Chapter 3
3.1
If o a were to increa, the bandgap energy would decrea and the material would begin to behave less like a miconductor and more like a metal. If o a were to decrea, the bandgap energy would increa and the
material would begin to behave more like an insulator.
_______________________________________ 3.2
Schrodinger's wave equation is:
()()()t x x V x
t x m ,,2222ψ⋅+∂ψ∂- ()t想和你永远在一起
t x j ∂ψ∂=,
Assume the solution is of the form:
()()⎥⎥⎦⎤
⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Region I: ()0=x V . Substituting the
assumed solution into the wave equation, we obtain:
()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧∂∂-t E kx j x jku x m exp 22 ()⎪⎭
⎪
⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp ()⎥⎥⎦
⎤
⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⋅⎪⎭⎫ ⎝⎛-=t E kx j x u jE j exp which becomes
()()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 22
2 ()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk
exp 2 ()⎪⎭
⎪
⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 22 ()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+=t E kx j x Eu exp This equation may be written as
()()()()02222
22
=+∂∂+∂∂+-x u mE
x x u x x u jk x u k
Setting ()()x u x u 1= for region I, the equation becomes:
()
()
()
()0212212
12=--+x u k dx
x du jk
dx x u d α where
2
22
mE
=
α Q.E.D. In Region II, ()O V x V =. Assume the same form of the solution:
()()⎥⎥⎦⎤
⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=ψt E kx j x u t x exp , Substituting into Schrodinger's wave equation, we find:
()()⎥⎥⎦⎤⎢⎢⎣
⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎩⎨⎧-t E kx j x u jk m exp 22
2 ()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u jk
exp 2 ()⎪⎭
⎪
⎬⎫⎥⎥⎦⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-∂∂+t E kx j x x u exp 2
2 ()⎥⎥⎦⎤
⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-+t E kx j x u V O exp ()⎥⎥⎦
⎤⎢⎢⎣⎡⎪⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-=t E kx j x Eu exp This equation can be written as:
()()()222
2x x u x x u jk x u k ∂∂+∂∂+- ()()0222
2=+-x u mE
x u mV O
Setting ()()x u x u 2= for region II, this equation becomes
()()dx x du jk dx
x u d 22
222+ ()0222
2
2=⎪⎪⎭
⎫ ⎝⎛+--x u mV k O α where again
222
mE
=α Q.E.D.
_______________________________________
3.3 We have ()()()
()0212
21212=--+x u k dx x du jk dx x u d α Assume the solution is of the form:
()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexp
The first derivative is
()()()[]x k j A k j dx
x du --=ααexp 1 ()()[]x k j B k j +-+-ααexp and the cond derivative becomes
()
()[]()[]x k j A k j dx x u d --=ααexp 2
2
12
()[]()[]x k j B k j +-++ααexp 2
Substituting the equations into the differential equation, we find
()()[]x k j A k ---ααexp 2
()()[]x k j B k +-+-ααexp 2
(){()[]x k j A k j jk --+ααexp 2
()()[]}x k j B k j +-+-ααexp ()
()[]{x k j A k ---ααexp 22 ()[]}0exp =+-+x k j B α Combining terms, we obtain
()()()[]
222222αααα----+--k k k k k ()[]x k j A -⨯αexp
()()()[]
骗取银行贷款罪222222αααα--++++-+k k k k k
()[]0exp =+-⨯x k j B α We find that
00= Q.E.D. For the differential equation in ()x u 2 and the propod solution, the procedure is exactly the same as above.
_______________________________________ 3.4
We have the solutions ()()[]x k j A x u -=αexp 1 ()[]x k j B +-+αexp for a x <<0 and
()()[]x k j C x u -=βexp 2
()[]x k j D +-+βexp for 0<<-x b .
