PROBLEM SESSIONS
Kvantfenomen och fysik 2013
November 27, 2013
Christian Spånslätt
Fysikum
Stockholm University
christian.spanslatt@fysik.su.se
08-5537 8742
Contents
Introduction
1
1 Problem Session 1: Atoms 1
4.15 Hot Atomic Gas . . . . . . . . . . . . . .
4.21 Emitted Balmer Series . . . . . . . . . .
4.27 Recoil in Photon-Atom Collisions . . . .
4.33 Isotope Shift of the Hydrogen Spectrum
4.42 Impact Parameter of the Gold Nucleus .
7.8 Structure of the Periodic Table . . . . . .
7.16 Sodium Energy Spectrum . . . . . . . .
7.18 How Does the Atomic Radius Depend on
2 Problem session 2: Atoms 2
6.26 Harmonic Oscillator Selection Rules . .
7.22 Spin-Orbit Interaction in the Hydrogen
7.26 Angular Momentum Calculations . . .
7.34 More Angular Momentum . . . . . . .
7.38 Magnetic Moment of Chlorine . . . . .
7.40 The Kα Line in a X-ray Spectrum . . .
7.43 Singlet and Triplet States . . . . . . .
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the Atomic Number?
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8
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11
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12
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14
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18
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4 Problem session 4: Time Dependent Perturbation Theory
Preliminary Theory . . . . . . . . . . . . . . . . . . . . . . . . . .
Problem 1, Hydrogen Atoms in an Electrical Field . . . . . . . . .
Problem 2, Perturbed Harmonic Oscillator . . . . . . . . . . . . . .
Problem 3, Quantum Atom-Photon Interaction . . . . . . . . . . .
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20
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5 Problem session 5: Quantum Statistics
9.5 Population of Hydrogen Rotational States . .
9.14 Neutron Beam Density . . . . . . . . . . . .
9.15 Pressures Of Distributions . . . . . . . . . .
9.20 Sunspots . . . . . . . . . . . . . . . . . . . .
9.33 Gas Cloud Properties . . . . . . . . . . . . .
9.37 Free Electron Gas . . . . . . . . . . . . . . .
9.49 Classical Limit of Bose Einstein Distribution
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26
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6 Problem session 6: Solid State Physics
10.2 Madelung Constant of NaCl . . . . . . .
10.3 Cohesive Energies . . . . . . . . . . . . .
10.5 The Joule-Thomson Effect . . . . . . . .
10.9 Return of the Free Electron Gas . . . . .
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30
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30
31
32
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Atom
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3 Problem session 3: Molecules
8.5 Molecule Rotation inertia . . . . . . . . . . .
8.7 Rotational Spectrum As an Isotope Identifier
8.9 Rotational Spectrum for HCl . . . . . . . . .
8.13 Correspondence Principle . . . . . . . . . .
8.15 Zero Point Energy of Hydrogen Isotopes . .
8.16 Zero Vibrational and Rotational Energy . .
8.18 Return of the Harmonic Oscillator . . . . .
8.21 Room Temperature Vibrations . . . . . . . .
2
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10.14 Band Structure of Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
