solution manual semiconductor physics and devices 4th edition neamen

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<p>Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
Chapter 2 
 
 
2.1 
     Sketch 
_______________________________________ 
 
2.2 
     Sketch 
_______________________________________ 
 
2.3 
     Sketch 
_______________________________________ 
 
2.4   
     From Problem 2.2, phase 
t
x





2
 
                                                           = constant 
     Then 
        




















2
,0
2
p
dt
dx
dt
dx
 
     From Problem 2.3, phase 
t
x





2
 
                                                         = constant 
     Then 
        




















2
,0
2
p
dt
dx
dt
dx
 
_______________________________________ 
 
2.5 
        
E
hc
hc
h
E







 
     Gold:  
90
.4

E
eV



19
10
6
.1
90
.4



 J 
     So, 
        






5
19
10
34
10
54
.2
10
6
.1
90
.4
10
3
10
625
.6










cm 
     or 
        


254
.0

m 
     Cesium:  
90
.1

E
eV



19
10
6
.1
90
.1



 J 
     So, 
        






5
19
10
34
10
54
.6
10
6
.1
90
.1
10
3
10
625
.6










cm 
     or 
        


654
.0

m 
_______________________________________ 
 
 
 
2.6  
     (a)  
9
34
10
550
10
625
.6







h
p
 
                     
27
10
205
.1



kg-m/s 
            
3
31
27
10
32
.1
10
11
.9
10
2045
.1








m
p

m/s 
       or  
5
10
32
.1 


cm/s 
     (b)   
9
34
10
440
10
625
.6







h
p
 
                
27
10
506
.1



kg-m/s 
             
3
31
27
10
65
.1
10
11
.9
10
5057
.1








m
p

m/s 
        or  
5
10
65
.1 


cm/s 
     (c) Yes 
_______________________________________ 
 
2.7 
(a) (i)  
 




19
31
10
6
.1
2
.1
10
11
.9
2
2





 mE
p
 
                   
25
10
915
.5



kg-m/s 
       
9
25
34
10
12
.1
10
915
.5
10
625
.6









p
h

m 
  or  
o
A
2
.
11


 
(ii)




19
31
10
6
.1
12
10
11
.9
2





p
 
                 
24
10
87
.1



kg-m/s 
      
10
24
34
10
54
.3
10
8704
.1
10
625
.6









m 
 or  
o
A
54
.3


 
(iii) 




19
31
10
6
.1
120
10
11
.9
2





p
 
                  
24
10
915
.5



kg-m/s 
      
10
24
34
10
12
.1
10
915
.5
10
625
.6









m 
 or  
o
A
12
.1


 
 
 
 
 
 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
(b) 
       




19
27
10
6
.1
2
.1
10
67
.1
2





p
 
                   
23
10
532
.2



kg-m/s 
       
11
23
34
10
62
.2
10
532
.2
10
625
.6









m 
  or  
o
A
262
.0


 
_______________________________________ 
 
2.8 
        


03885
.0
0259
.0
2
3
2
3









kT
Eavg
eV 
     Now 
        
avg
avg
mE
p
2

 
                




19
31
10
6
.1
03885
.0
10
11
.9
2





 
     or 
        
25
10
064
.1



avg
p
kg-m/s 
     Now 
        
9
25
34
10
225
.6
10
064
.1
10
625
.6









p
h

m 
     or 
        
o
A
25
.
62


 
_______________________________________ 
 
2.9 
        
p
p
p
hc
h
E

 

 
     Now 
        
m
p
E
e
e
2
2

 and  
2
2
1











e
e
e
e
h
m
E
h
p


 
     Set 
e
p E
E 
 and  
e
p


10

 
     Then 
        
2
2
10
2
1
2
1


















p
e
p
h
m
h
m
hc



 
     which yields 
        
mc
h
p
2
100


 
        
100
2
2
100
2
mc
mc
h
hc
hc
E
E
p
p






 
               



100
10
3
10
11
.9
2
2
8
31 



 
               
15
10
64
.1



J 
25
.
10

keV 
_______________________________________ 
 
 
2.10 
(a) 
10
34
10
85
10
625
.6







h
p
 
           
26
10
794
.7



 kg-m/s 
        
4
31
26
10
56
.8
10
11
.9
10
794
.7








m
p

m/s 
or     
6
10
56
.8 


cm/s 
       


2
4
31
2
10
56
.
8
10
11
.
9
2
1
2
1






m
E
 
           
21
10
33
.3



J 
or    
2
19
21
10
08
.2
10
6
.1
10
334
.3








E
eV 
(b) 


2
3
31
10
8
10
11
.
9
2
1




E
 
           
23
10
915
.2



J 
or    
4
19
23
10
82
.1
10
6
.1
10
915
.2








E
eV 
       



3
31
10
8
10
11
.9






m
p
 
           
27
10
288
.7



kg-m/s 
        
8
27
35
10
09
.9
10
288
.7
10
625
.6









p
h

m 
or     
o
A
909


 
_______________________________________ 
 
2.11 
(a) 



