Мартыненко Т.П. и др. Практический курс физики. Квантовая физика. Элементы физики твёрдого тела и ядерной физики
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121
( . .
min
~ 2d). ,
.
2) << . / >> 1, , / → ∞,
(4.63),
,
,
RT
5
3
U
1e
dxxT
3RT3UU
3
44
0m
0
3
3
0mm
θ
π
⋅+=
−
⋅
θ
⋅+=
∞
(4.64)
.
T
R
5
12
dT
dU
C
3
4
m
m
θ
π==
(4.69)
.
, <<
( ~
3
)
.
( . 4.14). <<
<< k / >>
min
.
– , ,
U
1
, U
2
.
,
.
:
0
0
( . 4.14).
, ,
U
3
,
. , - ,
,
k , .
k
(
F
).
U
3
= < >·N
A
,
.
E
kT
12
5
1E
5
3
NU
2
F
2
FAm3
π+⋅=
(4.70)
(4.70) ,
122
,
T2
RT
E
kT
2
R
E
Tk
6
5
NE
5
3
T
U
C
F
2
F
2
2
F
22
AF
m3
Me
π
=⋅
π
=
π
⋅⋅=
∂
∂
=
(4.71)
R = kN
– ,
F
=
F
/k (4.19).
(4.71)
,
; Z
,
.
T
T
2
RZ
E
kT
2
RZ
CZC
F
2
F
2
MeMeZ
⋅
π
=⋅
π
=⋅=
(4.71a)
<< T
3
,
, ( ~ T),
0
.
0
,
" ",
θ⋅
θ
=
F
0
T
Z
145,0T
.
,
, ,
.
,
,
,dT)T(cmQ
2
1
T
T
⋅=
(4.72)
m – , ( ) – ,
c( )=
m
( )/ .
.
4.1.
= 0 , ,
.
,
= 0 , – (4.12):
( )
.n3
m2
)0(EE
3/2
0
2
2
Fmax
π==
123
, , (4.11)
A0
N·
M
=n
, = 8,9·10
3
/
3
,
N
A
= 6,02·10
23
–1
, = 64·10
–3
/ :
.1083,0
1064
1002,6109,8
n
329
3
233
0
−
−
⋅=
⋅
⋅⋅⋅
=
.2,6109,9
101,92
)1083,014,33()1005,1(
E
19
31
3/2292234
max
=⋅=
⋅⋅
⋅⋅⋅⋅
=
−
−
−
, ,
.
),0(E
2
mv
F
2
max
=
./107,4
101,9
109,92
m
E2
v
7
31
19
F
max
⋅=
⋅
⋅⋅
==
−
−
4.2. ,
, ∆ =0,05
: 1)
1
= 290 ; 2)
2
= 50 .
- (4.17)
.
1kT/)EEexp(
1
)T,E(f
F
+−
=
,
, ∆ = –0,05 .
f( , ) ,
k
1
k
2
- :
,1025
106,1
1
2901038,1kT
3
19
23
1
−
−
−
⋅=
⋅
⋅⋅=
,103,4
106,1
1
501038,1kT
3
19
23
2
−
−
−
⋅=
⋅
⋅⋅=
,881,0
1)025,0/05,0exp(/1
1
)T,E(f
1
=
+
=
.99991,0
1)0043,0/05,0exp(/1
1
)T,E(f
2
=
+
=
124
, ∆ = +0,05
,
:
,119,0
1)025,0/05,0exp(
1
)T,E(f
1
=
+
=
.108,8
1)0043,0/05,0exp(
1
)T,E(f
6
2
−
⋅=
+
=
4.3.
F
,
.
<<
F
,
- ,
F
–
- .
F
,
F
= 6,2 ( 4.1). (4.19)
;107
/1038,1
106,12,6
T
4
23
19
F
⋅=
⋅
⋅⋅
=
−
−
,
<<
F
, ,
- .
4.4. ,
0
F
= 0
n
0
.
N 0
F
(4.16)
⋅=
F
E
0
dE)T,E(f)E(gN
(1)
(4.13)
.