10.26 The SQUID . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
7 Problem session 7: The Atomic Nucleus
11.2 Boron Mixtures . . . . . . . . . . . . . . .
11.14 Oxygen Beta Decay . . . . . . . . . . . .
11.17 Mass-Energy Relation . . . . . . . . . . .
11.18 Magnesium Atomic Mass . . . . . . . . .
11.22 Mirror Isobars . . . . . . . . . . . . . . .
11.23 More Mirror Isobars . . . . . . . . . . . .
11.29 Deuteron Model . . . . . . . . . . . . . .
8 Problem session 8: Nuclear Reactions
12.19 Upper Bound on the Age of Our Solar
12.21 Uranium Decay Chain . . . . . . . . .
12.22 More Uranium Decay . . . . . . . . .
12.24 Radium Alpha-Decay . . . . . . . . .
12.51 Nuclear Fusion . . . . . . . . . . . . .
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34
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38
System
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39
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42
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44
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9 Problem session 9: Elementary Particles
12.55 Nuclear Electrostatics . . . . . . . . . . .
12.61 Proton Energy Barrier . . . . . . . . . .
13.1 Virtual Photon Interaction . . . . . . . . .
13.5 Electron-Positron Pair Creation . . . . . .
13.22 Meson Properties . . . . . . . . . . . . .
13.26 The Weak Interaction . . . . . . . . . . .
10 Epilogue
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49
I
Introduction
These notes are a collection of solutions to problems presented in the course Kvantfenomen
och fysik, FK5014 during the fall 2013 at Stockholm University. The problems are taken
from the book Concepts of Modern Physics, 6th Edition by A. Beiser.
My purpose with these notes is to apply theory, presented in the lectures and the
book, to various problems in the microscopic world. I have aimed for a conceptual
approach to each problem, defining a strategy on how to solve it by pointing out key
concepts and lines of thought. The solution is then presented in a, hopefully, clear way
following that strategy. Further, some comment or discussion regarding the problem is
added.
Below, I have presented some general guidelines for solving problems in quantum
physics (of course they are applicable to most other subjects as well) that should be
useful.
• Spell out a strategy for the problem before you dive into calculations. Ask yourself:
What do I already know and what do I need to know in order to solve the problem?
• Use the book and formula collections. Know where to quickly find material you
need to solve the problem. Wolfram Alpha can be very handy sometimes to find
needed quantities.
• Solve the problems algebraically first (with letters for the relevant quantities) and
put in numbers in the end. This minimizes the risk of round-off errors and makes
calculations more structured and readable.
• Learn to use proper units. In quantum physics, SI-units are not very useful. Natural
constants are most conveniently expressed in eV, c, K.
• Taylor expansions are very useful to estimate and simplify formulas. Learn some
of them by heart or have them quickly accessible.
• Check your result. Do the dimensions agree? Does the result make any sort of
sense? (Beware that intuition and quantum physics are not always compatible
though.)
GOOD LUCK!
1
1
Problem Session 1: Atoms 1
4.15 Hot Atomic Gas
Problem formulation: What effect would you expect the rapid random motion of
the atoms of an excited gas to have on the spectral lines they produce?
Approach: What we know here is that in a hot atomic gas there is a rapid motion of
excited atoms in all directions at any instant. Therefore we will expect two effects. One
is that since many different levels are occupied, many different transitions will occur.
This implies that we will see many spectral lines corresponding to these transitions. The
other one is the relation between motion and frequency i.e. that of Doppler shift. We
need to understand how the Doppler shift affects the emitted frequencies observed.
Solution: Consider an atom in the hot gas move, with speed v c, either away or
towards us (i.e. we are stationary). The observed frequency ν of an emitted frequency
ν0 is given by
s
1 + v/c
≈ ν0 (1 + v/c) ⇒ ν − ν0 ≡ ∆ν ≈ ν0 v/c.
ν = ν0
1 − v/c
To see the magnitude of this shift, we need to know some sort of speed associated
with the moving atoms. It is well known from statistical physics that the average
kinetic energy of a particle with mass m in a temperature T , moving in one dimension
(relevant here since movement towards or from an observer is one dimensional) is given
by 12 m < v >2 = 12 kB T , where kB is Boltzmann’s constant. Thus we can extract the
q
speed as < v >= kBmT .
To get some numbers, we can calculate the temperature needed to overlap the sodium
doublet: 589.592 nm and 588.995 nm. We note that for the shorter wavelength line to
shift onto the larger one we need v/c = ∆ν/ν0 ≈ 0.001.
Solving for the temperature, we then get
T =
mN a v 2
22.98 · 931.49M eV /c2 · 0.0012 c2
≈ 2.5 · 108 K.