10
8
34
10
1
10
3
10
625
.6










hc
h
E
 
            
15
10
99
.1



J 
   Now 
        
19
15
10
6
.1
10
99
.1









e
E
V
V
e
E
 
        
4
10
24
.1 

V
V
4
.
12

kV 
(b) 



15
31
10
99
.1
10
11
.9
2
2





 mE
p
 
         
23
10
02
.6



kg-m/s 
   Then 
        
11
23
34
10
10
.1
10
02
.6
10
625
.6









p
h

m 
   or 
        
o
A
11
.
0


 
_______________________________________ 
 
 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
2.12 
    
6
34
10
10
054
.1







x
p

 
          
28
10
054
.1



kg-m/s 
_______________________________________ 
 
2.13 
(a) (i) 



 x
p
 
       
26
10
34
10
783
.8
10
12
10
054
.1








p
kg-m/s 
(ii)
p
m
p
dp
d
p
dp
dE
E















2
2
 
            
m
p
p
p
m
p





2
2
 
Now  
mE
p
2

 
         




19
31
10
6
.1
16
10
9
2





 
         
24
10
147
.2



kg-m/s 
so 



31
26
24
10
9
10
783
.8
10
1466
.2







E
 
           
19
10
095
.2



J 
or 
31
.1
10
6
.1
10
095
.2
19
19







E
eV 
(b) (i) 
26
10
783
.8



p
kg-m/s 
      (ii)




19
28
10
6
.1
16
10
5
2





p
 
              
23
10
06
.5



kg-m/s 
        



28
26
23
10
5
10
783
.8
10
06
.5







E
 
             
21
10
888
.8



J 
  or 
2
19
21
10
55
.5
10
6
.1
10
888
.8








E
eV 
_______________________________________ 
 
2.14 
    
32
2
34
10
054
.1
10
10
054
.1










x
p

kg-m/s 
      
1500
10
054
.1
32








m
p
m
p


 
      
36
10
7




m/s 
_______________________________________ 
 
 
 
 
 
 
 
2.15 
(a) 




t
E
 
        



16
19
34
10
23
.8
10
6
.1
8
.
0
10
054
.1








t
s 
(b) 
10
34
10
5
.1
10
054
.1








x
p

 
             
25
10
03
.7



kg-m/s 
_______________________________________ 
 
2.16 
(a) If 


t
x,
1
 and 


t
x,
2
 are solutions to  
       Schrodinger's wave equation, then 
 
   


 




t
t
x
j
t
x
x
V
x
t
x
m











,
,
,
2
1
1
2
1
2
2


 
   and 
   


 




t
t
x
j
t
x
x
V
x
t
x
m











,
,
,
2
2
2
2
2
2
2


 
   Adding the two equations, we obtain 
    






t
x
t
x
x
m
,
,
2
2
1
2
2
2







 
                    
 






t
x
t
x
x
V
,
,
2
1




 
                                 






t
x
t
x
t
j
,
,
2
1





 
 
   which is Schrodinger's wave equation.  So 
   




t
x
t
x
,
,
2
1



 is also a solution. 
 
(b) If  




t
x
t
x
,
,
2
1



 were a solution to 
Schrodinger's wave equation, then we 
could write 
   


 

2
1
2
1
2
2
2
2











x
V
x
m

 
                                           


2
1 





t
j
 
   which can be written as 
    
























x
x
x
x
m
2
1
2
1
2
2
2
2
2
1
2
2
2

 
            
 





















t
t
j
x
V
1
2
2
1
2
1

 
   Dividing by 
2
1 


, we find 



























x
x
x
x
m
2
1
2
1
2
1
2
1
2
2
2
2
2
2
1
1
2

            
 

















t
t
j
x
V
1
1
2
2
1
1

 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
   Since 
1  is a solution, then 
    
 
t
j
x
V
x
m















1
1
2
1
2
1
2
1
1
2


 
   Subtracting these last two equations, we have 
     





















x
x
x
m
2
1
2
1
2
2
2
2
2
2
1
2

 
                                                   
t
j







2
2
1

 
     Since 
2  is also a solution, we have 
   
 
t
j
x
V
x
m















2
2
2
2
2
2
2
1
1
2


 
   Subtracting these last two equations, we obtain 
     
  0
2
2
2
1
2
1
2














x
V
x
x
m

 
   This equation is not necessarily valid, which  
   means that 
2
1

 is, in general, not a solution  
   to Schrodinger's wave equation. 
_______________________________________ 
 
2.17 
          
1
2
cos2
3
1
2










dx
x
A

 
         


1
2
sin
2
3
1
2











x
x
A
 
         
1
2
1
2
3
2












 

A
 
     so  
2
1
2 
A
 
     or  
2
1

A
 
_______________________________________ 
 
2.18 
          


1
cos2
2
/
1
2
/
1
2




dx
x
n
A

 
         


1
4
2
sin
2
2
/
1
2
/
1
2












n
x
n
x
A
 
         






















2
1
1
4
1
4
1
2
2
A
A
 
      or  
2

A
 
_______________________________________ 
 
 
 
 
2.19 
     Note that 
1
0
*






dx
 
     Function has been normalized. 
(a) Now 
  
dx
a
x
a
P
o
a
o
o
2
4
0
exp
2
















 

