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Chapter 3
3.1
If ao were to increase, the bandgap energy
would decrease and the material would begin
to behave less like a semiconductor and more
like a metal. If ao were to
?2jk???u?x??E???exp?j?kx???t???? ?x?????????
???2u?x??E????????expjkx?t??????? ??x2??????????
???E???decrease, the ??Eu?x?exp?j?kx???t???? ????????? bandgap energy would increase and
the This equation may be written as material would begin to behave more like an ?u?x??2u?x?2mE2?ku?x??2jk??2u?x??0 insulator. ?x?x2?_______________________________________ 3.2 Setting u?x??u1?x? for region I, Schrodinger's wave equation is: the equation
22?????x,t? becomes: ?V?x????x,t? 22m?x
d2u1?x?du1?x?22?2jk?k??u1?x??0 2???x,t?dxdx ?j??t where
Assume the solution is of the form: 2mE ?2?2
???E????kx? ??x,t??u?x?exp?j???t?? ??Q.E.D. ????????? In Region II, V?x??VO. Assume
Region I: V?x??0. Substituting
the same
the
form of the solution:
assumed solution into the wave
???E???equation, we kx? ??x,t??u?x?exp?j???t???? ? obtain: ????????????2???E??? Substituting into Schrodinger's kx? ??t?? ?jku?x?exp?j???wave 2m?x?????????? equation, we find:
?????u?x??E????? ?exp?j?kx???2???t??E???2?? ???????jkuxexpjkx? ??t??x??????????? ?????2m??????????
????jE??E??????u?x??j??kx??E?????u?x?exp?j???t?? ????2jkexpjkx???t?????????????? ????x????????? which becomes
2??????E???2??kx? ??t??2u?x??E?????? ??jk?u?x?exp?j??????expjkx?t2m??????? ??????????2??x????????????----------------------------------------------------
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???E????VOu?x?exp?j?kx???t???? ?????????
d2u1?x? ???E????Eu?x?exp?j?kx???t???? ?????????dx2
2??j???k??Bexp??j???k?x?
??j???k??Aexp?j???k?x?
2 This equation can be written as:
?u?x??2u?x?2 ?ku?x??2jk ??x?x2 2mVO2mE???ux?u?x??0 ?2?2 Setting u?x??u2?x? for region II,
Substituting these equations into the
differential equation, we find
2 ????k?Aexp?j???k?x? ????k?Bexp??j???k?x?
2 ?2jk?j???k?Aexp?j???k?x? ?j???k?Bexp??j???k?x??
?k2??2?Aexp?j???k?x?
?? this
?Bexp??j???k?x???0 equation becomes
Combining terms, we obtain d2u2?x?du2?x? ?2jk dxdx2??2?2?k?k2?2k???k??k2??2
?Aexp?j???k?x? 2mVO??22?? ??k????ux?02?? ?2?????2?2?k?k2?2k???k??k2??2 where again
2mE ?Bexp??j???k?x??0
?2?2
? We find that
Q.E.D. 0?0 _______________________________________ Q.E.D. For the differential equation in u2?x? and the
proposed solution, the procedure
is exactly
the same as above. 3.3
_______________________________________ We have
3.4 d2u1?x?du1?x?22?2jk?k??u1?x??0 We have the solutions 2dxdx u1?x??Aexp?j???k?x?
Assume the solution is of the form:
u1?x??Aexp?j???k?x?
?Bexp??j???k?x?
for 0?x?a and ?Bexp??j???k?x?
u2?x??Cexp?j???k?x?
The first derivative is
du1?x? ?j???k?Aexp?j???k?x? ?Dexp??j???k?x?
dx for ?b?x?0.
The first boundary condition is ?j???k?Bexp??j???k?x?
u1?0??u2?0?
and the second derivative becomes
??????????????----------------------------------------------------
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which yields
A?B?C?D?0
The second boundary condition is
dudu 1 ?2dxx?0dxx?0 which yields
???k?A????k?B????k?C ????k?D?0
The third boundary condition is u1?a??u2??b?
which yields
Aexp?j???k?a??Bexp??j???k?a? ?Cexp?j???k???b??
?Dexp??j???k???b??
and can be written as
Aexp?j???k?a??Bexp??j???k?a? ?Cexp??j???k?b? ?Dexp?j???k?b??0
The fourth boundary condition is
dudu?2 1 dxx?adxx??b which yields
j???k?Aexp?j???k?a?
?j???k?Bexp??j???k?a? ?j???k?Cexp?j???k???b??
?j???k?Dexp??j???k???b??
and can be written as ???k?Aexp?j???k?a?
????k?Bexp??j???k?a? ????k?Cexp??j???k?b? ????k?Dexp?j???k?b??0
_______________________________________ 3.5
(b) (i) First point: ?a??