=
kB
8.617 · 10−5 eV /K
Discussion: We can conclude that the effect on the spectrum from Doppler shifts is
not relevant for low temperatures, but it will in fact broaden the spectrum lines a bit to
both higher and lower wavelengths/frequencies due to movement both away and towards
the observer.
4.21 Emitted Balmer Series
Problem formulation: A beam of electrons bombards a sample of hydrogen. Through
what potential difference must the electrons have been accelerated if the first line of the
Balmer series is to be emitted?
Approach: For this problem, we should identify what is meant by the first line in the
Balmer series. We must also understand what energy has to be supplied for this process
to happen. We should also understand how this energy is supplied by the electron beam.
2
Solution: The energy differences between states (spectral lines) in the hydrogen spectrum are given by
∆E = −13.6(
1
1
− 2 ) eV,
n2f
ni
where nf and ni is the final and initial state quantum number respectively.
After some reading in the book, one can conclude that the first line of the Balmer
series is given by n = 3 → n = 2. Thus, the hydrogen atoms must be excited to n = 3
from the ground state, n = 1, to make this process possible. The electrons need then to
supply the energy
∆E = −13.6(
1
1
− 2 ) eV ≈ 12.1 eV.
2
3
1
Discussion: When one accelerates a charged object in an electrical field, it will acquire
the energy E = qV , where q and V is the object’s charge and the applied voltage
respectively. We can then conclude that for an electron to receive the energy 12.1 eV
(so it can excite the atom to the correct level) it must be accelerated through a voltage
(potential difference) of 12.1 V.
4.27 Recoil in Photon-Atom Collisions
Problem formulation: When an excited atom emits a photon, the linear momentum
of the photon must be balanced by the recoil momentum of the atom. As a result, some
of the excitation energy of the atom goes into the kinetic energy of its recoil. (a) Modify
Eq. (4.16) to include this effect. (b) Find the ratio between the recoil energy and the
photon energy for the n = 3 → n = 2 transition in hydrogen, for which Ef − Ei = 1.9
eV. Is the effect a major one? A non relativistic calculation is sufficient here.
Approach: We should note that the key to this problem is the conservation of energy
and linear momentum which is true in elastic collisions. We can also always choose
a coordinate system such that the reaction takes place in one dimension. To use the
conservation laws, we must know the kinetic energy of the photon and the atom, as well
as the expressions for linear momentum. For the first part of the problem, we should
just be able to re-express the formula in the book with these conservations fulfilled. For
the second part, we have to find a suitable way to compare the recoil energy from the
first part to the photon energy.
Solution: The requirement for conservation of energy gives:
∆E ≡ Ei − Ef = hν +
2
m A vA
2
And conserving linear momentum yields
!
mA vA ≡ PA = Pν =
hν
hν
⇒ vA =
c
cmA
Combining these equations gives
∆E = hν +
h2 ν 2
2c2 mA
To simplify this expression a bit, let us define ν0 ≡ ∆E/h and νA ≡ mA c2 /h. Then
we can write
3
ν0 = ν +
and solve for ν. We get ν = −νA + νA
q
ν2
2νA
1+
ν0
2νA
, where we have chosen the positive
square root in order to have a positive frequency. At this point we should be able to
check the ratio ν0 /νA . We have from the problem statement that ν0 = 1.9eV /h and
νA = 1.008 · 931.49M eV /c2 · c2 /h which gives ν0 /νA ≈ 2 · 10−9 (here we have expressed
the hydrogen mass in eV/c2 ). It is very small and we can Taylor expand the square root
√
2
above ( 1 + x ≈ 1 + x2 − x8 ...) to second order yielding
ν ≈ ν0 (1 −
ν0
).
2νA
This is the correction to the photon frequency due to atomic recoil and we can see already
here that it is very small.
