ALL 15 CHAPTERS COVERED
SOLUTIONS MANUAL
,FOUNDATIONS OF
QUANTUM PHYSICS
SOLUTIONS MANUAL
9/27/08
Charles E. Burkhardt
Jacob J. Leventhal
, This is page 535
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Preface to the Solutions
Manual
This Solutions Manual has been prepared in an effort to save time for
the instructor. The solutions have been displayed in a manner that, it is
hoped, will eliminate the need to carry out tedious algebra. In addition to
the references in the text a helpful source that will be referred to as "the
handbook" in this solutions manual is:
References
[1] M. R. Spiegel, Mathematical Handbook of Formulas and Tables
(McGraw-Hill, New York, 1998)
Comments, criticism and suggestions are welcome ( or
).
, This is page 1
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Chapter 1
Introduction
1. Light of wavelength λ illuminates a metal surface and photoelectrons
having maximum kinetic energy of 1 eV are ejected. The light source
is replaced by a one which emits light of wavelength λ/2 and the
photoelectrons are observed to have a maximum kinetic energy of
4.28 eV. What is the work function of the metal? Find a table of
work functions and decide which metal it is.
Solution: Using
KE = hν − W
we have
hc
= KE + W
λ
so
hc
= 1+W
λ
hc
= 4.28 + W
(λ/2)
Dividing, we have
1 1+W W
= =⇒ 2.14 + =1+W
2 4.28 + W 2
So
W = 2.28 eV
The metal is sodium.
2. If the state of mercury atoms that is excited by electrons in the
Franck—Hertz experiment decays back to the state from which it was
excited by emitting light, what will be the wavelength of that light?
Solution:
λ(in nm)E(in eV) = 1, 240
so
1, 240
λ = nm
4.9
= 253.1 nm
3. Calculate the following:
,2 Chapter 1. Introduction
(a) The wavelengths in nm of the first three lines of the Lyman
series and the Balmer series.
(b) The series limit of the Lyman and Balmer series. The series
limit is defined as the shortest possible wavelength.
Solution:
(a) The Bohr energy is given by
−13.6 eV
En =
n2
so the energies of the appropriate transitions are
µ ¶
1 1
∆En→m = − 13.6 eV
n2 m2
which are listed in the second column of Table 1.1. Then using
the relation
λ(in nm)E(in eV) = 1, 240
Table 1.1: The first three lines of the Lyman series and the Balmer series.
(n → m) ∆En→m ( eV) Designation λ ( nm)
4→1 12.75 Lγ 97.3
3→1 9.06 Lβ 136.9
2→1 10.2 Lα 121.6
5→2 2.86 Hγ 433.6
4→2 2.55 Hβ 486.3
3→2 1.89 Hα 656.1
(b) To obtain the series limit we make the highest state the state
for which n → ∞. The energies of the series limit transitions
are therefore 13.6 eV (Lyman) and 13.6/22 = 3.4 eV. The wave-
lengths of photons of these wavelength are 91.2 nm and 364.7 nm.
4. It is possible to form a hydrogenlike atom with a proton and a nega-
tive μ-meson having mass mμ ≈ 200me . Find the radius of the first
Bohr orbit in terms of a0 , the velocity of the μ-meson in the first Bohr
orbit in terms of the same quantity for hydrogen, and the ionization
energy from the ground state in electron-volts.
Solution:
According to Equation 1.31 the Bohr radius is given by
~2
a0 = (4π 0 )
me e2
, Chapter 1. Introduction 3
The mass of the electron is actually the reduced mass of the pro-
ton/electron system (which is essentially me ) so we must find the
reduced mass, call it M to avoid confusion with the meson, of the
proton/μ-meson system. In terms of the electron mass it is
1 1 1
= + =⇒ M ≈ 180
M 2000me 200me
Then the first Bohr radius a0μ is
µ ¶
~2 me a0
a0μ = (4π 0 ) =
M e2 me 180
According to Equation 1.41 the velocity is independent of the mass
so it is the same as for a Bohr atom, αc.