 
         
dx
a
x
a
o
a
o
o








 

4
0
2
exp
2
 
         
4
0
2
exp
2
2
o
a
o
o
o
a
x
a
a







 





 

 
  or 
     







 


















 


2
1
exp
1
1
4
2
exp
1
o
o
a
a
P
 
  which yields 
     
393
.0

P
 
(b) 
      
dx
a
x
a
P
o
o
a
a
o
o
2
2
4
exp
2
















 

 
           
dx
a
x
a
o
o
a
a
o
o








 

2
4
2
exp
2
 
            
2
4
2
exp
2
2
o
o
a
a
o
o
o
a
x
a
a







 





 

 
   or 
        















 




2
1
exp
1
exp
1
P
 
  which yields 
        
239
.0

P
 
(c) 
        
dx
a
x
a
P
o
a
o
o
2
0
exp
2
















 

 
            
dx
a
x
a
o
a
o
o








 

0
2
exp
2
 
            
o
a
o
o
o
a
x
a
a
0
2
exp
2
2







 





 

 
            





1
2
exp
1




 
   which yields 
        
865
.0

P
 
_______________________________________ 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
2.20 
          
  dx
x
P
2

 
 
(a) 
dx
x
a
a













2
cos
2
2
4
/
0

 
       
4
/
0
4
2
sin
2
2
a
a
a
x
x
a




















���













 
        






































a
a
a


4
2
sin
2
4
2
 
         
  















4
1
8
2
a
a
a
 
     or  
409
.0

P
 
(b) 
dx
a
x
a
P
a
a












 

2
/
4
/
2
cos
2
 
           
2
/
4
/
4
2
sin
2
2
a
a
a
a
x
x
a


































 
           
 








































a
a
a
a
a




4
2
sin
8
4
sin
4
2
 
            











4
1
8
1
0
4
1
2
 
   or  
0908
.0

P
 
(c) 
dx
a
x
a
P
a
a












 



2
2
/
2
/
cos
2
 
           
2
/
2
/
4
2
sin
2
2
a
a
a
a
x
x
a




































 
          
 



























 















a
a
a
a
a




4
sin
4
4
sin
4
2
 
or  
1

P
 
_______________________________________ 
 
 
2.21 
(a) 
dx
a
x
a
P
a












 

2
sin
2
2
4
/
0
 
           
4
/
0
2
4
4
sin
2
2
a
a
a
x
x
a


































 
           
 


























a
a
a


8
sin
8
2
 
or  
25
.0

P
 
(b) 
dx
a
x
a
P
a
a












 

2
sin
2
2
2
/
4
/
 
           
2
/
4
/
2
4
4
sin
2
2
a
a
a
a
x
x
a


































 
          


 








































a
a
a
a
a




8
sin
8
8
2
sin
4
2
 
or  
25
.0

P
 
(c) 
dx
a
x
a
P
a
a












 



2
sin
2
2
2
/
2
/
 
           
2
/
2
/
2
4
4
sin
2
2
a
a
a
a
x
x
a




































 
       





























 















a
a
a
a
a




8
2
sin
4
8
2
sin
4
2
 
or  
1

P
 
_______________________________________ 
 
2.22 
(a) (i) 
4
8
12
10
10
8
10
8





k
p


m/s 
    or     
6
10

p

cm/s 
            
9
8
10
854
.7
10
8
2
2









k
m 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
    or     
o
A
54
.
78


 
(ii)



4
31 10
10
11
.9



 
m
p
 
                   
27
10
11
.9



kg-m/s 
       


2
4
31
2
10
10
11
.
9
2
1
2
1





m
E
 
           
23
10
555
.4



J 
or   
4
19
23
10
85
.2
10
6
.1
10
555
.4








E
eV 
(b) (i) 
4
9
13
10
10
5
.1
10
5
.1

���





k
p


m/s 
or  
6
10


p

cm/s 
       
9
9
10
19
.4
10
5
.1
2
2









k
m 
  or  
o
A
9
.
41


 
(ii)   
27
10
11
.9




p
kg-m/s 
        
4
10
85
.2



E
eV 
_______________________________________ 
 
2.23 
(a) 




t
kx
j
Ae
t
x





,
 
(b) 



2
19
2
1
10
6
.
1
025
.
0

m
E




 
                                     

 2
31
10
11
.
9
2
1




 
    so  
4
10
37
.9



m/s
6
10
37
.9 

cm/s 
For electron traveling in x
 direction, 
         
6
10
37
.9 



cm/s 
      



4
31
10
37
.9
10
11
.9







m
p
 
                   
26
10
537
.8




kg-m/s 
      
9
26
34
10
76
.7
10
537
.8
10
625
.6









p
h

m 
      
8
9
10
097
.8
10
76
.7
2
2









k
m 1
  
      



4
8
10
37
.9
10
097
.8







k
 
or  
13
10
586
.7



rad/s 
_______________________________________ 
 
2.24 
(a) 



