Мартыненко Т.П. и др. Практический курс физики. Квантовая физика. Элементы физики твёрдого тела и ядерной физики
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131
. –
F
= 1 – 0,5 = 0,5 .
:
.1006,2
1)20exp(
1
1
2901038,1
106,15,0
exp
1
)T,E(f
9
23
19
−
−
−
⋅=
+
=
+
⋅⋅
⋅⋅
=
, .
,
= hν = h / ∆
.
1)
1
= 1·10
–6
,
;24,1
/106,110
/103c1062,6c
hE
196
834
1
1
=
⋅⋅
⋅⋅⋅⋅
=
λ
=
−−
−
. . > 1 ,
.
2)
2
= 2·10
–6
;62,0
/106,1102
/103c1062,6c
hE
196
834
2
2
=
⋅⋅⋅
⋅⋅⋅⋅
=
λ
=
−−
−
. .
2
= 0,62 < 1 ,
.
4.14.
=1,76·10
–6
.
= 300 .
(4.46) (4.48) ,
;
kT2
E
e
2
kT2
E
0
∆
⋅ρ=
∂
ρ∂
∆
,
kT2
E
e
e
1
2
kT2
E
0
kT2
E
0
∆
−⋅ρ⋅
ρ
=α
∆
∆
= –∆ /2k
2
.
,
E
hc
∆
=λ
.
hc
E
λ
=∆
132
, ,
,
kT2
h
2
λ
−=α
.K047,0
1076,11091038,12
1031062,6
1
6423
834
−
−−
−
−=
⋅⋅⋅⋅⋅⋅
⋅⋅⋅
−=α
4.15. -
( )
1
= 0,65·10
-6
,
2
= 2,07·10
-6
.
,
,
=0 ( ). ,
,
,
.hE
1
1
λ
=
∆ ,
,
.hE
2
λ
=∆
, ,
2
=
1
– ,
,hc
c
hhE
21
12
21
2
λλ
λ
λ
=
λλ
=
-
-
.0510,2
1065,01007,2
10)65,007,2(1031062,6
E
19
66
6834
2
−
−−
−−
=
⋅⋅⋅
⋅−⋅⋅⋅
=
.
2
= –2,05⋅10
-19
= –1,31 .
=0
2
1
133
4.16.
, 0,001% ,
. ∆ = 0,75 ,
∆ = 0,015 (m
= m ).
(G ) ,
( s) – , ,
( . 4.12 ). = 300
,025,0
106,1
3001038,1
kT
19
23
=
⋅
⋅⋅
=
−
−
. . kT >
a
, , .
, 0,001%
, (4.11):
%,001,0N
M
n
A
Ge
Ge
s
⋅⋅
ρ
=
Ge
= 5,3·10
3
/
3
,
Ge
= 73·10
–3
/ , N
A
= 6,02·10
23
–1
,
.104,410
1073
1002,6103,5
n
3235
3
233
s
−−
−
⋅=⋅
⋅
⋅⋅⋅
=
,
, (4.50):
,e
kTm2
4
1
)T(n
kT2
E
2/3
2
e
Ge
∆
−
⋅
π
=
.104,7e
1005,114,3
106,1025,0101,92
4
1
)T(n
318
05,0
75,0
2/3
682
1931
Ge
−−
−
−
−−
⋅=⋅
⋅⋅
⋅⋅⋅⋅⋅
=
4.17.
0,72 , ,
,
, 300 ?
(4.50)
,
mk2
4
1
C
2/3
2
π
=
134
1
kT2
E
2/3
11
eTCn
∆
−
⋅⋅= (1)
= 300 .
2
=
1
+ ∆ ,
( )
.eTTCn
)TT(k2
E
2/3
12
1
∆+
∆
−
⋅∆+=
∆+
∆
⋅
∆∆+
==
)TT(
T
kT2
E
exp
T
TT
2
n
n
11
2/3
1
1
1
2
.