The Bohr energy is proportional to the mass, but it can also be writ-
ten in terms of the orbital radius which we have already calculated
(see Equation 1.32). The ionization energy from the ground state of
hydrogen is µ 2 ¶
e 1
E1H = −
4π 0 2a0
Make the substitution a0 → a0 /180 and we obtain
µ 2 ¶
e 1
E1μ = −
4π 0 2n2 (a0 /180)
= 180E1H ≈ 2450 eV
5. Show that a0 = ~/ (me cα) and that the speed of the electron in the
nth Bohr orbit is vn = αc/n.
Solution: ∙ ¸
e2
α=
(4π 0 ) }c
and
}c ~
a0 = (4π 0 ) ·
me e2 c
so
~
a0 =
me cα
Also s µ ¶
1 e2 1
vn =
me 4π 0 n2 a0
,4 Chapter 1. Introduction
Now substitute the above result for a0 to obtain
s µ 2 ¶ r
1 e 1 ~c me cα
vn = 2
· ·
me 4π 0 n ~c ~
α
= c
n
6. Using the definition of the fine structure constant α, (Equation 1.35),
and its known value, show that, in atomic units, the speed of light is
137 a.u. of length per a.u. of time.
Solution: ∙ ¸
e2 1
α= =
(4π 0 ) }c 137
1
But e = 1 = ~ = me = = 1 in atomic units so, in a.u. c = 137.
4π 0
7. Compton scattering experiments can be performed using protons
rather than electrons.
(a) Find the Compton wavelength of the proton in terms of the
Compton wavelength of the electron.
(b) If the apparatus is such that ∆λ/λ must be ∼ 0.03, what must
be the wavelength of the incident photon? In what region of the
electromagnetic spectrum are photons of this energy?
Solution:
(a) The Compton wavelength of the electron is given by Equation
1.15
h
λce = ≈ 2.43 × 10−12 m
me c
Multiplying the rhs by mp /mp we have
µ ¶
h mp
λce =
me c mp
µ ¶
mp
= λcp
me
Solving for λcp we have
µ ¶
me 1
λcp = λce ≈ λce
mp 2000
≈ 1.2 × 10−15 m
, Chapter 1. Introduction 5
(b) For ∆λ/λ = 0.03 the energy of the incident photon must be
0.03mp c2 = 28 MeV the wavelength of which is
1, 240
λp = = 44 × 10−7 nm
28 × 106 eV
They are still x rays, hard x rays, because the lower limit of
energy for γ rays is usually taken to be about 50k eV.
8. Show that fitting de Broglie waves to the circumference of the Bohr
orbits leads to the postulate that Bohr never made, that is, angular
momentum is quantized in units of ~.
Solution:
Assuming that there will be an integral number of deBroglie wave-
lengths λd in the circumference of each Bohr orbit we have
nλd = 2πrn
where rn is the radius of the nth Bohr orbit. Inserting the expression
for the deBroglie wavelength we have
h
n = 2πrn
pn
where pn is the linear momentum in the nth Bohr orbit. Recognizing
that the angular momentum Ln = pn rn for a circular orbit we have
h
Ln = pn rn = n
2π
, This is page 6
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Chapter 2
Elementary Wave Mechanics
1. A particle of mass m that is confined in a potential well is known
2 2
to be in an eigenstate having eigenfunction ψ (x) = Ae−α x /2 and
2
energy E = (α~) / (2m).
(a) Find the potential energy function U (x) that confines the par-
ticle.
(b) What is the force that confines the particle?
(c) Find the value of the constant A that is required to normalize
this eigenfunction.