4
31
10
5
10
11
.9






m
p
 
           
26
10
555
.4



kg-m/s 
       
8
26
34
10
454
.1
10
555
.4
10
625
.6









p
h

m 
 
       
8
8
10
32
.4
10
454
.1
2
2









k
m 1
  
      



4
8
10
5
10
32
.4



 

k
 
          
13
10
16
.2


rad/s 
(b) 



6
31 10
10
11
.9



p
 
           
25
10
11
.9



kg-m/s 
       
10
25
34
10
27
.7
10
11
.9
10
625
.6









m 
       
9
10
10
64
.8
10
272
.7
2






k
m 1
  
      



15
6
9
10
64
.8
10
10
64
.8





rad/s 
_______________________________________ 
 
2.25 
        




2
10
31
2
2
34
2
2
2
2
2
10
75
10
11
.
9
2
10
054
.
1
2










n
ma
n
En

 
        


21
2
10
0698
.1


 n
En
J 
     or 
          


19
21
2
10
6
.1
10
0698
.1





n
En
 
     or  


3
2
10
686
.6


 n
En
eV 
     Then 
           
3
1
10
69
.6



E
eV 
           
2
2
10
67
.2



E
eV 
           
2
3
10
02
.6



E
eV 
_______________________________________ 
 
2.26 
(a) 




2
10
31
2
2
34
2
2
2
2
2
10
10
10
11
.
9
2
10
054
.
1
2










n
ma
n
En

      


20
2
10
018
.6


 n
J 
or     




3761
.0
10
6
.1
10
018
.6
2
19
20
2
n
n
En






eV 
Then 
          
376
.0
1 
E
eV 
          
504
.1
2 
E
eV 
          
385
.3
3 
E
eV 
(b) 
E
hc



 
      



19
10
6
.1
504
.1
385
.3




E
 
            
19
10
01
.3



J 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
         



19
8
34
10
01
.3
10
3
10
625
.6







 
             
7
10
604
.6



m 
or      
4
.
660


nm 
_______________________________________ 
 
2.27 
(a) 
2
2
2
2
2ma
n
En



 
       




2
2
3
2
2
34
2
3
10
2
.
1
10
15
2
10
054
.
1
10
15










n
 
       


62
2
3
10
538
.2
10
15





n
 
    or  
29
10
688
.7


n
 
(b) 
15
1 

nE
mJ 
(c) No 
_______________________________________ 
 
2.28 
     For a neutron and 
1

n
: 
        




2
14
27
2
2
34
2
2
2
1
10
10
66
.1
2
10
054
.
1
2









ma
E

 
              
13
10
3025
.3



J 
     or 
        
6
19
13
1
10
06
.2
10
6
.1
10
3025
.3







E
eV 
     For an electron in the same potential well: 
        




2
14
31
2
2
34
1
10
10
11
.
9
2
10
054
.
1







E
 
              
10
10
0177
.6



J 
     or 
        
9
19
10
1
10
76
.3
10
6
.1
10
0177
.6







E
eV 
_______________________________________ 
 
2.29 
     Schrodinger's time-independent wave  
     equation 
        
 
 

   0
2
2
2
2





x
x
V
E
m
x
x



 
     We know that 
        
  0

x

 for 
2
a
x 
 and 
2
a
x


 
     We have 
        
  0

x
V
 for 
2
2
a
x
a




 
 
 
     so in this region 
        
 
  0
2
2
2
2




x
mE
x
x



 
     The solution is of the form 
        
 
kx
B
kx
A
x
sin
cos 


 
     where 
        
2
2

mE
k 
 
     Boundary conditions: 
        
  0

x

 at 
2
,
2
a
x
a
x




 
     First mode solution: 
        
 
x
k
A
x
1
1
1
cos


 
     where 
        
2
2
2
1
1
2ma
E
a
k






 
     Second mode solution: 
        
 
x
k
B
x
2
2
2
sin


 
     where 
        
2
2
2
2
2
2
4
2
ma
E
a
k






 
     Third mode solution: 
        
 
x
k
A
x
3
3
3
cos


 
     where 
        
2
2
2
3
3
2
9
3
ma
E
a
k






 
     Fourth mode solution: 
        
 
x
k
B
x
4
4
4
sin


 
     where 
        
2
2
2
4
4
2
16
4
ma
E
a
k






 
_______________________________________ 
 
2.30 
     The 3-D time-independent wave equation in  
     cartesian coordinates for 

 0
,
,

z
y
x
V
 is: 
        






2
2
2
2
2
2
,
,
,
,
,
,
z
z
y
x
y
z
y
x
x
z
y
x











 
                                           

 0
,
,
2
2


z
y
x
mE


 
     Use separation of variables, so let 
        


     
z
Z
y
Y
x
X
z
y
x

,
,

 
     Substituting into the wave equation, we  
     obtain 
 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
        
2
2
2
2
2
2
z
Z
XY
y
Y
XZ
x
X
YZ








 
                                                  
0
2
2


XYZ
mE

 
     Dividing by XYZ  and letting 
2
2
2

mE
k 
, we  
     find 
(1) 
0
1
1
1
2
2
2
2
2
2
2













k
z
Z
Z
y
Y
Y
x
X
X
 
     We may set 
        
0
1
2
2
2
2
2
2










X
k
x
X
k
x
X
X
x
x
 
     Solution is of the form 
        
 




x
k
B
x
k
A
x
X
x
x
cos
sin


 
     Boundary conditions: 
 
