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By D. A. Neamen Exercise Solutions
Chapter 1
Exercise Solutions
Ex 1.
(a) Number of atoms per unit cell
(b) Volume Density
3 83
- 25 10
−
a
22 = 5. 21 10 cm − 3
_______________________________________
Ex 1.
Intercepts of plane; p=1, q=2, s=
Inverse;
Multiply by lowest common denominator,
( 211 ) plane
_______________________________________
Ex 1.
(a) Number of atoms per (100) plane
Surface Density
2 82
- 25 10
−
a
15 = 1. 11 10 cm
− 2
(b) Number of atoms per (110) plane
Surface Density
− 82
aa 14 = 7. 83 10 cm
− 2
_______________________________________
Test Your Understanding Solutions
TYU 1.
Number of atoms per unit cell 1 8
Volume Density 3
a
8
- 92 10
− a = cm
o = 2. 92 A
Radius
o = r = a 2 = 1. 46 A
TYU 1.
(a) Number of atoms per (100) lattice plane
Surface Density
2 82
- 65 10
−
a
14
- 62 10 cm − 2
(b) Number of atoms per (110) lattice plane
Surface Density
− 82
aa 14 = 3. 27 10 cm − 2
(c) Number of atoms per (111) lattice plane
Lattice plane area bh 2
where b = a 2
2 1 /^2 2 2 2
h = a − a
1 / 2 2 2 a a = a
Then lattice plane area
2
2
a a = a
Surface Density
14
82
−
cm
− 2
_______________________________________
TYU 1.
(a) For (100) planes, distance
o = a = 4. 83 A (b) For (110) planes, distance
( )^ o
A
a
- 42 2
_______________________________________
By D. A. Neamen Exercise Solutions
TYU 1.
(a) 8 corner atoms (b) 6 face-centered atoms (c) 4 atoms totally enclosed
TYU 1.
Number of atoms in the unit cell
Volume Density ( )
3 83
- 43 10
−
a
22 = 5 10 cm − 3
_______________________________________
By D. A. Neamen Problem Solutions
(a) 4 Ga atoms per unit cell
Number density ( ) 83
- 65 10
−
Density of Ga atoms 22 = 2. 22 10 cm − 3
4 As atoms per unit cell
Density of As atoms
22 = 2. 22 10 cm − 3
(b) 8 Ge atoms per unit cell
Number density ( ) 83
- 65 10
−
Density of Ge atoms 22 = 4. 44 10 cm − 3
_______________________________________
From Figure 1.
(a) ( ) a
a d 0. 4330 2
=( )( )
o
- 4330 5. 65 d = 2. 447 A
(b) ( ) a
a d 2 0. 7071 2
( )( )
o = 0. 70715. 65 d = 3. 995 A
sin
a
a
_______________________________________
(a) Simple cubic:
o a = 2 r = 3. 9 A
(b) fcc:
o A
r a 5. 515 2
(c) bcc:
o A
r a 4. 503 3
(d) diamond:
( )^ o A
r a 9. 007 3
_______________________________________
(a) 2 ( 1. 035 ) 2 = 2 ( 1. 035 )+ 2 r B o rB = 0. 4287 A
(b) ( )
o a = 2 1. 035 = 2. 07 A
(c) A-atoms: # of atoms^1 8
Density ( ) 83
- 07 10
−
23 = 1. 13 10 cm − 3
B-atoms: # of atoms 3 2
Density ( ) 83
- 07 10
−
23 = 3. 38 10 cm − 3
_______________________________________
(a)
o a = 2 r = 4. 5 A
of atoms 1
8
Number density ( )
83
- 5 10
−
22 = 1. 097 10 cm
− 3
Mass density
( )
N A
N At. Wt.
