Modern Physics: October 28, 2013
Solutions for the Homework 4
Problem 4.1: The great majority of alpha particles pass through gases and thin metal foils with no
deflections. To what conclusion about atomic structure does this observation lead?
Solution: Atoms are point-like particles and the thin metal foils and gases have no charges.
Problem 4.4: Find the frequency of revolution of the electron in the classical model of the hydrogen
atom. In what region of the spectrum are electromagnetic waves of this frequency?
Solution: Follow the Section 4.2. From this process, we can get the relation between v and r
but in order to solve this problem, we need one more condition that is given in the Example 4.1.
So, use the result of the Example 4.1. Then,
ν =
v
= 6.6 × 1015 Hz,
2πr
which corresponds to ultraviolet region. (Figure 2.2)
Problem 4.7: In the Bohr model, the electron is in constant motion. How can such an electron have
a negative amount of energy?
Solution: When the electron is bounded to proton, electron must have negative total energy
even if the kinetic energy is positive. This means that the potential energy will be negative and
it’s natural because of the electron and proton have opposite sign. And from the result of Eq.
(4.15), it’s true.
Problem 4.8: Lacking de Broglie’s hypothesis to guide his thinking, Bohr arrived at his model by
postulating that the angular momentum of an orbital electron must be an integral multiple of
~. Show that this postulate leads to Eq. (4.13).
Solution: Suppose that the angular momentum of an electron is quantized by integral multiple
of ~. Then, it gives the relation as follows:
mvr = n~,
and from the classical result,
v = √
e
.
4π0 mr
Mix these two equations by eliminating v,
n~
e
= √
mrn
4π0 mrn
⇒
1
rn =
4π0 n2 ~2
0 n2 ~2
=
.
me2
πme2
Modern Physics: October 28, 2013
Problem 4.9: The fine structure constant is defined as α = e2 /20 hc. This quantity got its name
because it first appeared in a theory by the German physicist Arnold Sommerfeld that tried to
explain the fine structure in spectral lines (multiple lines close together instead of single lines) by
assuming that elliptical as well as circular orbits are possible in the Bohr model. Sommerfeld’s
approach was on the wrong track, but α has nevertheless turned out to be a useful quantity in
atomic physics. (a) Show that α = v1 /c, where v1 is the velocity of the electron in the ground
state of the Bohr atom. (b) Show that the value of α is very close to 1/137 and is a pure
number with no dimensionless. Because the magnetic behavior of a moving charge depends on
its velocity, the small value of α is representative of the relative magnitudes of the magnetic and
electric aspects of electron behavior in an atom. (c) Show that αa0 = λc /2π, where a0 is the
radius of the ground-state Bohr orbit and λc is the Compton wavelength of the electron.
Solution: (a) From the velocity with radius at level 1,
v1 = √
e
e2
e2
= αc
=
= p 2
20 h
4π0 me r1
40 h2
⇒
α =
v1
.
c
(b) From the result of (a),
α =
e2
= 7.30 × 10−3 ,
20 hc
and so 1/α = 137 up to 3 digits of significant figures.
(c) Since the compton wavelength is given by λC = h/mc from Eq. (2.22) and rn = n2 h2 /πme e2 ,
αa0 =
e2 h2 0
h
λC
=
=
.
20 hc πme e2
2πmc
2π
Problem 4.13: Compare the uncertainty in the momentum of an electron confined to a region of
linear dimension a0 with the momentum of an electron in a ground-state Bohr orbit.
Solution: Uncertainty in the position is given by ∆x = a0 and so,
∆p ≥
~
~
=
.
2∆x
2a0
Now, in order to get the momentum of an electron in a ground-state, use the quantized de Broglie
wavelength
λ = 2πa0
for n=1.
Then,
p =
h
~
=
,
λ
a0
which is twice of the uncertainty.
Problem 4.14: When radiation with a continuous spectrum is passed through a volume of hydrogen
gas whose atoms are all in the ground state, which spectral series will be present in the resulting
absorption spectrum?
Solution: Hydrogen gas will be excited from the ground state to n-th state because of the
hydrogen gas will absorb the light and it corresponds to Lyman series.
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Modern Physics: October 28, 2013
Problem 4.15: What effect would you expect the rapid random motion of the atoms of an excited
gas to have on the spectral lines they produce?
Solution: Rapid random motion leads that the atoms have high speed and it will make Doppler
effect. So, the frequency of emission spectral lines are slightly differed and it makes the spectral
line widen.
Problem 4.23: The longest wavelength in the Lyman series is 121.5nm and the shortest wavelength
in the Balmer series is 364.6nm. Use the figures to find the longest wavelength of light that
could ionize hydrogen.