Now we can compare the kinetic energies of the atom (the recoil energy) and the
photon by
Eν
=
EA
hν
2
m A vA
2
=
hν
h2 ν 2
2c2 mA
=
2mA c2
2νA
2νA
=
≈
≈ 109 .
hν
ν
ν0
Discussion: We conclude that the photon energy is tremendously larger than the atom
recoil energy and can safely be neglected in our calculations of transition energies. Note
that no potential energy of the photon is needed in the calculation. The photon does not
couple to the electromagnetic field. The potential energy of the atom is stored in the
electron levels, and it is this energy that is distributed into kinetic energy to the atom
and photon. A crude picture is tho view the the atom as an enormous massive cannon
and the photon is the cannon ball. The potential energy is the gunpowder in the loaded
cannon which in the firing (emission) event is consumed by firing the cannon ball and
rocking the cannon a little bit.
4.33 Isotope Shift of the Hydrogen Spectrum
Problem formulation: A mixture of ordinary hydrogen and tritium, a hydrogen
isotope whose nucleus is approximately 3 times more massive than ordinary hydrogen, is
excited and its spectrum observed. How far apart in wavelength will the Hα lines of the
two kinds of hydrogen be?
Approach: For this problem, we must understand how the nuclear mass affects the
hydrogen spectrum. The key here is the notion of reduced mass. We must obviously also
understand what is meant by the Hα line.
Solution: As is stated in the book, the reduced mass, m0H of hydrogen (with the proton
and electron mass M and m respectively) is given by:
m0H =
mM
.
m+M
3mM
For tritium, we can immediately state that its reduced mass is m0T = m+3M
. The energy
spectrum of hydrogen, when the reduced mass is taken into account, is given by
EnH =
m0H E1
,
m n2
4
where E1 = −13.6 eV.
We further note that from this follows that
EnT =
m0T E1
.
m0H n2
To extract the photon wavelength from a transition we use the relation E = hν =
hc/λ ⇒ λ = hc/E to write
λT =
hc
m0T
m0H Eα
,
where Eα is the energy between the levels in the relevant transition. The wavelength
difference between hydrogen and tritium is then given by
∆λ ≡ λT − λα =
hc
m0T
m0H Eα
−
hc
hc m0H
m0
=
( 0 − 1) = λα ( H
− 1).
Eα
Eα mT
m0T
Now we are ready to plug in some numbers. After some “research”, we should find
that the Hα line is the transition from n = 3 → n = 2 and has a wavelength of λα = 656.3
nm. We have also that
m0H
1 1 + 3M
m
≈ 0.999637,
=
m0T
3 1+ M
m
which gives us ∆λ = 656.3(0.999637 − 1)nm ≈ −0.238 nm.
Discussion: We can conclude that the Hα line is shifted to a shorter wavelength (called
a blue shift) which means a higher energy. This is because tritium has a larger reduced
mass than hydrogen implying that it binds the electron harder.
4.42 Impact Parameter of the Gold Nucleus
Problem formulation: What is the impact parameter of a 5.0-MeV alpha particle
scattered by 10◦ when it approaches a gold nucleus?
Approach: For this problem, we must understand what the Rutherford scattering
formula says and find the relation between impact parameter and scattering angle. Make
sure you understand the derivation (or at least the general steps) on pages 152 − 157.
Solution: From the Rutherford cross section derivation, it should be clear that the
relation between the impact parameter, b, and the scattering angle θ is given by
θ
4π0 EK
cot( ) =
b,
2
Ze2
where 0 , EK , Z and e is the electric vacuum permittivity, the incoming kinetic energy,
the atomic number of the target nucleus and the electron charge respectively.
1
The relevant numbers to use are 4π
= 8.988 · 109 N m2 /C 2 , EK = 5 MeV, Z = 79,
0
−19
◦
e = 1.602 · 10
C and θ = 10 .