2
VE)m2(
)E(g
32
2/12/3
π
⋅⋅
=
(2)
= 0 f(E,T) = 1,
(2) (1),
⋅
π
⋅
=
π
⋅⋅⋅⋅
=
F F
E
0
E
0
2/1
32
2/3
32
2/12/32/3
;dEE
2
V)m2(
2
dEVEm2
N
125
.
3
V)mE2(
32
EV)m2(2
N
32
2/3
F
E
0
32
2/12/3
F
π
⋅
=
⋅π
⋅⋅
=
n
0
( )
.
3
)mE2(
V
)E(N
)E(n
32
2/3
F
0
π
==
4.5. dN(E)
, dN( ) .
= 0 .
(4.21) :
.dE
1kT/)EEexp(
Em2
2
1
)T,E(dN
F
2/1
2/3
22
⋅
+−
⋅
π
=
(1)
m
2
p
=E
2
(2),
.dpp2
m2
1
m2
p
ddE
2
⋅⋅==
(3)
(2) (3) (1),
;
1kT/)Em2/pexp(
dpp2
m2
1
)m2(
pm2
2
1
)p(dN
F
22/1
2/3
22
+−
⋅
⋅⋅⋅
π
=
.dp
1kT/)Em2/pexp(
1p1
)p(dN
F
23
2
2
+−
⋅⋅
π
=
= 0
;1
1)kT/)Em2/pexp((
1
F
2
=
+−
.dp
p
)p(dN
32
2
⋅
π
=
4.6. ,
max
0,1
max
40
3
= 0 .
F
= 6,2 .
,
= 0 .
.dpp
1
)p(dn
2
32
⋅
π
=
(1)
126
∆n ,
(
max
–0,1
max
)
max
,
(1) :
,
3
p
dpp
1
)p(n
max
max
max
max
p
p9,0
p
p9,0
32
3
2
32
π
=
π
=∆
.
3
271,0
)9,01(
3
p
)p(n
32
3
max
3
32
3
max
π
⋅
=−
π
=∆
∆N V
.V
3
271,0
VnN
32
3
max
π
⋅
=⋅=∆
(2)
,
max
= 0
F
2
max
mE2p =
2/1
Fmax
)mE2(p =
,
(2) :
.V
3
)mE2(271,0
N
32
2/3
F
π
=∆
(3)
(3) m=9,1⋅10
-31
,
F
=6,2⋅1,6⋅10
-19
,
V = 4⋅10
-5
3
, = 1,05⋅10
-34
⋅ ,
,104
)1005,1(14,33
)106,12,6101,92(271,0
N
5
3342
2/31931
−
−
−−
⋅⋅
⋅⋅⋅
⋅⋅⋅⋅⋅⋅
=∆
∆N = 1,4 ⋅10
-35
.
4.7.
= 0 ,
?
dn
E + d = 0
(4.23)
.dEE
m2
2
1
)E(dN
2/1
2/3
22
⋅⋅
π
=
(1)
∆n
1
= 0 ,
, . . >
max
/2 =
F
/2,
(1)
F
/2
F
:
127
;
3
E2m2
2
1
dEE
m2
2
1
dnn
F
F
F
F
F
F
E
2/E
2/3
F
2/3
22
E
2/E
2/1
2/3
22
E
2/E
1
π
=
π
==∆
.
2
1
1E
m2
3
1
n
2/3
2/3
F
2/3
22
1
−⋅
π
=∆
(2)
= 0 ∆n
2
,
0
max
=
F
.E
m2
3
1
dEE
m2
2
1
dnn
2/3
F
2/3
22
E
0
2/1
2/3
22
E
0
2
FF
⋅
π
=
π
==∆
(3)
∆n
1
∆n
2
(2) (3) :
(
)
( )
%.65%1005,01%100
E
5,01E
n
n
2/3
2/3
F
2/32/3
F
2
1
≅⋅−=⋅
−
=
∆
∆
4.8.
1
3
0 , ,
.