0
0
0



B
X
 
     and  


a
n
k
a
x
X
x
x





0
 
     where 
....
3
,
2
,1

xn
 
     Similarly, let 
        
2
2
2
1
y
k
y
Y
Y





 and  
2
2
2
1
zk
z
Z
Z





 
     Applying the boundary conditions, we find 
        
a
n
k
y
y


,  
....
3
,
2
,1

yn
 
        
a
n
k
z
z


,  
...
3
,
2
,1

zn
 
     From Equation (1) above, we have 
        
0
2
2
2
2





k
k
k
k
z
y
x
 
     or 
        
2
2
2
2
2
2

mE
k
k
k
k
z
y
x




 
     so that 
        


2
2
2
2
2
2
2
z
y
x
n
n
n
n
n
n
ma
E
E
z
y
x






 
_______________________________________ 
 
2.31 
     (a) 





 0
,
2
,
,
2
2
2
2
2








y
x
mE
y
y
x
x
y
x




 
        Solution is of the form: 
          


y
k
x
k
A
y
x
y
x
sin
sin
,



 
       We find 
          


y
k
x
k
Ak
x
y
x
y
x
x
sin
cos
,




 
          


y
k
x
k
Ak
x
y
x
y
x
x
sin
sin
,
2
2
2




 
 
 
          


y
k
x
k
Ak
y
y
x
y
x
y
cos
sin
,




 
          


y
k
x
k
Ak
y
y
x
y
x
y
sin
sin
,
2
2
2




 
 
      Substituting into the original equation, we 
find: 
    (1)      
0
2
2
2
2





mE
k
k
y
x
 
     From the boundary conditions, 
          
0
sin

a
k
A
x
, where 
o
A
a 40

 
     So  
a
n
k
x
x


,  
...
,
3
,
2
,1

xn
 
     Also  
0
sin

b
k
A
y
, where 
o
A
b 20

 
     So  
b
n
k
y
y


,  
...
,
3
,
2
,1

yn
 
    Substituting into Eq. (1) above 
          










2
2
2
2
2
2
2
2
b
n
a
n
m
E
y
x
n
n
y
x



 
(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 
(b) 


2
1
2
2
2
2
2
2
n
n
ma
E





 
   For 
1
,2
1
2

 n
n
 
   Then 
        
2
2
2
2
3
ma
E




 
  (i) For 
o
A
a 4

 
        




2
10
27
2
2
34
10
4
10
67
.
1
2
10
054
.
1
3









E
 
               
22
10
155
.6



J 
   or  
3
19
22
10
85
.3
10
6
.1
10
155
.6








E
eV 
 
 
 
 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
  (ii) For 
5
.0

a
cm 
        




2
2
27
2
2
34
10
5
.0
10
67
.1
2
10
054
.1
3









E
 
                
36
10
939
.3



J 
     or 
        
17
19
36
10
46
.2
10
6
.1
10
939
.3








E
eV 
_______________________________________ 
 
2.33 
(a) For region II, 
0

x
 
        
 


  0
2
2
2
2
2
2





x
V
E
m
x
x
O 


 
  General form of the solution is 
        
 




x
jk
B
x
jk
A
x
2
2
2
2
2
exp
exp




 
  where 
        


OV
E
m
k


2
2
2

 
  Term with 
2B  represents incident wave and  
term with 
2A  represents reflected wave. 
  Region I, 
0

x
 
        
 
  0
2
1
2
2
1
2




x
mE
x
x



 
  General form of the solution is 
        
 




x
jk
B
x
jk
A
x
1
1
1
1
1
exp
exp




 
  where 
        
2
1
2

mE
k 
 
  Term involving 
1B  represents the 
transmitted wave and the term involving 
1A  
represents reflected wave:  but if a particle is 
transmitted into region I, it will not be 
reflected so that 
0
1 
A
. 
  Then 
        
 


x
jk
B
x
1
1
1
exp 


 
        
 




x
jk
B
x
jk
A
x
2
2
2
2
2
exp
exp




 
(b) 
  Boundary conditions: 
(1) 



0
0
2
1



x
x


 
(2) 
0
2
0
1







x
x
x
x


 
      Applying the boundary conditions to the 
solutions, we find 
        
2
2
1
B
A
B


 
        
1
1
2
2
2
2
B
k
B
k
A
k



 
 
 
      Combining these two equations, we find 
          
2
1
2
1
2
2
B
k
k
k
k
A









���


 
           
2
1
2
2
1
2
B
k
k
k
B











 
       The reflection coefficient is 
            
2
1
2
1
2
*
2
2
*
2
2












k
k
k
k
B
B
A
A
R
 
       The transmission coefficient is 
             

2
2
1
2
1
4
1
k
k
k
k
T
R
T





 
_______________________________________ 
 
2.34 
          
 


x
k
A
x
2
2
2
exp 


 
          
