Chapter 2 Solutions
Prob. 2.1
(a&b) Sketch a vacuum tube device. Graph photocurrent I versus retarding voltage V for
several light intensities.
I
light
intensity
Vo
V
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Note that Vo remains same for all intensities.
(c) Find retarding potential.
λ=2440Å=0.244μm
=4.09eV
1.24eV  μm
1.24eV  μm
Vo = hν - Φ =
-=
- 4.09eV = 5.08eV - 4.09eV  1eV
λ(μm)
0.244μm
Prob. 2.2
Show third Bohr postulate equates to integer number of DeBroglie waves fitting within
circumference of a Bohr circular orbit.
4o n 2
mq2
2
4o n 2
rn =
mq2
2
rn =
and
q2
mv2
=
and p = mvr
4πo r 2
r
n 2 2 4πo rn 2 n 2 2 rn
n2 2
=

=

=
mrB2 q2
mrn 2 mv2 m2 v2rn
m2 v2rn 2 = n 2 2
mvrn = n
p = n is the third Bohr postulate
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Prob. 2.3
(a) Find generic equation for Lyman, Balmer, and Paschen series.
ΔE =
hc
mq4
=
λ
32π2o2 n12
2
-
mq4
32π2o 2 n 22
2
mq4 (n 22 - n12 )
mq4 (n 22 - n12 )
hc
=
=
λ
32o 2 n12n 22 2 π2
8o 2 n12n 22 h2
8o 2 n12n 22h2  hc 8εo 2h3c n12n 22
=

=
mq4 (n 22 - n12 )
mq4 n 22 - n12
8(8.85

 10-12 mF )2 (6.63 1034 Js)
 3 2.998  108 ms n12n 22
=
 2 2
9.11  10-31kg  (1.60  10-19C)4
n 2 - n1
n12n 22
n12n 22
=
9.11
Å

n 22 - n12
n 22 - n12
n1 =1 for Lyman, 2 for Balmer, and 3 for Paschen
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 = 9.11108 m
(b) Plot wavelength versus n for Lyman, Balmer, and Paschen series.
n
2
3
4
5
n^2
4
9
16
25
LYMAN SERIES
n^2-1
n^2/(n^2-1)
3
1.33
8
1.13
15
1.07
24
1.04
LYMAN LIMIT
n
3
4
5
6
7
n^2
9
16
25
36
49
BALMER SERIES
n^2-4
4n^2/(n^2-4)
5
7.20
12
5.33
21
4.76
32
4.50
45
4.36
BALMER LIMIT
911*n^2/(n^2-1)
1215
1025
972
949
911Ǻ
911*4*n^2/(n^2-4)
6559
4859
4338
4100
3968
n
4
5
6
7
8
9
10
n^2
16
25
36
49
64
81
100
PASCHEN SERIES
n^2-9
9*n^2/(n^2-9)
7
20.57
16
14.06
27
12.00
40
11.03
55
10.47
72
10.13
91
9.89
PASCHEN LIMIT
911*9*n^2/(n^2-9)
18741
12811
10932
10044
9541
9224
9010
8199Ǻ
3644Ǻ
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Prob. 2.4
(a) Find Δpx for Δx=1Ǻ.
Δpx  Δx =
h
h
6.6310-34J  s
 Δpx =
=
= 5.03 10-25 kgsm
-10
4
4  Δx
4π 10 m
(b) Find Δt for ΔE=1eV.
h
h
4.14 10-15eV  s
ΔE  Δt =
 Δt =
=
= 3.30 10-16s
4
4  ΔE
4π 1eV
Prob. 2.5
Find wavelength of 100eV and 12keV electrons. Comment on electron microscopes compared to
visible light microscopes.
2 E
m
h
=
2 E m
E = 12 mv2  v =
h
h
=
=
p mv
For 100eV,
6.63 10-34J  s
 E- 2 = E- 2  4.9110-19J 2  m
1
1
1
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λ=
2  9.1110 kg
-31
λ = E- 2  4.9110-19J 2  m = (100eV 1.602 10-19 eVJ )- 2  4.9110-19J 2  m = 1.2310-10m = 1.23Å
1
1
For 12keV,
1
1
λ = E- 2  4.9110-19J 2  m = (1.2 104eV 1.602 10-19 eVJ )- 2  4.9110-19J 2  m = 1.12 10-11m = 0.112Å
1
1
1
1
The resolution on a visible microscope is dependent on the wavelength of the light which is
around 5000Ǻ; so, the much smaller electron wavelengths provide much better resolution.
Prob. 2.6
Which of the following could NOT possibly be wave functions and why? Assume 1-D in each
case. (Here i= imaginary number, C is a normalization constant)
A) Ψ (x) = C for all x.
B) Ψ (x) = C for values of x between 2 and 8 cm, and Ψ (x) = 3.5 C for values of x between 5
and 10 cm. Ψ (x) is zero everywhere else.
C) Ψ (x) = i C for x= 5 cm, and linearly goes down to zero at x= 2 and x = 10 cm from this peak
value, and is zero for all other x.
If any of these are valid wavefunctions, calculate C for those case(s). What potential energy for x
≤ 2 and x ≥ 10 is consistent with this?
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A) For a wavefunction (x) , we know Ρ =