1
1
k2
exp
T
T
12
1
1
2/3
1
+
∆
⋅
∆∆
+=
(2)
.
, ,
∆T <<T
1
. ,
∆T/
1
= .
(2) :
( )
.
1kT2
E
exp12
1
2/3
α+
α
⋅
∆
α+=
(3)
(3):
( )
,
1kT2
E
1ln
2
3
2ln
1
α+
α
⋅
∆
+α+=
<< 1,
( )
,
1
1ln
α+
α
≈α+
;
kT2/E2/3
2ln
11kT2
E
12
3
2ln
11
∆+
=
α+
α
α+
α
∆
+
α+
α
⋅=
;045,0
3001038,12
106,172,0
5,1
693,0
1
23
19
=
⋅⋅⋅
⋅⋅
+
=
α+
α
−
−
(4)
135
(4) , = 0,047,
∆ = 0,047⋅300 = 14, K
,
14,1
.
4.18. n -
L = 10 d = 6
= 0,1
. U = 250 ,
,
U = 8,8 . : 1) R , 2)
n
n
.
= 80 / .
U , (4.36): U = R jd.
djB
U
R
x
x
⋅⋅
=
(1).
j ,
j = ⋅E, –
. ,
= U/L,
L
U
j σ=
. (2)
(2) (1),
d
U
B
LU
R
x
x
⋅
σ
⋅
⋅
⋅
=
. (3)
U = 8,8·10
–3
, L = 0,1 ,
d = 6·10
–3
, = 0,1 , = 80 / , U = 250 ,
./1033,7
106802501,0
1,0108,8
R
35
3
3
x
−
−
−
⋅=
⋅⋅⋅⋅
⋅⋅
=
(4.38) (n -
)
n
x
n
1
8
3
R
⋅
π
=
,
(4)
136
– ( = 1,6·10
–19
), n
n
– .
(4)
eR8
3
n
x
⋅
π
=
, , R , π,
.10
106,11033,78
14,33
n
323
195
n
−
−−
=
⋅⋅⋅⋅
⋅
=
4.19. 100
20 50° 8300 . ,
,
.
, 20 50° ,
(4.72)
⋅=
2
1
T
T
m
dT)T(CmQ
(1)
∆ ,
M
C
V
m
=
,
,
V
= 3R, (2)
−==
2
1
T
T
12m
m
).TT(C
M
m
dT
M
C
mQ (3)
(3), Q ∆ ,
m
.
Q
TCm
M
m
∆⋅
=
,
,/109
103,8
3031,831,0
M
3
3
−
⋅=
⋅
⋅
⋅
⋅
=
= 9⋅10
–3
/ (
4
9
).
4.20.
,
:
1
= 0,054 /( ⋅ )
1
= 20
2
= 0,18 /( ⋅ )
2
= 30 .
137
<<
(4.69). :
3
4
m
T
R
5
12
θ
π=
,
(1)
- , - .
(1)
.
T
T
C
C
3
2
1
2m
1m
=
;3,0296,0
27
8
T
T
3
2
1
≅==
;3,0
18,0
054,0
C
C
2m
1m
≅=
C
m
∼
3
, . .
20 ≤ ≤30 , (1)
=⋅
π
=θ
1
3/1
m
4
T
C5
R12
.K5,40820
19,4054,05
31,814,312
3/1
4
=⋅
⋅⋅
⋅⋅
=
4.21. 15
1
= 10
2
= 20
Q = 1,065 .
, ,
1
2
. – M = 107,86·10
–3
/ .
,
(4.69)
3
3
4
m
T
R
5
12
θ
π
=
, (1)
. . ,
(4.72):
dT)T(C
M
m
dQQ
m
==
. (2)
(1) (2) ,
⋅θ
−π
=⋅
θ
π
⋅=
2
1
T
T
3
4
1
4
2
4
3
3
4
.