 


x
k
A
A
x
P
2
*
2
2
2
2
exp 



 
     where  


2
2
2

E
V
m
k
o 

 
             




34
19
31
10
054
.1
10
6
.
1
8
.
2
5
.
3
10
11
.9
2








 
        
9
2
10
286
.4


k
m 1
  
(a) For 
10
10
5
5




o
A
x
m 
             


x
k
P
2
2
exp 

 
                 





10
9
10
5
10
2859
.4
2
exp





 
                 
0138
.0

 
(b) For 
10
10
15
15




o
A
x
m 
          





10
9
10
15
10
2859
.4
2
exp





P
 
              
6
10
61
.2



 
(c) For 
10
10
40
40




o
A
x
m 
           





10
9
10
40
10
2859
.4
2
exp





P
 
               
15
10
29
.1



 
_______________________________________ 
 
2.35 
          


a
k
V
E
V
E
T
o
o
2
2
exp
1
16



















 
     where 


2
2
2

E
V
m
k
o 

 
             




34
19
31
10
054
.1
10
6
.1
1
.
0
0
.
1
10
11
.9
2








 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
     or        
2
k
9
10
860
.4


m 1
  
(a) For 
10
10
4



a
m 





10
9
10
4
10
85976
.4
2
exp
0
.1
1
.
0
1
0
.1
1
.
0
16


















T
    
0295
.0

 
(b) For 
10
10
12



a
m 





10
9
10
12
10
85976
.4
2
exp
0
.1
1
.
0
1
0
.1
1
.
0
16


















T
    
5
10
24
.1



 
(c) 

e
N
J
t

, where 
tN  is the density of  
                 transmitted electrons. 
       
1
.0

E
eV
20
10
6
.1



J 
                          

 2
31
2
10
11
.
9
2
1
2
1





 m
 
        
5
10
874
.1



m/s
7
10
874
.1


cm/s 
          



7
19
3
10
874
.1
10
6
.1
10
2
.1






tN
 
         
8
10
002
.4


tN
 electrons/cm 3  
     Density of incident electrons, 
      
10
8
10
357
.1
0295
.0
10
002
.4




iN
cm 3
  
_______________________________________ 
 
2.36 
          


a
k
V
E
V
E
T
O
O
2
2
exp
1
16



















 
(a) For 

 om
m
067
.0

 
      


2
2
2

E
V
m
k
O 

 







2
/
1
2
34
19
31
10
054
.1
10
6
.1
2
.
0
8
.
0
10
11
.9
067
.0
2
















 
     or 
          
9
2
10
027
.1


k
 m 1
  
     Then 
          














8
.0
2
.
0
1
8
.0
2
.
0
16
T
 
                      





10
9
10
15
10
027
.1
2
exp





 
     or 
          
138
.0

T
 
(b) For 

 om
m
08
.1

 
 
 
 
 
 
          
2
k = 
     







2
/
1
2
34
19
31
10
054
.1
10
6
.1
2
.
0
8
.
0
10
11
.9
08
.1
2















 
     or 
          
9
2
10
124
.4


k
 m 1
  
     Then 
          














8
.0
2
.
0
1
8
.0
2
.
0
16
T
 
                          





10
9
10
15
10
124
.4
2
exp





 
     or 
          
5
10
27
.1



T
 
_______________________________________ 
 
2.37 
          


a
k
V
E
V
E
T
o
o
2
2
exp
1
16



















 
    where  


2
2
2

E
V
m
k
o 

 
       





34
19
6
27
10
054
.1
10
6
.1
10
1
12
10
67
.1
2










 
       
14
10
274
.7


m 1
  
     (a) 
  





14
14 10
10
274
.7
2
exp
12
1
1
12
1
16

















T
 
      


548
.
14
exp
222
.1


 
      
7
10
875
.5



 
     (b) 
     



7
10
875
.5
10



T
 
                     




a
14
10
274
.7
2
exp
222
.1



 
          










6
14
10
875
.5
222
.1
ln
10
274
.7
2
a
 
     or  
14
10
842
.0



a
m 
_______________________________________ 
 
2.38 
        Region I 
0

x
, 
0

V
; 
        Region II 

a
x 

0
, 
OV
V 
 
        Region III 

a
x 
, 
0

V
 
(a) Region I: 
          
 




x
jk
B
x
jk
A
x
1
1
1
1
1
exp
exp




 
                           (incident)       (reflected) 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
      where 
              
2
1
2

mE
k 
 
     Region II: 
          
 




x
k
B
x
k
A
x
2
2
2
2
2
exp
exp




 
     where 
          


2
2
2

E
V
m
k
O 

 
     Region III: 
          
 




x
jk
B
x
jk
A
x
1
3
1
3
3
exp
exp




 
     (b) 
        In Region III, the 
3B  term represents a  
     reflected wave.  However, once a particle  
     is transmitted into Region III, there will  
     not be a reflected wave so that 
0
3 
B
. 
     (c) Boundary conditions: 
          At 
0

x
: 


2
1 

 
                     
2
2
1
1
B
A
B
A



 
                          


dx
d
dx
d
2
1


 
                    
