Semiconductor physics and devices basic principles 4th edition neamen solutions manual PDF

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Semiconductor Physics And Devices Basic Principles 4th Edition Neamen Solutions Manual Full Download: http://alibabadownload.com/product/semiconductor-physics-and-devices-basic-principles-4th-edition-neamen-s Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 2 2.1 Sketch _______________________________________

2.6 (a) p 

h





6.625  10 34

550  10 9  27  1.205 10 kg-m/s p 1.2045 10 27  1.32 10 3 m/s   m 9.11 1031

2.2 Sketch _______________________________________

or   1.32 10 cm/s h 6.625 10 34 (b) p    440  10 9 27  1.506  10 kg-m/s p 1.5057 10 27  1.65  10 3 m/s   m 9.11 10 31 5

2.3 Sketch _______________________________________ 2.4 From Problem 2.2, phase 

2 x



 t

or   1.65 10 cm/s (c) Yes _______________________________________ 5

= constant Then 2 dx    dx     0,    p      dt dt  2  2 x  t From Problem 2.3, phase 

2.7 (a) (i)

= constant

o

or   11.2 A



hc E  h     E









 1.87  10 kg-m/s 6.625 1034  3.54 10 10 m   1.8704 10 24





o





or   3.54 A



l



d ll h





t

19



lib b d



 5.915  10  24 kg-m/s 34 6.625  10  1.12  10 10 m   24 5.915  10 o



t



(iii) p  2 9.1110 31 120 1.6 10 19

Cesium: E  1.90 eV  1.90  1.6  10  J So,  6.625 10 34 3 10 10  6.54 10  5 cm  1.90 1.6  10 19 or   0.654  m _______________________________________





24

Gold: E  4.90 eV  4.90  1.6  10 19 J So, 6.625 10 34 3 10 10  2.54 10 5 cm  4.90  1.6  10 19 or   0. 254  m

D

 

(ii) p  2 9.1110 31 12 1.6 10 19

hc

l



 5.915 10 kg-m/s h 6.625  10  34  1.12 10  9 m   p 5.915  10  25

2.5

l



25

Then 2 dx    dx     0,    p      dt dt  2  _______________________________________

Thi



p  2 mE  2 9.11 10 31 1.2  1.6 10 19



l

or   1.12 A

d





Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)



p  2 1.67 10

2.10

1.2 1.6 10 

27

19

(a) p 

 2.532  10 23 kg-m/s

6.625  10  34  2.62 10 11 m 23 2.532  10 



o

or   0.262 A _______________________________________

or

3  3 kT    0.0259  0.03885 eV 2  2

or

Now pavg  2mEavg

(b)







 2 9.11 10 31 0.03885 1.6 10 19



or

or

p avg  1.064  10  25 kg-m/s

6.625 1034

85 10 10  7.794 10 26 kg-m/s p 7.794 10  26  8.56 10 4 m/s   31 m 9.11 10   8.56 10 6 cm/s 1 1 2 E  m  2  9.11 10  31 8.56  10 4 2 2  3.33  10 21 J 3.334  10 21  2.08 10  2 eV E 1.6 1019 1 2 E  9.11 10 31 8 10 3 2  23  2.915 10 J 2.915  10 23 E  1.82 10 4 eV 1.6 10 19 p  m  9.11 10 31 8 10 3



6.625  1034

h  6.225 10 9 m  p 1.064 10 25







Now









2.8

E avg 

h





 7.288 10





 27

kg-m/s h 6.625 10  35  9.09  10  8 m   p 7.288 10  27

or o

  62 .25 A _______________________________________

o

or   909 A _______________________________________

2.9

E p  h p 

2.11

hc

p

(a) E  h 

Now

Ee 

p e2 2m

and p e 

h

e

 Ee 

1  h   2 m   e 

2

2

1  h 1  10h       p 2m   e  2m   p  which yields 100h p  2mc hc hc 2mc 2 E p E    2 mc   p 100h 100 hc







2

2 9.11 10 31 3 108 100  1.64  10 15 J  10.25 keV _______________________________________ 



 1.99  10

2

Set E p  E e and  p  10  e Then

hc

6.625 10 3 10  34



8

10 1 10

15

J

Now

E 1.99  10 15  e 1.6 10  19 V  1.24 10 4 V  12.4 kV E  e V  V 





31 1.99 10 15 (b) p  2 mE  2 9.11 10



 23

 6.02  10 kg-m/s Then  h 6.625 10 34  1.10  10 11 m    p 6.02  10 23 or o

  0.11 A _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.12

1.054 10 34  p   x 10  6  1.054  10 28 kg-m/s _______________________________________ 2.13 (a) (i) p x  

p 

1.054  10 34  8.783 10  26 kg-m/s 12  10 10

d  p2     p (ii)  E   p  dp  2 m  dp p p 2p   p  2m m dE

Now p  2mE



 