Solution:
(a) The TISE is
∙ ¸
~2 d2
− + U (x) ψ (x) = Eψ (x)
2m dx2
so, cancelling out the A we have
~2 d2 ³ −α2 x2 /2 ´ 2 2 2 2
− e + U (x) e−α x /2 = Ee−α x /2
2m dx2
The derivatives are
d ³ −α2 x2 /2 ´ 2 2
e = −α2 xe−α x /2
dx
d2 ³ −α2 x2 /2 ´ 2
³
−α2 x2 /2 2 2
´
e = −α e − α x
dx2
2 2
After dividing through by the common factor e−α x /2
(which
can never vanish) the TISE becomes
~2 2 ~2 4 2
α − α x + U (x) = E
2m 2m
or µ ¶
~2 2 ~2 4 2
U (x) = E − α + α x
2m 2m
2
But we are given that E = (α~) / (2m) so
~2 α4 2
U (x) = x
2m
, Chapter 2. Elementary Wave Mechanics 7
(b) The force on the particle is the negative
¡ gradient
¢ of the potential
energy so F = −dU (x) /dx = − ~2 α4 /m x and we see that we
have a restoring force (due to the minus sign) that is propor-
tional to the displacement. This is a Hooke’s law force so the
potential energy is that of a harmonic oscillator.
(c) Find the value of the constant A for normalization we require
the definite integral
Z ∞ r
−ay 2 π
e dy =
−∞ a
We have
Z ∞
2 2
|A|2 e−α x
dx = 1
−∞
√
2 π
|A| = 1
α
p √
so A = α/ π and the normalized eigenfunction is
√
α 2 2
ψ (x) = 1/4 e−α x /2
π
2. The normalized wave function of a particle of mass m is given by
√
α 2 2
Ψ (x, t) = 1/4 e−α x /2 ei(kx−ωt)
π
(a) What is the probability of finding the particle between x and
x + dx at time t?
(b) What is the probability of finding the particle in the range −∞ <
x < ∞?
Solution:
(a) The probability of finding the particle between x and x + dx,
P (x) dx is the probability density Ψ∗ (x, t) Ψ (x, t) multiplied by
dx so
α 2 2
P (x) dx = √ e−α x dx
π
(b) The probability of fiinding the particle in the range −∞ < x <
∞, Ptotal is
Z ∞
Ptotal = P (x) dx
−∞
Z ∞
α 2 2
= √ e−α x dx
−∞ π
SOLUTIONS MANUAL
,FOUNDATIONS OF
QUANTUM PHYSICS
SOLUTIONS MANUAL
9/27/08
Charles E. Burkhardt
Jacob J. Leventhal
, This is page 535
Printer: Opaque this
Preface to the Solutions
Manual
This Solutions Manual has been prepared in an effort to save time for
the instructor. The solutions have been displayed in a manner that, it is
hoped, will eliminate the need to carry out tedious algebra. In addition to
the references in the text a helpful source that will be referred to as "the
handbook" in this solutions manual is:
References
[1] M. R. Spiegel, Mathematical Handbook of Formulas and Tables
(McGraw-Hill, New York, 1998)
Comments, criticism and suggestions are welcome ( or
).
, This is page 1
Printer: Opaque this
Chapter 1
Introduction
1. Light of wavelength λ illuminates a metal surface and photoelectrons
having maximum kinetic energy of 1 eV are ejected. The light source
is replaced by a one which emits light of wavelength λ/2 and the
photoelectrons are observed to have a maximum kinetic energy of
4.28 eV. What is the work function of the metal? Find a table of
work functions and decide which metal it is.
Solution: Using
KE = hν − W
we have
hc
= KE + W
λ
so
hc
= 1+W
λ
hc
= 4.28 + W
(λ/2)
Dividing, we have
1 1+W W
= =⇒ 2.14 + =1+W
2 4.28 + W 2
So
W = 2.28 eV
The metal is sodium.
2. If the state of mercury atoms that is excited by electrons in the
Franck—Hertz experiment decays back to the state from which it was
excited by emitting light, what will be the wavelength of that light?
Solution:
λ(in nm)E(in eV) = 1, 240
so
1, 240
λ = nm
4.9
= 253.1 nm
3. Calculate the following:
,2 Chapter 1. Introduction
(a) The wavelengths in nm of the first three lines of the Lyman
series and the Balmer series.
(b) The series limit of the Lyman and Balmer series. The series
limit is defined as the shortest possible wavelength.