The first boundary condition is ()()0021u u =
which yields 0=--+D C B A
The cond boundary condition is 0201===x x dx du dx du
好的耳机品牌which yields
()()()C k B k A k --+--βαα ()0=++D k β The third boundary condition is ()()b u a u -=21 which yields
()[]()[]a k j B a k j A +-+-ααexp exp ()()[]b k j C --=βexp
()()[]b k j D -+-+βexp and can be written as
()[]()[]a k j B a k j A +-+-ααexp exp ()[]b k j C ---βexp
()[]0exp =+-b k j D β The fourth boundary condition is
b
x a x dx du
dx du -===21
which yields
()()[]a k j A k j --ααexp ()()[]a k j B k j +-+-ααexp ()()()[]b k j C k j ---=ββexp
()()()[]b k j D k j -+-+-ββexp and can be written as
()()[]a k j A k --ααexp ()()[]a k j B k +-+-ααexp ()()[]b k j C k ----ββexp
()()[]0exp =+++b k j D k ββ
_______________________________________ 3.5
(b) (i) First point: πα=a
Second point: By trial and error, πα729.1=a (ii) First point: πα2=a
Second point: By trial and error, πα617.2=a
_______________________________________
3.6 (b) (i) First point: πα=a Second point: By trial and error,
πα515.1=a (ii) First point: πα2=a
Second point: By trial and error, πα375.2=a _______________________________________
3.7 ka a a
a P cos cos sin =+'ααα Let y ka =, x a =α Then
y x x
x P cos cos sin =+' Consider dy d of this function. ()[]{}
y x x x P dy d sin cos sin 1
-=+⋅'- We find
()()()⎭⎬⎫⎩
⎨⎧⋅+⋅-'--dy dx x x dy dx x x P cos sin 11
2
y dy
dx
x sin sin -=- Then
y x x x x x P dy dx sin sin cos sin 12
-=⎭
⎬⎫⎩⎨⎧-⎥⎦⎤
⎢⎣⎡+-' For πn ka y ==, ...,2,1,0=n 0sin =⇒y So that, in general,
()()dk d ka d a d dy dx
αα===0 And 2
2 mE
=α So
dk dE
m mE dk d ⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛=-22
/122221 α This implies that
dk dE dk d =
=0α for a
n k π= _______________________________________ 3.8 (a) πα=a 1 π=⋅a E m o 2
1
2 ()(
)
(
)(
)
210312
34222
21102.41011.9210054.12---⨯⨯⨯==ππa m E o 19104114.3-⨯=J From Problem 3.5
πα729.12=a
π729.122
2
=⋅a E m o ()(
德国旅行)
(
)(
)
210312
3422102.41011.9210054.1729.1---⨯⨯⨯=πE 18100198.1-⨯=J 12E E E -=∆
1918104114.3100198.1--⨯-⨯= 19107868.6-⨯=J
or 24.4106.1107868.619
19
三年级竖式计算题
=⨯⨯=∆--E eV
(b) πα23=a
π222
3=⋅a E m o
()()(
)()
2
1031
2
3423102.410
11.9210054.12---⨯⨯⨯=
πE
18103646.1-⨯=J
From Problem 3.5, πα617.24=a
π617.222
4=⋅a E m o
()()(
)()
2
1031
2
3424102.410
11.9210054.1617.2---⨯⨯⨯=
πE
18103364.2-⨯=J 34E E E -=∆
1818103646.1103364.2--⨯-⨯= 1910718.9-⨯=J
or 07.6106.110718.919
19
=⨯⨯=∆--E eV
_______________________________________
3.9 (a) At π=ka , πα=a 1
π=⋅a E m o 2
12 ()()(
)
()
2
1031
2
3421102.410
11.9210054.1---⨯⨯⨯=
πE
19104114.3-⨯=J
At 0=ka , By trial and error, πα859.0=a o ()
()()()
2
1031
2
342
102.41011.9210054.1859.0---⨯⨯⨯=
πo
E
19105172.2-⨯=J
o E E E -=∆1
1919105172.2104114.3--⨯-⨯= 2010942.8-⨯=J