( ) ( ) 23
22
= 0. 228 gm/cm
3
(b)
o A
r a 5. 196 3
of atoms 1 2
8
Number density ( )
83
- 196 10
−
22 = 1. 4257 10 cm
− 3
Mass density
( ) ( ) 23
22
= 0. 296 gm/cm
3
_______________________________________
From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,
Volume = 0.74 cm
3
By D. A. Neamen Problem Solutions
(b)
o a = 1. 8 + 1. 0 = 2. 8 A
(c) Na: Density
( )
( )
83
- 8 10
−
22 = 2. 28 10 cm
− 3
Cl: Density
22 = 2. 28 10 cm
− 3
(d) Na: At. Wt. = 22. Cl: At. Wt. = 35. So, mass per unit cell
( ) ( ) 23 23
− =
Then mass density
( )
21
8 10
85 10 83
23
−
−
grams/cm
3
_______________________________________
(a) ( ) ( )
o a 3 = 22. 2 + 21. 8 = 8 A
Then
o a = 4. 62 A Density of A:
( )
22 8 3
−
cm
− 3
Density of B:
( )
22 8 3
−
cm − 3
(b) Same as (a) (c) Same material
( ) ( )^ o a 4. 619 A 3
(a) For 1.12(a), A-atoms
Surface density ( )
2 82
- 619 10
−
a
14 = 4. 687 10 cm − 2
For 1.12(b), B-atoms:
o a = 4. 619 A
Surface density 14 2
a
cm − 2
For 1.12(a) and (b), Same material
(b) For 1.12(a), A-atoms;
o a = 4. 619 A
Surface density
2 a
14 = 3. 315 10 cm − 2
B-atoms; Surface density
14 2
a
cm − 2
For 1.12(b), A-atoms;
o a = 4. 619 A
Surface density
2 a
14 = 3. 315 (^10) cm − 2
B-atoms; Surface density
14 2
a
cm
− 2
For 1.12(a) and (b), Same material
(a) Vol. Density 3
a o
Surface Density 2
2 a o
(b) Same as (a)
(i) (110) plane (see Figure 1.10(b))
(ii) (111) plane (see Figure 1.10(c))
(iii) (220) plane , (^1 ,^1 ,^0 ) 2
Same as (110) plane and [110] direction
(iv) (321) plane ( 2 , 3 , 6 ) 1
Intercepts of plane at p = 2 , q = 3 , s = 6
[321] direction is perpendicular to (321) plane
_______________________________________
By D. A. Neamen Problem Solutions
(c) (111) plane:
Surface density
82
- 43 10
−
14 = 7. 83 10 cm − 2
( )^ o
A
r a 6. 703 2
(a) #/cm 3
3 83
- 703 10
−
a
22 = 1. 328 10 cm − 3
(b) #/cm
2
2
2 a
− 82
14 = 3. 148 10 cm − 2
(c)
( )^ o
A
a d 4. 74 2
(d) # of atoms 2 2
Area of plane: (see Problem 1.19) o b = a 2 = 9. 4786 A
o A
a h 8. 2099 2
Area
( )( ) 8 8
- 4786 10 8. 2099 10 2
= bh =
15
- 8909 10 − = cm 2
#/cm 2 15
- 8909 10
−
14 5.^14 ^10 cm
− 2
( )^ o
A
a d 3. 87 3
_______________________________________
Density of silicon atoms
22 = 5 10 cm
− 3 and 4 valence electrons per atom, so
Density of valence electrons 23 = 2 10 cm − 3
_______________________________________
Density of GaAs atoms
22 8 3
−
cm − 3
An average of 4 valence electrons per atom, So Density of valence electrons 23 = 1. 77 10 cm
− 3
_______________________________________
(a) 100 % 10 % 5 10
22
17 − =
(b) 100 % 4 10 % 5 10
22
15 − =
_______________________________________
(a) Fraction by weight
7 22
16
- 542 10 5 10 28. 06
(b) Fraction by weight
5 22
18
- 208 10 5 10 28. 06
_______________________________________
Volume density
16 3
d
cm
− 3
So
6
- 684 10 − d =^ cm
o d = 368. 4 A
We have
o ao = 5. 43 A
Then 67. 85
- 43
a o
d
_______________________________________
Volume density
15 3
d
cm − 3
So
6
- 30 10
− d = cm
o d = 630 A
We have
o a (^) o = 5. 43 A
Then 116
- 43
a o
d
_______________________________________
By D. A. Neamen Exercise Solutions
Chapter 2
Exercise Solutions
Ex. 2.