Solution:
(a) The Bohr energy is given by
−13.6 eV
En =
n2
so the energies of the appropriate transitions are
µ ¶
1 1
∆En→m = − 13.6 eV
n2 m2
which are listed in the second column of Table 1.1. Then using
the relation
λ(in nm)E(in eV) = 1, 240
Table 1.1: The first three lines of the Lyman series and the Balmer series.
(n → m) ∆En→m ( eV) Designation λ ( nm)
4→1 12.75 Lγ 97.3
3→1 9.06 Lβ 136.9
2→1 10.2 Lα 121.6
5→2 2.86 Hγ 433.6
4→2 2.55 Hβ 486.3
3→2 1.89 Hα 656.1
(b) To obtain the series limit we make the highest state the state
for which n → ∞. The energies of the series limit transitions
are therefore 13.6 eV (Lyman) and 13.6/22 = 3.4 eV. The wave-
lengths of photons of these wavelength are 91.2 nm and 364.7 nm.
4. It is possible to form a hydrogenlike atom with a proton and a nega-
tive μ-meson having mass mμ ≈ 200me . Find the radius of the first
Bohr orbit in terms of a0 , the velocity of the μ-meson in the first Bohr
orbit in terms of the same quantity for hydrogen, and the ionization
energy from the ground state in electron-volts.
Solution:
According to Equation 1.31 the Bohr radius is given by
~2
a0 = (4π 0 )
me e2
, Chapter 1. Introduction 3
The mass of the electron is actually the reduced mass of the pro-
ton/electron system (which is essentially me ) so we must find the
reduced mass, call it M to avoid confusion with the meson, of the
proton/μ-meson system. In terms of the electron mass it is
1 1 1
= + =⇒ M ≈ 180
M 2000me 200me
Then the first Bohr radius a0μ is
µ ¶
~2 me a0
a0μ = (4π 0 ) =
M e2 me 180
According to Equation 1.41 the velocity is independent of the mass
so it is the same as for a Bohr atom, αc.
The Bohr energy is proportional to the mass, but it can also be writ-
ten in terms of the orbital radius which we have already calculated
(see Equation 1.32). The ionization energy from the ground state of
hydrogen is µ 2 ¶
e 1
E1H = −
4π 0 2a0
Make the substitution a0 → a0 /180 and we obtain
µ 2 ¶
e 1
E1μ = −
4π 0 2n2 (a0 /180)
= 180E1H ≈ 2450 eV
5. Show that a0 = ~/ (me cα) and that the speed of the electron in the
nth Bohr orbit is vn = αc/n.
Solution: ∙ ¸
e2
α=
(4π 0 ) }c
and
}c ~
a0 = (4π 0 ) ·
me e2 c
so
~
a0 =
me cα
Also s µ ¶
1 e2 1
vn =
me 4π 0 n2 a0
,4 Chapter 1. Introduction
Now substitute the above result for a0 to obtain
s µ 2 ¶ r
1 e 1 ~c me cα
vn = 2
· ·
me 4π 0 n ~c ~
α
= c
n
6. Using the definition of the fine structure constant α, (Equation 1.35),
and its known value, show that, in atomic units, the speed of light is
137 a.u. of length per a.u. of time.
Solution: ∙ ¸
e2 1
α= =
(4π 0 ) }c 137
1
But e = 1 = ~ = me = = 1 in atomic units so, in a.u. c = 137.
4π 0
7. Compton scattering experiments can be performed using protons
rather than electrons.
(a) Find the Compton wavelength of the proton in terms of the
Compton wavelength of the electron.
(b) If the apparatus is such that ∆λ/λ must be ∼ 0.03, what must
be the wavelength of the incident photon? In what region of the
electromagnetic spectrum are photons of this energy?