2
2
2
2
1
1
1
1
B
k
A
k
B
jk
A
jk



 
          At 
a
x 
: 


3
2 

 
                    




a
k
B
a
k
A
2
2
2
2
exp
exp


 
                                               


a
jk
A
1
3 exp

 
                          


dx
d
dx
d
3
2


 
                    




a
k
B
k
a
k
A
k
2
2
2
2
2
2
exp
exp


 
                                            


a
jk
A
jk
1
3
1
exp

 
     The transmission coefficient is defined as 
          
*
1
1
*
3
3
A
A
A
A
T 
 
     so from the boundary conditions, we want  
     to solve for 
3A  in terms of 
1A .  Solving  
     for 
1A  in terms of 
3A , we find 
  









a
k
a
k
k
k
k
k
jA
A
2
2
2
1
2
2
2
1
3
1
exp
exp
4





 
             






a
k
a
k
k
jk
2
2
2
1
exp
exp
2



 
                                       


a
jk1
exp

 
     We then find 
 
 
 
 
 
 
 
 
        








a
k
k
k
k
k
A
A
A
A
2
2
1
2
2
2
2
1
*
3
3
*
1
1
exp
4


 
                                           

2
2
exp
a
k


 
                   





 
2
2
2
2
2
2
1
exp
exp
4
a
k
a
k
k
k



 
     We have 
          


2
2
2

E
V
m
k
O 

 
     If we assume that 
E
VO 
, then a
k2
 will  
     be large so that 
          




a
k
a
k
2
2
exp
exp


 
     We can then write 
          








2
2
2
1
2
2
2
2
1
*
3
3
*
1
1
exp
4
a
k
k
k
k
k
A
A
A
A


 
                                       



 
2
2
2
2
2
1
exp
4
a
k
k
k

 
     which becomes 
          






a
k
k
k
k
k
A
A
A
A
2
2
1
2
2
2
2
1
*
3
3
*
1
1
2
exp
4


 
     Substituting the expressions for 1k  and  
     
2
k
, we find 
          
2
2
2
2
1
2

O
mV
k
k


 
     and 
          
















2
2
2
2
2
1
2
2


mE
E
V
m
k
k
O
 
                    

 E
E
V
m
O 







2
2
2

 
                    


 E
V
E
V
m
O
O
















1
2
2
2

 
     Then 
          


 






























E
V
E
V
m
a
k
mV
A
A
A
A
O
O
O
1
2
16
2
exp
2
2
2
2
2
2
*
3
3
*
1
1


 
                    


a
k
V
E
V
E
A
A
O
O
2
*
3
3
2
exp
1
16



















 
     Finally, 
      


a
k
V
E
V
E
A
A
A
A
T
O
O
2
*
1
1
*
3
3
2
exp
1
16




















 
_____________________________________ 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
2.39 
     Region I: 
0

V
 
          
 
  




0
2
1
2
2
1
2
x
mE
x
x



 
          
 




x
jk
B
x
jk
A
x
1
1
1
1
1
exp
exp




 
                          incident             reflected 
     where 
          
2
1
2

mE
k 
 
     Region II: 
1V
V 
 
          
 


  





0
2
2
2
1
2
2
2
x
V
E
m
x
x



 
          
 




x
jk
B
x
jk
A
x
2
2
2
2
2
exp
exp




 
                          transmitted        reflected 
     where 
          


2
1
2
2

V
E
m
k


 
     Region III: 
2V
V 
 
          
 


  





0
2
3
2
2
2
3
2
x
V
E
m
x
x



 
          
 


x
jk
A
x
3
3
3
exp


 
                          transmitted 
     where 
          


2
2
3
2

V
E
m
k


 
     There is no reflected wave in Region III. 
     The transmission coefficient is defined as: 
 
          
*
1
1
*
3
3
1
3
*
1
1
*
3
3
1
3
A
A
A
A
k
k
A
A
A
A
T






 
     From the boundary conditions, solve for 
3A   
     in terms of 
1A .  The boundary conditions are: 
     At 
0

x
:  


2
1 

 
                   
2
2
1
1
B
A
B
A



 
                       






x
x
2
1


 
                   
2
2
2
2
1
1
1
1
B
k
A
k
B
k
A
k



 
     At 
a
x 
:  


3
2 

 
                   




a
jk
B
a
jk
A
2
2
2
2
exp
exp


 
                                                 


a
jk
A
3
3 exp

 
 
 
 
 
 
 
 
                       






x
x
3
2


 
                   




a
jk
B
k
a
jk
A
k
2
2
2
2
2
2
exp
exp


 
                                                  


a
jk
A
k
3
3
3
exp

 
     But 

 
n
a
k
2
2
 
          



 1
exp
exp
2
2



a
jk
a
jk
 
     Then, eliminating 
1B , 
2A  , 
2B
 from the  
     boundary condition equations, we find 
          