(a)
( )( ) 8
34 10
−
−
hc E h
17
- 9875 10
− = J
or 124
- 6 10
19
17
−
− E eV
(b)
( )( ) 8
34 10
−
−
hc E
19
- 417 10
− = J
or 2. 76
- 6 10
19
19
−
− E eV
_______________________________________
Ex 2.
(a) p = 2 mE
( )( )( )
31 3 19 1 /^2 2 9. 11 10 12 10 1. 6 10
− − − = 26
- 915 10 − = kg-m/s
8 26
34
- 12 10
- 915 10
−
− =
p
h m
or
o
= 112 A
(c) 10
34
−
−
h p
26
- 915 10 − = kg-m/s
( ) 31
2 262
−
−
m
p E
21
- 952 10 − J
or 2 19
21
- 97 10
- 6 10
−
− =
(^) E = eV
_______________________________________
Ex 2.
(a) 2
2 2 2
2 ma
n En
( )
( )( ) 31 102
342 2 2
− −
−
n
20 2
- 179 10 n
− = J
or 2 19
20 2
- 261
- 6 10
n
n En =
−
− eV
Then E 1 = 0. 261 eV, E 2 = 1. 045 eV, E 3 = 2. 351 eV
(b) 2
2 2 2
2 ma
n En
( )
( )( ) 27 102
342 2 2
− −
−
n
23 2
- 28 10 n
− = J
or 19
23 2
−
−
n En
4 2
- 425 10 n
− = eV
Then 4 1 1.^42510
− E = eV
4 2 5.^7010
− E = eV
3 3 1.^2810
− E = eV
Ex 2.
( )( ) 2 31 52
- 11 10 10 2
E = m =
21
- 555 10 − = J
Now
( V E )
m k = (^) o − (^22)
Set Vo = 3 E
Then
k 2 m ( 2 E )
2
( ) ( ) ( ) 34
31 21 1 /^2
−
− −
or 9 k 2 = 1. 222 10 m − 1
P =exp − 2 k 2 d
(a) 10 10 10 10 − = =
o d A m
( ) ( )( ) 9 10 exp 21. 222 10 10 10 − P = −
or P = 0. 0868 8. 68 %
(b) 10 100 100 10 − = =
o d A m
( ) ( )( )
9 10 exp 21. 222 10 100 10
− P = −
or 11 9
- 43 10 2. 43 10 − − P = %
_______________________________________
By D. A. Neamen Problem Solutions
Chapter 2
Sketch
Sketch
Sketch
From Problem 2.2, phase t
x
= constant Then
p dt
dx
dt
dx
From Problem 2.3, phase t
x
= constant Then
p dt
dx
dt
dx
_______________________________________
E
hc hc E = h = =
Gold: E = 4. 90 eV (^ )^ (^ )
19
- 6 10
− = J
So,
( )( )
( ) ( )
5 19
34 10
- 54 10
- 6 10
−
− =
= cm
or = 0. 254 m
Cesium: E = 1. 90 eV ( ) ( )
19
- 6 10
− = J
So,
( )( )
( ) ( )
5 19
34 10
- 54 10
- 6 10
−
− =
= cm
or = 0. 654 m
(a) 9
34
−
−
h p
27
- 205 10
− = kg-m/s
3 31
27
- 32 10
- 11 10
−
−
m
p
m/s
or
5 = 1. 32 10 cm/s
(b) 9
34
−
−
h p
27
- 506 10
− = kg-m/s
3 31
27
- 65 10
- 11 10
−
−
m
p
m/s
or
5 = 1. 65 10 cm/s (c) Yes
(a) (i)
( ) ( ) ( )
31 19 2 29. 11 10 1. 21. 6 10
− − p = mE =
25
