半导体物理与器件第四版课后习题答案2

2.21

a/4(a) P????2?2?0?a??sin?2?x??a??dx ??4?x??a ???2??sin???x?a??/4??a????2??4??2???

?a????0?? ???2???asin??????a????8??

??8???a?????or P?0.25

a/2(b) P???2??sin2?2?x?a?/4?a???a??dx ??4 ???2??xsin??x????a/2?a????a????2??4??2???a??

???a/4

?????2???asin?2???a?sin??????a????4?????? ??8???a??8??8??????a????or P?0.25

?a/2(c) P???2??sin2??2?x???a?/2?a??a?dx ?sin??4?x???a/2 ???2???a??a??x?????2???2??

?4???a?????a/2

????2????asin?2????a?sin??2????a????4???8???4??8?? ????????a????a????or P?1

_______________________________________ 2.22

(a) (i) ??8?1012p?k?8?108?104m/s or ?p?106cm/s

??2?k?2?8?108?7.854?10?9m

or ??78.54Ao (ii)p?m???9.11?10?31??104?

?9.11?10?27kg-m/s

E?1m?2?12?9.11?10?31??104?22

?4.555?10?23J

E?4.555?10?23or 1.6?10?19?2.85?10?4eV (b) (i) ??1.5?10134p?k??1.5?109??10m/s or ?p??106cm/s

??2?k?2?1.5?109?4.19?10?9m or ??41.9Ao

(ii) p??9.11?10?27kg-m/s

E?2.85?10?4eV

_______________________________________ 2.23

(a) ??x,t??Ae?j?kx??t? (b) E??0.025??1.6?10?19??12m?2 ?12?9.11?10?31??2 so

??9.37?104m/s?9.37?106cm/s

For electron traveling in ?xdirection, ???9?.37?106cm/s

p?m??9.11?10?31???9.37?104?

??8.537?10?26kg-m/s

?h6.625p??10?34?8.537?10?26?7.76?10?9m

k?2???2?7.76?10?9?8.097?108m?1 ??k????8.097?108??9.37?104?

or ??7.586?1013rad/s

_______________________________________ 2.24

(a) p?m???9.11?10?31??5?104?

?4.555?10?26kg-m/s

??h?6.625?10?34?1.454?10?8p4.555?10?26m

k?2??2??4.32?108m?1?1.454?10?8 ??k???4.32?108??5?104?

?2.16?1013rad/s (b) p??9.11?10?31??106?

?9.11?10?25kg-m/s

??6.625?10?34 9.11?10?25?7.27?10?10m k?2??8.64?109m?17.272?10?10 ?

??8.64?109??106??8.64?1015rad/s

_______________________________________ 2.25

2222E?n???34?22n?n12ma2?.054?10?2?9.11?10?31??75?10?10?2

E2n?n?1.0698?10?21?J

or

En2 ?1.0698?10?21 ?n??1.6?10?19 or E2n?n6.686?10?3?eV Then

E1?6.69?10?3eV

E2?2.67?10?2eV

E3?6.02?10?2eV

_______________________________________ 2.26

?34(a) E??2n2?2n21.054?10?2?2n2ma2?29.11?10?3110?10?10 ?n2???6.018?10?20????2J

or

n2?6.018?10?20E?n?1.6?10?19?n2?0.3761?eV Then

E1?0.376eV E2?1.504eV

E3?3.385eV

(b) ??hc?E ?E??3.385?1.504??1.6?10?19?

?3.01?10?19J

???6.625?10?34??3?108 ?3.01?10?19

?6.604?10?7m or ??660.4nm

_______________________________________ 2.27

?2n2?2(a) En?2ma2

2 15?10?3?n21.054?10?34?22??15?10?31.2?10?2

15?10?3?n2????2.538?10?62??2 or n?7.688?1029 (b) En?1?15mJ (c) No

_______________________________________ 2.28

For a neutron and n?1:

E?2?21.054?10?34?2 1?2ma2?2??1.66?10?27???210?14?2

?3.3025?10?13J

or

E3.3025?10?1361?1.6?10?19?2.06?10eV For an electron in the same potential well:?3422 E??1.054?10??12?9.11?10?31??10?14?2

?6.0177?10?10J or

E6.0177?10?10 91?1.6?10?19?3.76?10eV _______________________________________ 2.29

Schrodinger's time-independent wave equation

?2??x?2m?x2??2?E?V?x????x??0

We know that

??x??0 for x?a?a2 and x?2

We have

V?x??0 for ?a2?x??a2

so in this region

?2??x?2mE?x2??2??x??0 The solution is of the form ??x??Acoskx?Bsinkx where

k?2mE?2 Boundary conditions: ??x??0 at x??a2,x??a2 First mode solution:

?1?x??A1cosk1x where

k??2?2 1?a?E1?2ma2

Second mode solution: ?2?x??B2sink2x where

k2?4?2?2 2?a?E2?2ma2 Third mode solution: ?3?x??A3cosk3x where

3?9?2?2 k3?a?E3?2ma2 Fourth mode solution: ?4?x??B4sink4x where

k4?16?2?2 4?a?E4?2ma2 _______________________________________

2.30

The 3-D time-independent wave equation in

cartesian coordinates for V?x,y,z??0 is:

?2??x,y,z??2??x,y,z??2??x,y,z??x2??y2??z2 ?2mE?2??x,y,z??0 Use separation of variables, so let? ??x,y,z??X?xY?y?Z?z?

