Chapter 3
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 decrease, the
bandgap energy would increase and the
material would begin to behave more like an insulator.
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Schrodinger's wave equation is:
??2?2??x,t?2m?x2?V?x????x,t? ?j????x,t??t Assume the solution is of the form: ??x,t??u?x?exp????j?kx??E???????????t???? ?? Region I: V?x??0. Substituting the
assumed solution into the wave equation, we obtain:
??2??2m?x??jku?x?exp????j?kx???E??t????????????? ? ??u?x??xexp???j??kx?E???????????????t??????? ? ?j????jE???????u?x?exp???j?kx??E???????????t????? ? which becomes
??2?2??2m??jk?u?x?exp?j?????kx????E?????t??????? ? ?2jk?u?x??xexp????E????j???kx???????t????? ? ??2u?x??x2exp???j?????kx???E?????t???????????? ? ??Eu?x?exp????j?kx??E???????????t????? ? This equation may be written as
?k2u?x??2jk?u?x??2u?x?x???x2?2mE?2u?x??0
Setting u?x??u1?x? for region I,
the equation becomes:
d2u1?x?dx2?2jkdu1?x?dx??k2??2?u1?x??0 where
?2?2mE?2
In Region II, V?x??VO. Assume the
same
form of the solution:
??x,t??u?x?exp????j?kx??E???????????t????? ? Substituting into Schrodinger's wave
equation, we find:
??2???2m???jk?2u?x?exp??j???kx????E?????t?????? ?? ?2jk?u?x??xexp???j??kx???E??????????t????? ? ??2u?x????E?????x2exp??j????kx??????t?????? ???? ?V?exp???j?Ou?x???kx????E???????t????? ? ?Eu?x?exp???j????kx????E?????t??????? ? This equation can be written as:
?k2u?x??2jk?u?x??2u?x??x??x2 ?2mVO2mE?2u?x???2u?x??0
Setting u?x??u2?x? for region II,
this
equation becomes
d2u2?x?du2?x? ?2jkdxdx2 ?Bexp??j???k?x?
2mV?2?2O ??? for 0?x?a and ?k????u2?x??0 2??? u2?x??Cexp?j???k?x?
where again
?Dexp??j???k?x?
2mE ?2? for ?b?x?0.
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We have the solutions u1?x??Aexp?j???k?x?
?2We have
d2u1?x?dx2?2jkdu1?x?dx??k2??2?u1?x??0 Assume the solution is of the form: u1?x??Aexp?j???k?x? ?Bexp??j???k?x?
The first derivative is
du1?x?dx?j???k?Aexp?j???k?x?
?j???k?Bexp??j???k?x? and the second derivative becomes
d2u1?x?2dx2??j???k??Aexp?j???k?x? ??j???k??2Bexp??j???k?x? Substituting these equations into the
differential equation, we find
????k?2Aexp?j???k?x? ????k?2Bexp??j???k?x?
?2jk?j???k?Aexp?j???k?x?
?j???k?Bexp??j???k?x?? ??k2??2??Aexp?j???k?x? ?Bexp??j???k?x???0 Combining terms, we obtain???
??2?2?k?k2?2k???k??k2??2 ??Aexp??j???k?x???? ???2?2?k?k2??2k???k???k2??2??
?Bexp??j???k?x??0
We find that
0?0
For the differential equation in u2?x?
and the
proposed solution, the procedure
is exactly
the same as above.
The first boundary condition is
u1?0??u2?0?
which yields A?B?C?D?0
The second boundary condition is du1dx?du2dx x?0x?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
du1dx?du2 x?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 _______________________________________
(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?
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(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?
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P?sin?a?a?cos?a?coska
Let ka?y, ?a?x
Then
P?sinxx?cosx?cosy
Consider ddy of this function.
ddy??P???x??1sinx??cosx???siny
We find
P??????1??x??2sinx?dxdy??x??1cosx?dx?dy??
?sinxdxdy??siny
Then
dx??P????1cosx?2sinx??sinx?dy??xx??????siny For y?ka?n?,
n?0,1,2,...?siny?0
So that, in general,
dxd??a?d?dy?0?d?ka??dk And
??2mE?2 So
?1/2 d?1??dk?2?2mE?????2m?dE2???2??dk This implies that
d?dEn?dk?0?dk for k?a _______________________________________
(a) ?1a??
2moE1?2?a?? E?2???210?341??22m2?oa2??1.054??29.11?10?31??4.2?10?10?2 ?3.4114?10?19J From Problem ?2a?1.729?
2moE2?2?a?1.729?
E.729??2??1.054?10?34?22??129.11?10?31??4.2?10?10?2
?1.0198?10?18J ?E?E2?E1
?1.0198?10?18?3.4114?10?19 ?6.7868?10?19J
6.7868?10?19or ?E?1.6?10?19?4.24eV (b) ?3a?2?
2moE3?2?a?2?
半导体物理与器件第四版课后习题答案