- 915 10 − = kg-m/s
9 25
34
- 12 10
- 915 10
−
− =
p
h m
or
o = 11. 2 A
(ii) ( ) ( ) ( )
31 19 2 9. 11 10 121. 6 10
− − p =
24
- 87 10 − = kg-m/s
10 24
34
- 54 10
- 8704 10
−
− =
= m
or
o = 3. 54 A
(iii) ( ) ( ) ( )
31 19 2 9. 11 10 1201. 6 10
− − p =
24
- 915 10 − = kg-m/s
10 24
34
- 12 10
- 915 10
−
− =
= m
or
o = 1. 12 A
By D. A. Neamen Problem Solutions
(b)
( ) ( ) ( )
27 19 2 1. 67 10 1. 21. 6 10
− − p =
23
- 532 10
− = kg-m/s
11 23
34
- 62 10
- 532 10
−
− =
= m
or
o = 0. 262 A
( 0. 0259 ) 0. 03885 2
Eavg = kT = eV
Now
p (^) avg = 2 mEavg
( ) ( ) ( ) 31 19 2 9. 11 10 0. 038851. 6 10 − − =
or 25
- 064 10 − pavg = kg-m/s
Now
9 25
34
- 225 10
- 064 10
−
− =
p
h m
or o
= 62. 25 A
_______________________________________
p
p p
hc E h
Now
m
p E
e e 2
2 =^ and
2
e
e e
e
h
m
E
h p
Set (^) E (^) p = Ee and p = 10 e
Then 2 2 10
2
p e p
h
m
h
m
hc
which yields
mc
h p 2
2 mc mc h
hc hc E E p
p = = = =
( )( )
100
31 82 =
−
15
- 64 10 − = J = 10. 25 keV
(a) 10
34
−
−
h p
26
- 794 10
− = kg-m/s
4 31
26
- 56 10
- 11 10
−
−
m
p
m/s
or
6 = 8. 56 10 cm/s
( )( ) 2 31 42
- 11 10 8. 56 10 2
− E m
21
- 33 10
− = J
or 2 19
21
- 08 10
- 6 10
−
− =
E = eV
(b) ( )( ) 31 32
- 11 10 8 10 2
− E
23
- 915 10
− = J
or 4 19
23
- 82 10
- 6 10
−
− =
E = eV
( )( )
31 3 = = 9. 11 10 8 10
−
p m
27
- 288 10
− = kg-m/s
8 27
35
- 09 10
- 288 10
−
− −
p
h m
or
o = 909 A
(a)
( )( ) 10
34 8
−
−
hc E h
15
- 99 10 − = J
Now
19
15
−
−
e
E
E eV V
4 V = 1. 24 10 V = 12. 4 kV
(b) ( )( ) 31 15 2 29. 11 10 1. 99 10 − − p = mE =
23
- 02 10
− = kg-m/s Then
11 23
34
- 10 10
- 02 10
−
− =
p
h m
or o
= 0. 11 A
_______________________________________
By D. A. Neamen Problem Solutions
Since 1 is a solution, then
( ) t
V x j m (^) x
1
2
1
2
1
2 1 1
2
Subtracting these last two equations, we have
m x x x
1 2
1 2
2
2
2
2
2 1 2
2
t
j
2
2
Since 2 is also a solution, we have
( ) t
Vx j m x
2
2
2
2
2
2 1 1
2
Subtracting these last two equations, we obtain
( ) 0
1 2
1 2
2 − =
Vx m x x
This equation is not necessarily valid, which means that 1 2 is, in general, not a solution
to Schrodinger's wave equation.