Second point: By trial and error,
?a?1.729?
(ii) First point: ?a?2?
Second point: By trial and error,
?a?2.617?
_______________________________________ 3.6
(b) (i) First point: ?a??
Second point: By trial and error,
?a?1.515?
(ii) First point: ?a?2? Second point: By trial and error,
?a?2.375?
_______________________________________ 3.7
sin?a P??cos?a?coska
?a Let ka?y, ?a?x Then
sinx?cosx?cosy xd Consider of this function.
dy P?
d?1P???x?sinx?cosx??siny dy???? We find
?dxdx??2?1P????1??x?sinx???x?cosx??
dydy??
dx?sinx??siny
dy Then
dx???1cosx??sin?P??2sinx?dy??xx???x???siny ?----------------------------------------------------
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For y?ka?n?,
n?0,1,2,...?siny?0
So that, in general,
d??a?d?dx ?0??dyd?ka?dk And
?? So
d?1?2mE??? ?dk2??2??1/2 E3?29.11?10??2??2?1.054?10?34?2?31??4.2?10??102
?1.3646?10?18J From Problem 3.5, ?4a?2.617?
2moE4?2?a?2.617?
2mE ?2?2m?dE ?2????dk E4??2.617??2?1.054?10?34?229.11?10??31??4.2?10??102
This implies that
d?dEn? for k? ?0?dkdka_______________________________________ 3.8
(a) ?1a?? E1? ?2.3364?10?18J ?E?E4?E3
?2.3364?10?18?1.3646?10?18
?9.718?10?19J
9.718?10?19 or ?E??6.07eV
1.6?10?19_______________________________________ 3.9
(a) At ka??, ?1a??
2moE1?2?a??
2moE1?2?a??
??222moa2?1.054?10??2?9.11?10??4.2?10????2?342?31?102
?3.4114?10 From Problem 3.5 ?2a?1.729?
2moE2?2?19J
E1?29.11?10?314.2?10?10????2?1.054?10?34?2???2
?a?1.729?
E2??1.729??2?1.054?10?34?229.11?10??31??4.2?10??18?102
?3.4114?10?19J
At ka?0, By trial and error,
?oa?0.859? Eo??0.859??2?1.054?10?34?229.11?10?314.2?10?10 ?1.0198?10 ?E?E2?E1
J
????2
?1.0198?10?18?3.4114?10?19 ?6.7868?10?19J
6.7868?10?19or ?E??4.24eV
1.6?10?19(b) ?3a?2?
2moE3?2?a?2?
?2.5172?10?19J ?E?E1?Eo
?3.4114?10?19?2.5172?10?19 ?8.942?10?20J
8.942?10?20 or ?E??0.559eV ?191.6?10(b) At ka?2?, ?3a?2?
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2moE3?2?a?2?
E3?29.11?10??2??2?1.054?10?34?2?31??4.2?10??102
?4.4186?10?19J
4.4186?10?19 or ?E??2.76eV
1.6?10?19(b) ?3a?2?
2moE3?2?a?2?
?1.3646?10?18J
At ka??. From Problem 3.5, ?2a?1.729?
2moE2?2?a?1.729?
E3?29.11?10??2??2?1.054?10?34?2?31??4.2?10??102
E2??1.729??2?1.054?10?34?229.11?10 ?1.3646?10?18J
From Problem 3.6, ?4a?2.375?
2moE4?2??31??4.2?10??102?a?2.375?
?1.0198?10?18J ?E?E3?E2
?1.3646?10?18?1.0198?10?18 ?3.4474?10?19J
3.4474?10?19 or ?E??2.15eV
1.6?10?19_______________________________________
3.10
(a) ?1a??
2moE1?2 E4??2.375??2?1.054?10?34?229.11?10?314.2?10?10????2
?1.9242?10?18J ?E?E4?E3
?a??
E1?29.11?10?314.2?10????2?1.054?10?34?2?1.9242?10?18?1.3646?10?18 ?5.597?10?19J
5.597?10?19 or ?E??3.50eV
1.6?10?19_____________________________________
3.11
(a) At ka??, ?1a??
2moE1?2???102?
?a??
?3.4114?10?19J
From Problem 3.6, ?2a?1.515?
2moE2?2?a?1.515?
E1?29.11?10????2?1.054?10?34?2?31??4.2?10??102
E2??1.515??2?1.054?10?34?229.11?10??31??4.2?10??102
?3.4114?10?19J
At ka?0, By trial and error, ?oa?0.727?
2moEo?2?a?0.727?
?7.830?10?19J ?E?E2?E1
?7.830?10?19?3.4114?10?19
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半导体物理与器件第四版课后习题答案