Solution: The longest wavelength in the Lyman series obtained when the hydrogen excited from
n = 1 to n = 2 and the shortest wavelength in the Balmer series obtained when the hydrogen
excited from n = 2 to n = ∞. Combine these two result and then, we can obtain the wavelength
of photon excited from n = 1 to n = ∞, which corresponds to ionization, as follows:
1
E1 1
1
E1 1
1
E1 1
1
1
1
= −
− 2
= −
− 2 −
− 2
=
+
.
2
2
2
λI
ch 1
∞
ch 1
2
ch 2
∞
λL λB
From the above relation,
λI =
1
1
+
λL λB
−1
= 91.13nm.
Problem 4.28: Of the following quantities, which increase and which decrease in the Bohr model as
n increases? Frequency of revolution, electron speed, electron wavelength, angular momentum,
potential energy, kinetic energy, total energy.
Solution: Remember that rn ∼ n2 . When n increase, then the radius will increase and it makes
the kinetic energy, electron speed and frequency of revolution to decrease. But the potential
energy, total energy and angular momentum will increase.
−E1 2
e
fn =
,
vn = √
,
Ln = mvn rn = n~,
h
n3
4π0 me rn
KEn =
e2
,
8π0 rn
P En = −
e2
,
4π0 rn
T En = −
e2
E1
= 2.
8π0 rn
n
Problem 4.29: Show that the frequency of the photon emitted by a hydrogen atom in going from
the level n + 1 to the level n is always intermediate between the frequencies of revolution of the
electron in the respective orbits.
Solution: From the Eq. (4.19), the frequency of revolution at level n is given by
−E1 2
fn =
.
h
n3
Now, the frequency of the photon emitted between two given levels is
E1
1
1
−E1 (n + 1)2 − n2
ν =
−
=
.
h (n + 1)2 n2
h
(n + 1)2 n2
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Modern Physics: October 28, 2013
Then,
n(n + 1/2)
−E1 (n + 1)2 − n2
=
f
< fn ,
n
h
(n + 1)2 n2
(n + 1)2
(n + 1/2)(n + 1)
−E1 (n + 1)2 − n2
= fn+1
> fn+1 .
h
(n + 1)2 n2
n2
From these relation, the frequency of the photon emitted between two levels is always intermediate between the frequencies of revolution of the electron in the respective orbits as follows:
fn+1 < ν < fn .
Problem 4.32: Compare the ionization energy in positronium with that in hydrogen.
Solution: In this problem, we need to consider the reduced mass in order to find the ionization
energy. For each case,
0
EH
= −
m0
E1 ≈ −E1
me
for Hydrogen case,
m0
E1
for positronium case,
E1 = −
me
2
where the reduced mass of hydrogen is m0 = me mp /(me + mp ) ≈ me and positronium is m0 =
m2e /2me = me /2.
Ep0 = −
Problem 4.36: For laser action to occur, the medium used must have at least three energy levels.
What must be the nature of each of these levels? Why is three the minimum number?
Solution: One level is the ground state, and the other one is an excited state, the last one is
metastable state (Figure 4.26). The medium used to make an excited atom metastable, rapidly.
If there is only two levels, we can not make population inversion, because of the optical pumping
not only produce excited atoms but also atoms of ground state. Therefore, we need extra state
to make population inversion.
Problem 4.37: A certain ruby laser emits 1.00J pulses of light whose wavelength is 694nm. What
is the minimum number of Cr3+ ions in the ruby?
Solution: Since the energy of photon is E = hc/λ, the number of photons are
n =
1.00J
1.00J × λ
=
= 3.94 × 1018 ,
E
hc
and so there are same number of Cr3+ ions, at least.
Problem 4.38: Steam at 100◦ C can be thought of as an excited state of water at 100◦ C. Suppose
that a laser could be built based upon the transition from steam to water, with the energy lost
per molecule of steam appearing as a photon. What would the frequency of such a photon be?
To what region of the spectrum does this correspond? The heat of vaporization of water is
2260kJ/kg and its molar mass is 18.02kg/kmol.
Solution: The energy of vaporization of water per mol is 2260kJ/kg × 18.02kg/kmol = 4.07 ×
104 J/mol. So, each molecule emit the energy E = 6.76 × 10−20 J and the frequency of the
emission light is ν = 1.02 × 1014 Hz which corresponds to the infrared light (Figure 2.2).
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Modern Physics: October 28, 2013
Problem 4.39: The Rutherford scattering formula fails to agree with the data at very small scattering
angles. Can you think of a reason?
Solution: Rutherford scattering formula considered only the nucleus effect not the electrons, so
when alpha particle scattered far from the nucleus it leads to very small scattering angle in this
model. However, the alpha particle is not scattered in face because of the electron screened the
charge of nucleus and so it looks like a neutral charge at the long distance from the nucleus.
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