cos
2
3
1
2 =
−
dx
x A
( ) 1 2
sin
2
3
1
2 =
x x A
A −
so 2
A =
or 2
A =
_______________________________________
cos( ) 1 2
1 / 2
1 / 2
2
−
A n x dx
( ) 1 4
sin 2
2
1 / 2
1 / 2
2
−
n
x n x A
A A
or A = 2
Note that 1 0
=
dx
Function has been normalized. (a) Now
dx a
x
a
P
a o
o o
2 4
0
exp
dx a
x
a
a o
o o
4
0
exp
4
0
exp 2
2 o a
o
o
o a
a x
a
or
( )
1 1 exp 4
1 exp o
o a
a P
which yields P = 0. 393 (b)
dx a
x
a
P
o
o
a
a o o
2 2
4
exp
dx a
x
a
o
o
a
o (^) a o
2
4
exp
2
4
exp 2
2 o
o
a
o a
o
o a
a x
a
or
( ) ( )
P 1 exp 1 exp
which yields P = 0. 239 (c)
dx a
x
a
P
a o
o o
2
0
exp
dx a
x
a
a o
o o
0
exp
ao
o
o
o a
a x
a 0
exp 2
=( − 1 ) exp ( − 2 )− 1
which yields P = 0. 865
By D. A. Neamen Problem Solutions
P ( ) x dx
2
(a) dx
x
a
a
(^2)
cos
(^2 )
/ 4
0
/ 4
sin
a
a
a
x
x
a
a
a
a
sin
( )( )
2 a a
a
or P = 0. 409
(b) dx a
x
a
P
a
a
/ 2
/ 4
2 cos
/ 2
sin
a
a a
a
x
x
a
( )
a
a
a
a
a
sin
sin
4
or P = 0. 0908
(c) dx a
x
a
P
a
a
−
/ 2
/ 2
cos
/ 2
/ 2
sin
a
a a
a
x
x
a
( ) ( )
a
a
a
a
a
sin
4 4
sin
4
or P = 1
(a) dx a
x
a
P
a
sin
/ 4
0
/ 4
0
sin
a
a
a
x
x
a
( )
a
a
a
sin
8
or P = 0. 25
(b) dx a
x
a
P
a
a
sin
/ 2
/ 4
/ 2
/ 4
sin
a
a a
a
x
x
a
( ) ( )
a
a
a
a
a
sin
8 8
sin 2
4
or P = 0. 25
(c) dx a
x
a
P
a
a
−
sin
/ 2
/ 2
/ 2
/ 2
sin
a
a a
a
x
x
a
( ) ( )
a
a
a
a
a
sin 2
8 4
sin 2
4
or P = 1
(a) (i) 4 8
12 10 8 10
k
p
(^) m/s
or 6 p = 10 cm/s
9 8
k
m
By D. A. Neamen Problem Solutions
( )( ) 19
34 8
−
−
7
- 604 10 − = m
or = 660. 4 nm
_______________________________________
(a) 2
2 2 2
2 ma
n En
( )
( )( )
3 2 2
2 342 2 3
215 10 1. 2 10
- 054 10 15 10 − −
− −
=
n
( ) 3 2 62 15 10 2. 538 10 − − = n
or 29 n = 7. 688 10
(b) En + 1 15 mJ
(c) No
For a neutron and n = 1 :
( )
( )( ) 27 142
342 2
2
2 2
1
- 66 10 10
−
ma
E
13
- 3025 10 − = J
or
6 19
13
1 2.^0610
- 6 10
−
− E eV
For an electron in the same potential well:
( )
( )( ) 31 142
342 2
1
- 11 10 10
− −
−
E
10
- 0177 10 − = J
or
9 19
10
1 3.^7610
- 6 10
−
− E eV
_______________________________________
Schrodinger's time-independent wave equation
( ) ( ( )) ( ) 0
2 2
2
E V x x
m
x
x
We know that
( ) x = 0 for 2
a x and 2
a x
We have
V ( ) x = 0 for 2 2
a x
a +
so in this region
( ) ( ) 0
2 2
2
x
mE
x
x
The solution is of the form ( ) x = A cos k x + B sin k x
where
2
mE k =
Boundary conditions:
( ) x = (^0) at 2
a x
a x
First mode solution: 1 ( ) x = A 1 cos k 1 x
where
2
2 2
1 1 2 ma
E
a
k
Second mode solution: 2 ( ) x = B 2 sin k 2 x
where
2
2 2
2 2 2
ma
E
a
k
Third mode solution: 3 ( ) x = A 3 cos k 3 x
where
2
2 2
3 3 2
ma
E
a
k
Fourth mode solution: 4 ( ) x = B 4 sin k 4 x
where
2
2 2
4 4 2
ma
E
a
k
_______________________________________
The 3-D time-independent wave equation in
cartesian coordinates for V (^ x ,^ y , z )=^0 is:
( ) ( ) ( ) 2
2
2
2
2
2 , , , , , ,
z
x yz
y
x yz
x
xy z
( , , ) 0
2
mE Use separation of variables, so let ( x , y , z ) = X ( ) ( xY y ) Z ( ) z
Substituting into the wave equation, we obtain
By D. A. Neamen Problem Solutions
2
2
2
2
2
2
z
Z
XY
y
Y
XZ
x
X
YZ
2
+ XYZ =
mE
Dividing by XYZ and letting 2
mE k = , we
find
2
2
2
2
2
2
k z
Z
y Z
Y
x Y
X
X
We may set
2
2 2 2
2
k X x
X
k x
X
X
x x
Solution is of the form X ( ) x = A sin( kx x ) + B cos( kx x )
Boundary conditions: X ( ) 0 = 0 B = 0
and ( ) a
n X x a k x x
= = 0 =
where nx = 1 , 2 , 3 ....