 2 9 10 31 16 1.6 10 19



24

 2.147  10 kg-m/s 24 26 2.1466 10  8.783 10  so E  31 9 10  19  2.095 10 J 2.095  10 19 or E   1.31 eV 19 1.6  10  (b) (i) p  8.783 10 26 kg-m/s







 

(ii) p  2 5 10 28 16 1.6 10 19





 5.06 10  23 kg-m/s 23 26 5.06 10  8.783 10  E  28 5  10   21  8.888  10 J 8.888  10  21 or E   5.55  10 2 eV 1.6  10 19 _______________________________________







2.14

p 

 1.054  10 34   1.054  10 32 kg-m/s x 10  2

p  m    

p 1.054  10 32  m 1500

   7  10 36 m/s _______________________________________

2.15 (a) Et    1.054  10 34 t   8.23 10 16 s 0.8 1.6 10  19



(b) p 



34

1.054 10   x 1.5 1010

 7.03  10  25 kg-m/s _______________________________________

2.16 (a) If 1  x, t  and 2  x, t are solutions to Schrodinger's wave equation, then

 x, t    2  2 1 x , t   V  x 1  x, t  j 1  2m t x 2 and  2  x, t   2  2 2 x ,t   V x 2 x, t   j  2 t 2m x Adding the two equations, we obtain  2  2   1  x, t  2  x, t 2 m x 2  V x 1 x , t   2 x , t   1  x, t   2 x, t  t which is Schrodinger's wave equation. So 1  x , t    2 x , t  is also a solution.  j

(b) If 1 x, t  2 x , t  were a solution to Schrodinger's wave equation, then we could write 2   2  1  2   V x 1  2  2m x 2   j  1  2  t which can be written as   2   2 1  2   22 2 1   1  2 2 2 x x  2m  x x  2 1   V x  1   2  j   1  2 t t   Dividing by 1  2 , we find

 2 2m

 1  2 2 1  2 1 2  1  2        2 2 1 2  x x  1 x   2 x  1 2 1  1   V x   j   1  t   2 t

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Since 1 is a solution, then

2.19

 1  1 1 1  V x   j      2 1 t 2 m 1 x Subtracting these last two equations, we have 2   2  1  2 2 1  2      2 1 2 x x  2 m   2 x 1 2  j   2  t 2



2

0

 1 1  V x   j     t 2 2 m 2 x 2 Subtracting these last two equations, we obtain  2  2 2  V  x  0   1 2m  1 2 x x This equation is not necessarily valid, which means that 1 2 is, in general, not a solution to Schrodinger's wave equation. _______________________________________

2

ao

P

 2   x   dx exp    a o  ao 

4

 0

ao

2  ao

2

 2 2

*

Function has been normalized. (a) Now

Since 2 is also a solution, we have 2

    dx  1

Note that

2 ao



1

2

0

  2x  ao 4   ao    exp     2   ao  0

   2 ao P  1 exp    4a o which yields P  0.393 (b) ao

A

 x  cos 2  dx  1  2 



P

ao

 x sin x  1 A 2  2  1 2

2.18

 2   x   dx exp    a o   a o

4 ao

2  ao



or A  2 _______________________________________

ao

4

2 4

  1   P  1exp 1  exp    2   which yields P  0.239 (c) ao

P

 0

2

1  1   1 A2        1  A2   2 4  4 

  2x   dx o 

 exp a

or

1 / 2

 x sin 2 nx 1 / 2 1 A   4n   1 / 2 2

2

  2x  a o 2   ao    exp    ao  2   ao  ao

1 / 2

A 2 cos2 n x dx  1

   1  1   1  exp    2   

2 2

3

 3   1  A 2      1  2  2  1 so A 2  2 1 or A  2 _______________________________________

  2x  dx o 

 exp a

or

2.17 3

4



2

 2   x   dx exp    ao   a o

2 ao

ao

  2 x dx  o 

exp  a 0

  2x  ao   ao    exp     2   ao  0   1exp  2   1 which yields P  0.865 _______________________________________ 2  ao

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.21

2.20

a/ 4

2

P   x  dx



a/ 4

(a)

 0

(a) P 

a        sin    24   2       4    a 2      a     2   a 1 a        a   8 4  or P  0.409 a/ 2

(b) P 

2

  a  cos

   sin     2 sin    a a  2           a   4  4  8  4          a    a   1 1 1  2  0    8 4  4 or P  0.0908 a / 2

(c) P 

 2

  a  cos

a / 2

      a 2 sin          a   8  8       a   or P  0.25 a/ 2

(b) P 

2

 x   dx  a 

  2 x   a / 2 sin    a   2x     4    a2    a / 2   a   

             a a 2 sin sin            a   4  4   4   4          a    a   or P  1 _______________________________________