  (x)(x)dx = 1
*
-

Ρ =   (x)(x)dx = c
*
2
-

 dx
-
0 c = 0
 Ρ= 
 (x) cannot be a wave function
 c  0
B) For 5  x  8 , (x) has two values, C and 3.5C. For c  0 , (x) is not a function

and for c = 0 : Ρ =  * (x) (x)dx = 0  (x)cannot be a wave function.
-
 iC
 3  x-2 2  x  5
C) (x)= 
 iC  x-10 5  x  10
 5

5
10 2
c2
c
2
2
Ρ =   (x)(x)dx =   x-2 dx +   x-10 dx
-
2 9
5 25
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*
10
c2
c2
3 5

=
(x-2)  2 +
(x-10)3 5
3×9
3×25
 27 125  8c2
= c2  +
=
3
 27 3×25 
8c2
Ρ=1 
=1  c=0.612  (x) can be a wave function
3
Since (x) = 0 for x  2 and x  10 , the potential energy should be infinite in these two
regions.
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Prob. 2.7
A particle is described in 1D by a wavefunction:
Ψ = Be-2x for x ≥0 and Ce+4x for x<0, and B and C are real constants. Calculate B and C to make
Ψ a valid wavefunction. Where is the particle most likely to be?
A valid wavefunction must be continuous, and normalized.
For (0) = C = B


To normalize  ,
2
dx = 1
-
0

2 8x
2 -4x
 C e dx +  C e dx = 1
-
0

C
1
e  + C2   e-4x   1

0
8
4
C2 C2
8
+
=1  C=
8
4
3
8x 0
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2
Prob. 2.8
The electron wavefunction is Ceikx between x=2 and 22 cm, and zero everywhere else. What is
the value of C? What is the probability of finding the electron between x=0 and 4 cm?
 = Ceikx
22
-1
1
cm
20
  dx = C (20) = 1  C =
*
2
2
2
1
 1 
Probability =   dx = 
  2 = 10
 20 
0
4
2
Prob. 2.9
Find the probability of finding an electron at x<0. Is the probability of finding an electron at
x>0 zero or non-zero? Is the classical probability of finding an electron at x>6 zero or non?
The energy barrier at x=0 is infinite; so, there is zero probability of finding an electron at
x<0 (|ψ|2 =0). However, it is possible for electrons to tunnel through the barrier at 5<x<6;
so, the probability of finding an electron at x>6 would be quantum mechanically greater
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than zero (|ψ|2 >0) and classical mechanically zero.
Prob. 2.10
Find 4  px2  2  pz 2  7mE for ( x, y, z, t )  A  e j (10x3 y-4t ) .


j(10x+3y-4t)
dx
 j x  A  e
- A  e


=
= 100 

2 - j(10x+3y-4t) j(10x+3y-4t)
e
dx
A e
-

- j(10x+3y-4t) 
-

- j(10x+3y-4t) 
 A e
*
E =
2
2

j(10x+3y-4t)
dz
 j z  A  e
- A  e


=
=0

2 - j(10x+3y-4t) j(10x+3y-4t)
e
dz
A e
*
pz 2
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px 2
2
- j(10x+3y-4t) 
*
-



j(10x+3y-4t)
dt
  j t  A  e


= 4
2 - j(10x+3y-4t) j(10x+3y-4t)
A e
e
dt
-
4  px 2 +2  pz 2 +7mE = 400
2
 28(9.1110-31kg)
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Prob. 2.11
Find the uncertainty in position (Δx) and momentum (Δρ).
2
 πx 
 sin    e-2πjEt/h and
L
L
Ψ(x,t) =
L
x =  *  x dx =
0
L
Δx =
p 
x2 - x
2
  dx = 1
*
0
2
x 
x  sin 2   dx = 0.5L (from problem note)

L0
 L 
x 2 =  *  x dx =
0
L
L
2L 2
x 
x  sin 2   dx = 0.28L2 (from problem note)

L0
 L 
= 0.28L2 - (0.5L)2 = 0.17L
h
h
= 0.47 
4π  Δx
L
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Prob. 2.12
Calculate the first three energy levels for a 10Ǻ quantum well with infinite walls.
En =
n 2  π2  2
(6.63 10-34 )2
=
 n 2 = 6.03 10-20  n 2
2  m  L2
8  9.111031  (109 )2
E1 = 6.03 10-20J = 0.377eV
E2 = 4  0.377eV = 1.508eV
E3 = 9  0.377eV = 3.393eV
Prob. 2.13
Show schematic of atom with 1s22s22p4 and atomic weight 21. Comment on its reactivity.
nucleus with
8 protons and
13 neutrons
2 electrons in 1s
2 electrons in 2s
This atom is chemically reactive because
the outer 2p shell is not full. It will tend
to try to add two electrons to that outer
shell.
4 electrons in 2p
= proton
= neturon
= electron
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