.033 0.58 1.00 0.49 0.19 0.20
Molar C J/mol K
24.3 25.7 24.5 ... 25.6 26.4 24.9 24.8 25.2 28.3 111 75.2 36.9 ... ...
− 1) 2
CV
= 3NkB
f
E
(
ℏω
kBT
)
kBθE = ℏω
CV
=
3NkB
(θ E
T
)2
eθ E (eθE /T
/T
−
1)2
Diamond
Cp J/mol K 01 2 3 4 5 6
0 0.2 0.4 0.6 0.8 1.0 T/θE
3. Debye model
q 1 、2
ω = CLq
ω = CT q
...
Rb Sr
Y
Zr Nb Mo Tc
Ru
Rh
Pd
Ag Cd
In
Sn
Sb
Te
I
Xe
56 147 280 291 275 450
...
600 480 274 225 209 108 200 211 153 ...
64
0.58 ... 0.17 0.23 0.54 1.38 0.51 1.17 1.50 0.72 4.29 0.97 0.82 0.67 0.24 0.02 ...
3 L
+
2 CT3
)
ω →0~∞
∫0∞ ρ(ω)dω
∫ωm
0
ρ (ω )dω
=
3N
。 ωm ωm
ωm
ωm
=
C[6π2
(
N
1
)]3
V
( ℏω ) e2 ℏω / kBT
∫ CV
=
3kBV
2π 2C 3
ωm
0
kBT (eℏω / kBT
−1)2
ω 2dω
( ℏω )2 eℏω / kBT
CV
=
9
R( 1
)
→
9 Nk B
(T
θD
)3
∫0∞
ξ3
(eξ −
1)
dξ
=
12π
5
4
NkB
(T
θD
)3
T3
Li
Be
344 1440
0.85 2.00
Debye temperature and thermal conductivity Low temperature limit of Debye temperature in Kelvin
T
)
fD
(θD T
)
=
3( T θD
)3
∫θD T 0
ξ4eξ (eξ −1)2
dξ
CV J/mol K 0 5 10 15 20 25
Yb
0 0.2 0.4 0.6 0.8 1.0 T/θD
θD
100 105 110 115 120 125
In
0 5 10 15 20 25
TK
CV
(T
/θ
D
1. Dulong—Petit
Cv
=
(∂E ∂T
)v
KBT
E = 3NkBT
Dulong—Petit law
Cv = 3NkB
Ge Si
Cp J/mol K 01 2 3 4 5 6
0 100 200 300 TK
(ni + 12)ℏωi
ω ω Ei
=
1ℏ 2
∑ n ℏ e−niℏωi / kBT ii
ni
+ ∑ e i
−niℏωi / kBT
ni
β= 1
kBT
Ei
=
1 2
ℏωi
−
∂
∂β
ln ∑ e−niβℏωi
ni
Ei
=
1 2
ℏω
i
+
ℏωi
eβℏωi −1
E
=
3∑N Ei
i=1
=3∑N
i =1
(1 2
ℏωi
+
e
ℏωi
− βℏωi
) 1
ρω
∫ωm
0
ρ (ω )dω
=
3N
E
=
∫ωm
0
ℏω
eℏω / kBT
: 3纵7采采采3纵8
Substance
Aluminum Bismuth Copper Brass Gold Lead Silver Tungsten Zinc Mercury Alcohol(ethyl) Water Ice (-10 C) Granite Glass
c in J/gm K
0.900 0.123 0.386 0.380 0.126 0.128 0.233 0.134 0.387 0.140 2.4 4.186 2.05 .790 .84
...
...
...
K
Ca
Sc
Ti
V
Cr Mn Fe
Co
Ni
Cu Zn Ga Ge
wenku.baidu.com
As
Se
Br Kr
91 230 360 420 380 630 410 470 445 450 343 327 320 374 282 90
...
72
1.02 ... 0.16 0.22 0.31 0.94 0.08 0.80 1.00 0.91 4.01 1.16 0.41 0.6 0.50 0.02 ...