 2

 2x  dx a 

  a  sin 

a/ 4

2

  4 x   a / 2 sin    a   2 x      2    a2 4  a / 4   a   

 x    dx  a

  2x   a / 2 sin    a   2x        a2 4   a/ 4   a  

 2x  dx a 

  4 x   a / 4 sin    2x  a      2    a2 4   0   a   

2

a/ 4

2

0

2  2  x  dx   cos  a   2

  2x   a / 4 sin    2x  a         a2 4   0   a  

 2

  a  sin 

   2a     a4   or P  0.25 a / 2

(c) P 

 2

 sin 2   a  sin      8   8  8        a   a 

 2 x   dx a 

  a  sin 

a / 2

2

  4x   a / 2 sin    2 x  a       2    a2 4   a/ 2   a         2  a sin 2    a  sin  2           8    a  4  8   4         a   a  or P  1 _______________________________________ 2.22

or



8  10 4  10 m/s k 8  108  p  10 6 cm/s

(a) (i)  p 





12

2 2   7.854 10 9 m 8 k 8  10

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

2 1   4.32  108 m  1.454 10 8   k  4.32 10 8 5 10 4 

o

  78.54 A

or



 

(ii) p  m  9.11  10 31 10 4 27

 9.11 10 kg-m/s 2 1 1 E  m 2  9.11 10  31 10 4 2 2  4.555 10  23 J 23 4.555  10  2.85 104 eV or E  19 1.6  10  1.5  1013 4 (b) (i)  p    10 m/s k  1.5  109 or  p  10 6 cm/s





 

2 2   4.19 10  9 m 9 k 1.5  10

 2.16 1013 rad/s



25

 9.11  10 kg-m/s 34 6.625  10  7.27  10 10 m  9.11 10 25 2 1 k  8.64 10 9 m 10  7.272 10   8.64 10 9 10 6  8.64 10 15 rad/s _______________________________________



En 

p   9.11  10 27 kg-m/s

E  2.85  10 4 eV _______________________________________

19



1

2

m



2





For electron traveling in x direction,

  9.37 10 6 cm/s p  m  9.11 10 31  9.37 10 4    8.537  10 26 kg-m/s  h 6.625 10 34  7.76  10 9 m   p 8.537 10  26

2 8 1  8.097 10 m  7.76 10 9   k   8.097 10 8 9.37 10 4 k

2









or   7.586 10 rad/s _______________________________________ 13

2.24 (a) p  m  9.11 10 31 5 10 4



 4.555 10





 26





kg-m/s

h 6.625 10 34  1.454  10 8 m  p 4.555 10 26



2









2

E n  n 2 1.0698  10 21 J

or





n 2 1.0698 1021 19 1.6  10 or E n  n 2 6.686 10 3 eV Then E1  6.69 10  3 eV



1

9.11 10 31  2 2 6 4 so   9.37 10 m/s  9.37 10 cm/s 



n 2 1.054  10 34  2  2 n 2 2  2ma 2 2 9.1110  31 75  10 10

En 

2.23 j kx  t (a) x ,t   Ae    



 

2.25

or   41.9 A

(b) E  0.025 1.6 10

 

(b) p  9.11 10 31 10 6

o

(ii)

2

k



E 2  2.67  10 2 eV E 3  6.02  10 2 eV _______________________________________

2.26 (a) E n 











2

 6.018 10   n 0.3761 eV

 n 2 6.018  10 20 J

En 

or



n2 1.054  10 34  2  2 n 2 2  2 31 10 2ma 2 9.1110 10  10 n

2

20

2

1.6 10 

19

Then E1  0.376 eV E 2  1.504 eV E 3  3.385 eV

hc E E  3.385 1.504  1.6 10 19

(b)  



 3.01  10

19

J





2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

6.625 10 310  34



8

19

3.01 10  6.604 10 7 m   660 .4 nm or _______________________________________ 2.27 (a) E n 

 2 n 2 2 2ma 2

3 15 10 

15 10 3





2

 n 2 1.054 10 34 



2

1.2 10   n 2.538 10  2 15 10

2 2

3

62

2

or n  7.688 10 29 (b) E n 1  15 mJ (c) No _______________________________________ 2.28 For a neutron and n  1 :

where

 



34 2

1.054  10     E1  2 ma 2 2 1.66 10  27 10 14 2

2



2



2

 3.3025  10 13 J

or

E1 

3.3025 10 13

 2.06 10 6 eV 1.6 10  19 For an electron in the same potential well:

1.054 10   2 9.11 10 10   34 2

E1 

31

2

14 2

 6.0177  10 10 J

or

E1 

so in this region  2  x 2mE  2   x  0  x 2 The solution is of the form   x  A cos kx  B sin kx where 2 mE k 2 Boundary conditions: a a  x   0 at x  ,x 2 2 First mode solution:  1 x   A1 cos k 1 x where   2 2 k1   E1  a 2ma 2 Second mode solution:  2  x  B 2 sin k 2 x

6.0177 1010

3.76 10 9 eV 1.6 10 19 _______________________________________ 2.29 Schrodinger's time-independent wave equation  2  x 2m  2  E  V x  x  0 x 2  We know that a a  x   0 for x  and x  2 2 We have a a V x   0 for  x 2 2

2 4 2  2  E2  a 2 ma 2 Third mode solution:  3  x  A3 cos k 3 x where 2 2 3 9...