Substituting into the wave equation, we obtain

?2X?2 YZ?x2?XZY?2Z?y2?XY?z2

?2mE?2XYZ?0 Dividing by XYZ and letting

k2?2mE?2, we

find

(1)

1X??2X1?2Y1?2Z?x2?Y??y2?Z??z2?k2?0 We may set

1?2 X?X2?2X2?x2??kx??x2?kxX?0 Solution is of the form

X?x??Asin?kxx??Bcos?kxx? Boundary conditions: X?0??0?B?0 and X?x?a??0?knx?x?a where nx?1,2,3.... Similarly, let

1??2Y21?2Z2Y?y2??ky and Z??z2??kz

Applying the boundary conditions, we find

kny?y?a, ny?1,2,3....

knz?z?a, nz?1,2,3... From Equation (1) above, we have

?k2x?k22y?kz?k2?0 or

k2222x?ky?kz?k?2mE?2 so that

E?E?2?2?222nxnynz?2ma2nx?ny?nz? _______________________________________ 2.31 (a)

?2??x,y??2?x2???x,y??y2?2mE?2???x,y??0 Solution is of the form:

??x,y??Asinkxx?sinkyy

We find

???x,y??x?Akxcoskxx?sinkyy ?2 ??x,y?2?x2??Akxsinkxx?sinkyy

???x,y??y?Akysinkxx?coskyy

?2??x,y??y2??Ak2ysinkxx?sinkyy

Substituting into the original equation, we find:

(1) ?k222mEx?ky??2?0

From the boundary conditions, Asinkxa?0, where a?40Ao So knx?x?a, nx?1,2,3,... Also Asinkoyb?0, where b?20A So kny?y?b, ny?1,2,3,... Substituting into Eq. (1) above

22 E?2?n?n2x?n2y???xny?2m? ?a2?b2??(b)Energy is quantized - similar to 1-D result. There can be more than one quantum state

per given energy - different than 1-D result.

_______________________________________ 2.32

(a) Derivation of energy levels exactly the

same as in the text

?2(b) ?E??2?n222ma22?n1? For n2?2,n1?1 Then

?E?3?2?2 2ma2 (i) For a?4Ao ?E?31.054?10?34?2?22??1.67?10?27??4?10?10?2

?6.155?10?22J

6.155?10?22 or ?E??31.6?10?19?3.85?10eV

(ii) For a?0.5cm

?E?3?1.054?10?34?2?22?1.67?10?27??0.5?10?2?2

?3.939?10?36J or

?3.939?10?36?E?171.6?10?19?2.46?10eV _______________________________________ 2.33

(a) For region II, x?0

?2?2?x??x2?2m?2?E?VO??2?x??0

General form of the solution is

?2?x??A2exp?jk2x??B2exp??jk2x? where

k2m2??2?E?VO?

Term with B2 represents incident wave and

term with A2 represents reflected wave. Region I, x?0

?2 ?1?x?2mE?x2??2?1?x??0 General form of the solution is

?1?x??A1exp?jk1x??B1exp??jk1x? where

k2mE1??2

Term involving B1 represents the

transmitted wave and the term involving A1represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that A1?0. Then

?1?x??B1exp??jk1x?

?2?x??A2exp?jk2x??B2exp??jk2x? (b)

Boundary conditions: (1) ?1?x?0???2?x?0?

(2) ??1??2?x? x?0?xx?0 Applying the boundary conditions to the solutions, we find

B1?A2?B2

k2A2?k2B2??k1B1

Combining these two equations, we find A?k2?k1?2????k2?k?1???B2

B?2k2?1????k?k???B2

21? The reflection coefficient is

R?A*22A2BB*???k?2?k1??k?k?? 2221? The transmission coefficient is

T?1?R?T?4k1k2?k?k2 12?_______________________________________

2.34

?2?x??A2exp??k2x? P???x?2A*?exp??2k2x?

2A2 where k2m?Vo?E?2??2

?2?9.11?10?31??3.5?2.8??1.6?10?19?1.054?10?34

k2?4.286?109m?1

(a) For x?5Ao?5?10?10m P?exp??2k2x? ? ?

?exp?24.2859?109??5?10?10?? ?0.0138

(b) For x?15Ao?15?10?10m ?

P?exp?2?4.2859?109??15?10?10??

?2.61?10?6 (c) For x?40Ao?40?10?10m

P?exp??2?4.2859?109??40?10?10??

?1.29?10?15

_______________________________________ 2.35

T?16???E?????V1?E??exp??2k2a? o????Vo?? where k2m?Vo?E?2??2

?31?29.11?10???1.0?0.1??1.6?10?19?1.054?10?34

or k2?4.860?109m?1

(a) For a?4?10?10m

T?16??0.1??1.0?????1?0.1?1.0??exp??2?4.85976?109??4?10?10?? ?0.0295

(b) For a?12?10?10m