Similarly, let
2 2
2 1 ky y
Y
Y
and 2 2
2 1 kz z
Z
Z
Applying the boundary conditions, we find
a
n k
y y
= , ny = 1 , 2 , 3 ....
a
n k z z
= , nz = 1 , 2 , 3 ...
From Equation (1) above, we have
0 2 2 2 2 − k (^) x − ky − kz + k =
or
2
mE k (^) x + ky + kz = k =
so that
( )
2 2 2 2
2 2
n nn nx ny nz ma
E E
x yz
_______________________________________
(a)
( ) ( ) ( , ) 0
2 2
2
2
2
xy
mE
y
x y
x
xy
Solution is of the form:
( x , y )= A sin kx x sin ky y
We find ( ) Ak k x k y x
x y x cos x sin y
( ) Ak kx k y x
x y x sin x sin y
2
2 =−
( ) Ak kx k y y
x y y sin x cos y
( ) Ak k x k y y
x y y sin x sin y
2
2 =−
Substituting into the original equation, we find:
(1)^0
2
2 2 − − + =
mE k (^) x ky
From the boundary conditions,
A sin kxa = 0 , where
o a = 40 A
So a
n k x x
= , nx = 1 , 2 , 3 ,...
Also A sin kyb = 0 , where
o b = 20 A
So b
n k
y y
= , ny = 1 , 2 , 3 ,...
Substituting into Eq. (1) above
2
2 2
2
2 2 2
(^2) b
n
a
n
m
E
x y n (^) xny
^
(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.
(a) Derivation of energy levels exactly the same as in the text
(b) ( )
2 1
2 2 2
2 2
n n ma
E = −
For n 2 =^2 ,^ n 1 =^1
Then
2
2 2
ma
E
(i) For
o a = 4 A
( )
( )( ) 27 102
342 2
− −
−
E
22
- 155 10
− = J
or 3 19
22
- 85 10
- 6 10
−
− =
E = eV
By D. A. Neamen Problem Solutions
or k 2 9 = 4. 860 10 m − 1
(a) For 10 4 10 − a = m
( )( ) 9 10 exp 24. 85976 10 4 10
- 0
− − ^
T
(b) For
10 12 10 − a =^ m
( )( ) 9 10 exp 24. 85976 10 12 10
- 0
− − ^
T
5
- 24 10 − =
(c) J = Nt e , where Nt is the density of
transmitted electrons.