4M5
)TT(Rm12
dTT
M
mR
5
12
Q
(3)
(3)
138
;
MQ
)TT(Rm6,0
3
4
1
4
2
4
−π⋅
=θ
(
)
.K212
065,11086,107
1020101531,814,36,0
14,3
3
3
443
=
⋅⋅
−⋅⋅⋅⋅
⋅=θ
−
−
4.22. ,
NaCl 50 ∆ = 2 : 1)
1
= , 2)
2
= 2 . NaCl θ = 320 ,
M = 58,5·10
–3
/ .
,
(4.53).
, (4.72).
=
2
1
T
T
m
.dT)T(C
M
m
Q
(1)
,
(4.65)
.
1e
T
3
1e
dxxT
12R3C
T/
0
T/x
3
3
m
−
θ
−
−
⋅
θ
=
θ
θ
(2)
(2)
kT
h
x
ν
= , ν –
;
k
h
max
ν
=θ
, ν
max
–
.
:
1
= ;
1=
T
1
; ,
θ
=
−
=
−
1
T/
0
1
0
x
3
x
3
.225,0
1e
dxx
dx
1e
x
(3)
(2)
1
= (3):
139
.R805,2
17,2
3
225,0112R3C
m
=
−
−⋅⋅=
(4)
(4) (1), Q
1
:
∆+θ
θ
∆+θ
θ
∆⋅===
T T
m1
,TR805,2
M
m
dTR805,2
M
m
dT)T(C
M
m
Q
.85,39231,8805,2
105,58
05,0
Q
3
1
=⋅⋅⋅
⋅
=
−
2
= 2 << ,
(4.69), (1),
Q
2
:
⋅θ
−π
⋅=
θ
π
=
2
1
T
T
3
4
1
4
2
4
3
34
2
,
45
)TT(R12
M
m
5
dTRT12
M
m
Q
.10025,3
43205
)24(31,814,312
105,58
05,0
Q
3
3
444
3
2
−
−
⋅=
⋅⋅
−⋅
⋅
⋅
=
4.23.
.
E
kT
12
5
1E
5
3
E
2
F
2
F
π
+=
= 300
n = 6⋅10
22
–3
.
> 9 ( . .4), (4.53)
.m
= 3R.
dT
dU
C
e
e
=
U
e
– U
e
= < > N
A
:
;T
E
kR
4
NE
5
3
N
E
)kT(
12
5
E
5
3
E
5
3
U
2
F
2
AFA
2
F
22
FFe
π
+=
π
+=
;
)n3(
m2kT
6C
C
,
E
kT
R
2dT
dU
C
3/222
2
m,
e
F
2
e
e
π
⋅
⋅
π
=⋅⋅
π
==
.106,7
)10614,33()1005,1(6
101,923001038,114,3
C
C
3
3/2282234
31232
m,
e
−
−−
−−
⋅=
⋅⋅⋅⋅⋅
⋅⋅⋅⋅⋅⋅
=
140
4.24. ,
θ = 339 , – 6,2
600 .
(4.71 ), . .
F
2
Z,Me
E
kTZ
2
R
C
⋅
⋅
π
=
, Z – ,
, Z = 1 (
),
F
2
Z,Me
E2
RkT
C
π
=
. (1)
= 600 ,
Cu
= 339 .
(4.53),
.m
= 3R. (2)
(1) (2)
.
E6
kT
R3E2
RkT
C
C
F
2
F
2
m.
Z,Me
π
=
⋅
π
=
k = 1,38·10
–23
/ , = 600 ,
F
= 6,2·1,6⋅10
–19
,
.077,0
106,162,6
6001038,114,3
C
C
19
232
m.
Z,Me
=
⋅⋅
⋅⋅⋅
=
−
−
4.25.
,
374 .
,
3·10
–10
.
(4.62):
hν
max
= k . (1)
(1)