Solution:
(a) The Compton wavelength of the electron is given by Equation
1.15
h
λce = ≈ 2.43 × 10−12 m
me c
Multiplying the rhs by mp /mp we have
µ ¶
h mp
λce =
me c mp
µ ¶
mp
= λcp
me
Solving for λcp we have
µ ¶
me 1
λcp = λce ≈ λce
mp 2000
≈ 1.2 × 10−15 m
, Chapter 1. Introduction 5
(b) For ∆λ/λ = 0.03 the energy of the incident photon must be
0.03mp c2 = 28 MeV the wavelength of which is
1, 240
λp = = 44 × 10−7 nm
28 × 106 eV
They are still x rays, hard x rays, because the lower limit of
energy for γ rays is usually taken to be about 50k eV.
8. Show that fitting de Broglie waves to the circumference of the Bohr
orbits leads to the postulate that Bohr never made, that is, angular
momentum is quantized in units of ~.
Solution:
Assuming that there will be an integral number of deBroglie wave-
lengths λd in the circumference of each Bohr orbit we have
nλd = 2πrn
where rn is the radius of the nth Bohr orbit. Inserting the expression
for the deBroglie wavelength we have
h
n = 2πrn
pn
where pn is the linear momentum in the nth Bohr orbit. Recognizing
that the angular momentum Ln = pn rn for a circular orbit we have
h
Ln = pn rn = n
2π
, This is page 6
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Chapter 2
Elementary Wave Mechanics
1. A particle of mass m that is confined in a potential well is known
2 2
to be in an eigenstate having eigenfunction ψ (x) = Ae−α x /2 and
2
energy E = (α~) / (2m).
(a) Find the potential energy function U (x) that confines the par-
ticle.
(b) What is the force that confines the particle?
(c) Find the value of the constant A that is required to normalize
this eigenfunction.
Solution:
(a) The TISE is
∙ ¸
~2 d2
− + U (x) ψ (x) = Eψ (x)
2m dx2
so, cancelling out the A we have
~2 d2 ³ −α2 x2 /2 ´ 2 2 2 2
− e + U (x) e−α x /2 = Ee−α x /2
2m dx2
The derivatives are
d ³ −α2 x2 /2 ´ 2 2
e = −α2 xe−α x /2
dx
d2 ³ −α2 x2 /2 ´ 2
³
−α2 x2 /2 2 2
´
e = −α e − α x
dx2
2 2
After dividing through by the common factor e−α x /2
(which
can never vanish) the TISE becomes
~2 2 ~2 4 2
α − α x + U (x) = E
2m 2m
or µ ¶
~2 2 ~2 4 2
U (x) = E − α + α x
2m 2m
2
But we are given that E = (α~) / (2m) so
~2 α4 2
U (x) = x
2m
, Chapter 2. Elementary Wave Mechanics 7
(b) The force on the particle is the negative
¡ gradient
¢ of the potential
energy so F = −dU (x) /dx = − ~2 α4 /m x and we see that we
have a restoring force (due to the minus sign) that is propor-
tional to the displacement. This is a Hooke’s law force so the
potential energy is that of a harmonic oscillator.
(c) Find the value of the constant A for normalization we require
the definite integral
Z ∞ r
−ay 2 π
e dy =
−∞ a
We have
Z ∞
2 2
|A|2 e−α x
dx = 1
−∞
√
2 π
|A| = 1
α
p √
so A = α/ π and the normalized eigenfunction is
√
α 2 2
ψ (x) = 1/4 e−α x /2
π
2. The normalized wave function of a particle of mass m is given by
√
α 2 2
Ψ (x, t) = 1/4 e−α x /2 ei(kx−ωt)
π
(a) What is the probability of finding the particle between x and
x + dx at time t?
(b) What is the probability of finding the particle in the range −∞ <
x < ∞?
Solution:
(a) The probability of finding the particle between x and x + dx,
P (x) dx is the probability density Ψ∗ (x, t) Ψ (x, t) multiplied by
dx so
α 2 2
P (x) dx = √ e−α x dx
π
(b) The probability of fiinding the particle in the range −∞ < x <
∞, Ptotal is
Z ∞
Ptotal = P (x) dx
−∞
Z ∞
α 2 2
= √ e−α x dx
−∞ π