E = 0. 1 eV
20
- 6 10
− = J
( ) 2 31 2
- 11 10 2
− = m =
5 = 1. 874 10 m/s
7 = 1. 874 10 cm/s
( )( ) 3 19 7
- 2 10 = 1. 6 10 1. 874 10 − − N t 8 Nt = 4. 002 10 electrons/cm
3
Density of incident electrons,
10
8
- 357 10
- 0295
Ni = cm
− 3
_______________________________________
( k a ) V
E
V
E T O O
(^16 1) exp− (^22)
−
(a) For m =( 0. 067 ) m o
( ) (^22)
mV E k
O −
( ) ( ) ( ) ( )
( )
1 / 2
342
31 19
−
− −
or 9 k 2 = 1. 027 10 m
− 1
Then
T 16
( )( ) 9 10 exp 21. 027 10 15 10
− −
or T = 0. 138
(b) For m =( 1. 08 ) mo
k 2 =
( ) ( ) ( ) ( )
( )
1 / 2
342
31 19
−
− −
or 9 k 2 = 4. 124 10 m − 1
Then
T 16
( )( )
9 10 exp 24. 124 10 15 10
− −
or 5
- 27 10 − T =
( k a ) V
E
V
E
T
o o
(^16 1) exp− (^22)
where
( ) (^22)
mV E k
o −
( ) ( ) ( ) 34
27 6 19
−
− −
14 = 7. 274 (^10) m − 1
(a)
( )( ) 14 14 exp 27. 274 10 10 12
− −^
T
= 1. 222 exp − 14. 548 7
- 875 10
− =
(b)
( ) ( )
7 10 5. 875 10
− T =
( ) a
14 = 1. 222 exp− 27. 274 10
( )
14
- 875 10
2 7. 274 10 a ln
or
14
- 842 10 − a =^ m
Region I ( x 0 ), V = 0 ;
Region II ( 0 x a ), V = VO
Region III ( x a ), V = 0
(a) Region I: 1 ( ) x = A 1 exp ( jk 1 x ) + B 1 exp( − jk 1 x )
(incident) (reflected)
By D. A. Neamen Problem Solutions
where
(^12)
mE k =
Region II: 2 ( ) x = A 2 exp ( k 2 x ) + B 2 exp( − k 2 x )
where
( ) (^22)
mV E k
O −
Region III: 3 ( ) x = A 3 exp ( jk 1 x ) + B 3 exp( − jk 1 x )
(b) In Region III, the B 3 term represents a
reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected wave so that B 3 = 0.
(c) Boundary conditions: At (^) x = 0 : 1 = 2
A 1 (^) + B 1 = A 2 + B 2
dx
d
dx
d 1 2
jk 1 (^) A 1 − jk 1 B 1 = k 2 A 2 − k 2 B 2
At x = a : 2 = 3
A (^) 2 exp ( k 2 a ) + B 2 exp( − k 2 a )
= A (^) 3 exp( jk 1 a )
dx
d
dx
d 2 3
k (^) 2 A 2 exp ( k 2 a ) − k 2 B 2 exp( − k 2 a )
= jk (^) 1 A 3 exp( jk 1 a )
The transmission coefficient is defined as
1 1
3 3
A A
AA
T =
so from the boundary conditions, we want to solve for A 3 in terms of A 1. Solving
for A 1 in terms of A 3 , we find
( k (^) k ) ( (^) ka ) ( (^) k a ) kk
jA A (^) 2 2
2 1
2 2 12
3 1 exp exp 4
− 2 jk (^) 1 k 2 exp ( k 2 a ) +exp( − k 2 a )
exp( jk 1 a )
We then find
( )
( k k ) ( k a ) kk
AA
A A 2
2 1
2 2 2 1 2
( )
2 −exp − k 2 a
( ) ( )
2 2 2
2 2
2
- 4 k 1 (^) k exp ka +exp− ka
We have
( ) (^22)
mV E k
O −
If we assume that VO E , then k (^) 2 a will
be large so that exp ( k (^) 2 a ) exp( − k 2 a )
We can then write
( )
( ) ( )
2 2
2 1
2 2 2 1 2
k k ka kk
AA
A A = −
( )
2 2
2 2
2
which becomes
( )
( k (^) k ) ( (^) k a ) kk
AA
A A 2
2 1
2 2 2 12
Substituting the expressions for k 1 and
k 2 , we find
2
2 2
2 1
mV O k + k =
and
( )
2 2
2 2
2 1
mV E mE k k
O
( V E )( E )
m O −
2
2
( ) ( E ) V
E
V
m
O
O
2
2
Then
( )
( )
−
=
E V
E V
m
ka
mV AA
AA
O
O
O
1
2 16
exp 2
2
2
2
2
2
2
3 3
1 1
( k a ) V
E
V
E
AA
O O
2
3 3
16 1 exp− 2
Finally,
( k a ) V
E
V
E
AA
AA
T
O O
*^2 1 1
3 3 16 1 exp− 2
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