2
3
1
3
1
2
3
1
2
1
1
3
4
4
k
k
k
k
k
k
k
k
k
T





 
_______________________________________ 
 
2.40 
(a) Region I: Since 
E
VO 
, we can write 
          
 ���


  0
2
1
2
2
1
2





x
E
V
m
x
x
O



 
       Region II: 
0

V
, so 
          
 
  0
2
2
2
2
2
2




x
mE
x
x



 
       Region III: 
0
3 




V
 
     The general solutions can be written,  
     keeping in mind that 
1
 must remain  
     finite for 
0

x
, as 
          
 


x
k
B
x
1
1
1
exp


 
          
 




x
k
B
x
k
A
x
2
2
2
2
2
cos
sin



 
          
  0
3

x

 
     where 
          


2
1
2

E
V
m
k
O 

 and  
2
2
2

mE
k 
 
     (b) Boundary conditions 
     At 
0

x
:  


2
1 

2
1 B
B 
 
                       
2
2
1
1
2
1
A
k
B
k
x
x









 
     At 
a
x 
: 


3
2 

 
                         



 0
cos
sin
2
2
2
2


a
k
B
a
k
A
 
     or 
          


a
k
A
B
2
2
2
tan


 
     (c) 
          
1
2
1
2
2
2
1
1
B
k
k
A
A
k
B
k











 
     and since 
2
1 B
B 
, then 
          
2
2
1
2
B
k
k
A









 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
     From 


a
k
A
B
2
2
2
tan


, we can write 
          


a
k
B
k
k
B
2
2
2
1
2
tan










 
     or 
          


a
k
k
k
2
2
1
tan
1










 
     This equation can be written as 
          













a
mE
E
E
VO
2
2
tan
1

 
     or 
          












a
mE
E
V
E
O
2
2
tan

 
     This last equation is valid only for specific 
values of the total energy E .  The energy 
levels are quantized. 
_______________________________________ 
 
2.41 
        


2
2
2
4
2
4
n
e
m
E
o
o
n





(J) 
              


2
2
2
3
2
4
n
e
m
o
o





(eV) 
              









2
2
34
2
12
3
19
31
10
054
.
1
2
10
85
.
8
4
10
6
.
1
10
11
.
9
n











 
     or 
           
2
58
.
13
n
En


 (eV) 
         
58
.
13
1
1 



E
n
eV 
        
395
.3
2
2 



E
n
eV 
         
51
.1
3
3 



E
n
eV 
         
849
.0
4
4 



E
n
eV 
_______________________________________ 
 
2.42 
     We have 
          







 










o
o
a
r
a
exp
1
1
2
/
3
100


 
     and 
          
*
100
100
2
4


 r
P 
 
              







 











o
o
a
r
a
r
2
exp
1
1
4
3
2


 
     or 
 
 
 
          









 


o
o
a
r
r
a
P
2
exp
4
2
3
 
     To find the maximum probability 
          
 
0

dr
r
dP
 
                     











 











 

o
o
o
a
r
r
a
a
2
exp
2
4
2
3
 
                                                   











 

oa
r
r
2
exp
2
 
     which gives 
          
o
o
a
r
a
r





1
0
 
     or 
oa
r 
 is the radius that gives the greatest  
     probability. 
_______________________________________ 
 
2.43 
     
100

 is independent of   and  , so the wave  
     equation in spherical coordinates reduces to 
        
 


0
2
1
2
2
2
















r
V
E
m
r
r
r
r
o

 
     where 
         
 
r
a
m
r
e
r
V
o
o
o
2
2
4







 
     For 
        







 










o
o
a
r
a
exp
1
1
2
/
3
100


 
     Then 
        







 







 












o
o
o
a
r
a
a
r
exp
1
1
1
2
/
3
100


 
     so 
        







 













o
o
a
r
r
a
r
r
exp
1
1
2
2
/
5
100
2


 
     We then obtain 
        
2
/
5
100
2
1
1





















oa
r
r
r


 
                     















 
















 

o
o
o
a
r
a
r
a
r
r
exp
exp
2
2
 
     Substituting into the wave equation, we have 
 
 
 
 
 
Full file at http://testbank360.eu/solution-manual-semiconductor-physics-and-devices-4th-edition-neamen 
 
 
    















 








 










o
o
o
o
a
r
a
r
a
r
r
a
r
exp
exp
2
1
1
2
2
/
5
2 
 
               








r
a
m
E
m
o
o
o
2
2
2


 
                         
0
exp
1
1
2
/
3








 


















o
o
a
r
a

 
     where 
        


2
2
2
2
4
1
2
2
4
o
o
o
o
a
m
e
m
E
E









 
     Then the above equation becomes 
        



























 









o
o
o
o
a
r
r
a
r
a
r
a
2
2
2
/
3
2
1
exp
1
1

 
                      
0
2
2
2
2
2
















r
a
m
a
m
m
o
o
o
o
o



 
     or 
        















 









o
o
a
r
a
exp
1
1
2
/
3

 
                    
0
2
1
1
2
2
2























r
a
a
a
r
a
o
o
o
o
 
     which gives 0 = 0 and shows that 
100

 is  
     indeed a solution to the wave equation. 
_______________________________________ 
 
2.44 
     All elements are from the Group I column of  
     the periodic table.  All have one valence  
     electron in the outer shell. 
_______________________________________ 
 
 
</p>