EnEE
n
1i
i
==
=
(1), n –
1
3
,
E
–
. = 0
(4.29)
.
dE)E(g
dE)E(gE
E
F
F
E
0
E
0
⋅
=
(2)
(2) g( ), (4.13)
,E
m2
2
V
)E(g
2/1
2/3
22
⋅
π
=
,
.E
5
3
dEE
dEE
E
F
E
0
2/1
E
0
2/3
F
F
==
128
E
F
= 6,2 ( . 4.1) , ,
< >= 3,7 . n (4.11):
n = N
A
/ . = 8,9⋅10
3
/
3
,
Cu
= 64⋅10
-3
/
.108,0108,0
1064
1002,6109,8
n
323329
3
233
−−
−
⋅=⋅=
⋅
⋅⋅⋅
=
1
3
(1), < > n:
.108,4103103108,07,3E
4142323
⋅=⋅=⋅=⋅⋅=
4.9. < v >
v
max
, ,
.
= 0 (4.30)
.
dv)v(g
dv)v(gv
v
max
max
v
0
v
0
⋅
⋅⋅
=
(1)
(4.32) (4.12)
.n3
m
m2
)n3(2
m
E2
v
3
2
2
3/222
F
max
π=
π
==
(2)
,
, (4.14), :
,E
m2
2
V
)E(g
2/1
2/3
22
⋅
π
=
, d ,
,dEE
m2
2
V
dE)E(g
2/1
2/3
22
⋅⋅
π
=
(3)
g( )d
g(v)dv.
,
2
mv
E
2
=
d = m⋅v⋅dv
,
2
m
vE =
(3),
.dvvm
1
2
dvvmmvm2
2
1
dv)v(g
23
32
2/3
22
⋅⋅
π
=
⋅⋅⋅⋅
π
=
(1) ,
129
.v
4
3
dvv
dvv
v
max
v
0
2
v
0
3
max
max
==
(2) (4.11)
,N
M
n
A
ρ
=
,M/N3
m4
3
v
3
A
2
ρπ⋅=
./102,9
1094,6
1002,653414,33
101,9
1005,1
4
3
v
5
3
3
232
31
34
⋅=
⋅
⋅⋅⋅⋅
⋅
⋅
⋅=
−−
−
4.10. ∆E
i
( )
1
3
, 0,1 5 = 0 .
(4.1)
V
.
E)m2(V4
)2(
E
i
2/3
3
π
π
=∆
(1)
(1) = 1,05⋅10
-34
⋅ , V = 10
-6
3
,
m = 9,1⋅10
-31
,
1
= 0,1 ,
2
= 5 (1 = 1,6⋅10
-19
),
,
106,1106,11,0)101,92(1014,34
)1005,114,32(
E
19192/3316
334
1
−−−−
−
⋅⋅⋅⋅⋅⋅⋅⋅
⋅⋅⋅
=∆
,
106,1106,15)101,92(1014,34
)1005,114,32(
E
19192/3316
334
2
−−−−
−
⋅⋅⋅⋅⋅⋅⋅⋅
⋅⋅⋅
=∆
∆E
1
= 4,7⋅10
–22
, ∆E
2
= 6,6⋅10
–23
.
4.11. 4,5·10
28
/
3
. 300 2·10
9
.
0,4 0,2
2
/ ⋅ ).
.
130
.1025,2
10
2
105,4
n
319
9
28
p,n
−
⋅=
⋅
⋅
=
(4.33)
=
n,
= n
n,
⋅ (
n
+
p
) = 2,25⋅10
19
⋅ 1,6⋅10
–19
(0,4+0,2) = 2,16 / .
4.12. ,
L
R
0
R.
∆n
n
∆n ,
,
n
p
. ∆n
n
= ∆n .
∆n
n
= ∆n , = (
n
+
p
)∆n, ,
, = –
0
(1), –
,
0
–
.
S
L
R ρ=
, ,
1/
0
– 1/
0
,
.
RR
RR
L
l
R
1
R
1
S
l11
0
0
2
00
−
⋅=−=
ρ
−
ρ
=σ
(1) (2), ∆n
.
LRR)(e
RR
n
RR
RR
L
1
n)(e
0pn
0
0
0
pn
µ+µ
=∆⋅=∆µ+µ
--
4.13.
1 .
= 290 .
, :
1)
1
= 1 2)
2
= 2 ?
;
F
= 1 /2= 0,5 .
- (4.17):
1e
1
)T,E(f
kT/)EE(
F
+
=
−