or 559.010
6.110942.819
20
=⨯⨯=∆--E eV (b) At π2=ka , πα23=a
π222
3=⋅a E m o
()
()()()
2
1031
2
342
3
102.41011.9210054.12---⨯⨯⨯=
πE
18103646.1-⨯=J
At π=ka . From Problem 3.5,
πα729.12=a
π729.122
2=⋅a E m o
()()()()
2
10
312历史语言学
3422102.41011.9210054.1729.1---⨯⨯⨯=
πE
18
10
0198.1-⨯=J
23E E E -=∆
1818100198.1103646.1--⨯-⨯= 19104474.3-⨯=J
or 15.2106.1104474.319
19
=⨯⨯=∆--E eV
_______________________________________
3.10 (a) πα=a 1
π=⋅a E m o 2
12
()()(
)()
2
1031
2
3421102.410
11.9210054.1---⨯⨯⨯=
πE
19104114.3-⨯=J
From Problem 3.6, πα515.12=a
π515.122
2=⋅a E m o
()()(
)()
2
1031
2
3422102.410
11.9210054.1515.1---⨯⨯⨯=
πE
1910830.7-⨯=J 12E E E -=∆
1919104114.310830.7--⨯-⨯= 19104186.4-⨯=J
or 76.2106.1104186.419
19
=⨯⨯=∆--E eV
(b) πα23=a
π222
3=⋅a E m o
()()(
)()
2
1031
2
3423102.410
11.9210054.12---⨯⨯⨯=
πE
18103646.1-⨯=J
From Problem 3.6, πα375.24=a
π375.222
4=⋅a E m o
()()(
)
()
2
1031
2
3424102.410
11.9210054.1375.2---⨯⨯⨯=
πE
18109242.1-⨯=J 34E E E -=∆
1818103646.1109242.1--⨯-⨯= 1910597.5-⨯=J
or 50.3106.110597.519
19
=⨯⨯=∆--E eV
_____________________________________
主题党日主持词3.11 (a) At π=ka , πα=a 1
π=⋅a E m o 2
12
()()(
)()
2
1031
2
3421102.410
11.9210054.1---⨯⨯⨯=
πE
19104114.3-⨯=J
At 0=ka , By trial and error, πα727.0=a o
π727.022
=⋅a E m o o
()()(
)()
2
1031
2
342102.410
11.9210054.1727.0---⨯⨯⨯=
πo E
19
108030.1-⨯=J
o E E E -=∆1
1919108030.1104114.3--⨯-⨯= 19106084.1-⨯=J
or 005.110
6.1106084.119
19
=⨯⨯=∆--E eV (b) At π2=ka , πα23=a
π222
3=⋅a E m o
()
()()()
2
1031
2
342
3102.41011.9210054.12---⨯⨯⨯=
πE
18103646.1-⨯=J
At π=ka , From Problem 3.6,
πα515.12=a
π515.122
2=⋅a E m o
()()(
)()
2
1034
2
3422102.410
11.9210054.1515.1---⨯⨯⨯=
πE
1910830.7-⨯=J
23E E E -=∆
191810830.7103646.1--⨯-⨯= 1910816.5-⨯=J
or 635.3106.110816.519
19
=⨯⨯=∆--E eV
_______________________________________
3.12
For 100=T K, ()()⇒+⨯-
=-1006361001073.4170.12
4
g
E
164.1=g E eV 200=T K, 147.1=g E eV 300=T K, 125.1=g E eV 400=T K, 097.1=g E eV 500=T K, 066.1=g E eV
600=T K, 032.1=g E eV
_______________________________________ 3.13
The effective mass is given by
1
222*
1-⎪⎪⎭
⎫ ⎝⎛⋅=dk E d m
We have
()()B curve dk
E d A curve dk E d 222
2> so that ()()B curve m A curve m **<
_______________________________________ 3.14
The effective mass for a hole is given by 1
222*
1-⎪⎪⎭
⎫ ⎝
⎛⋅
=dk E d m p We have that
()()B curve dk
E
d A curv
e dk E d 2222>
so that ()()B curve m A curve m p p *
*<
_______________________________________ 3.15
Points A,B: ⇒<0dk dE
velocity in -x direction
Points C,D: ⇒>0dk
dE
四年级下册科学教学计划
velocity in +x direction
Points A,D: ⇒<02
2dk
E
d negativ
e effective mass
Points B,C: ⇒>02
2dk
E
d positiv
e effective mass _______________________________________