Re-arranging the formula we get
5
θ 1
1
b = cot( )
Ze2
2 4π0
EK
≈11.43 · 8.988 · 109 N m2 /C 2 · 79 · e · 1.602 · 10−19 C
≈2.6 · 10−13 m.
1
5 · 106 eV
Here, we have used that N/C ⇔ V /m which can be derived from the definition of
electric field.
Discussion: We can compare this distance to the Bohr radius a0 ≈ 5 · 10−11 m and
conclude that the gold nucleus should be smaller than this distance which is reasonable.
7.8 Structure of the Periodic Table
Problem formulation: What is true in general of the properties of elements in the
same period of the periodic table. Of elements in the same group?
Approach: This problem contains no calculation and is more of a reasoning exercise.
We should understand what is meant by a period and a group and find common properties
of their elements. Relevant information is found in section 7.4 of the book. Wikipedia
has a huge amount of nice material as well.
Solution/Discussion: A period of the table is a horizontal line. Elements in a period
have the same number of electron shells. When a new period is started, a new shell
begins to fill up and closes in the noble gas group.
A group in the table is a vertical line. All elements in a group have their valence
(outer) electrons in the same configuration. Due to this similarity, the elements have
common properties. For example in the group 18 (called 8 sometimes), the noble gases,
all elements have a closed valence shell which makes them inert.
7.16 Sodium Energy Spectrum
Problem formulation: The effective nuclear charge that acts on the outer electron
in the sodium atom is 1.84e. Use this figure to calculate the ionization energy of sodium.
Approach: We should in this problem understand the electron configuration of Sodium
(Natrium in Swedish) and use the fact that the ionization energy is the energy to remove
the outermost electron (that is making N a+ ). We should also understand why we can
not simply use Z = 11 for the hydrogen problem.
Solution: Sodium has the electron configuration 1s2 2s2 2p6 3s1 . The energy to ionize
(that is make N a+ ) should therefore be the energy to remove the 3s-electron. Naively
one would guess then that
1
1
11
− 2 ) = 13.6eV ( )2 ≈ 183 eV.
∞2
n
3
This is not even close to the measured value. The reason for this is that we have not
accounted for the reduced nuclear charge that comes from the concept of screening. The
3s-electron in the outer rims of the atom sees the nucleus accompanied by all the inner
En = E1 Z 2 (
6
shell electrons forming a chunk with charge 1.84 e. This screening effect is important
and with the reduced nuclear charge we get:
En = −E1
Z2
1.84 2
= 13.6eV (
) ≈ 5.12 eV.
2
n
3
Discussion: The actual ionization energy is 5.14 eV so our approximation with the
screening accounted for is quite good. It should be said that electron-electron interactions
are really hard to include in calculations, but for the alkali metals (to which Na belongs)
this “effective hydrogen atom” approach works quite well.
7.18 How Does the Atomic Radius Depend on the Atomic Number?
Problem formulation: Account for the general trends of the variation of atomic
radius with atomic number shown in Fig. 7.11 (in Beiser ).
Approach: This is also a more reasoning kind of problem rather than pure calculations.
Solution/Discussion: From the diagram we see that the atom radii peak at the alkali
metals. Why is this? It is because these elements can be viewed as inert gases with one
more electron and proton added. The outer electron sees Z − 1 inner shell electrons
shielding the Z protons, effectively reducing the nuclear charge to eZ ≈ 1e. This makes
the ionization energy low which corresponds to a large radius.
When going from left to right in the diagram, the elements “pick up” one more electron
and proton. And thus the effective charge increases by 1e for each step increasing the
ionization energy a bit, shrinking the radius.
7
2
Problem session 2: Atoms 2
6.26 Harmonic Oscillator Selection Rules
Problem formulation: The selection rule for transitions between states in a harmonic
oscillator is ∆n = ±1. (a) Justify this rule on classical grounds. (b) Verify from the
relevant wave functions that the n = 1 → n = 3 transition in a harmonic oscillator is
forbidden, whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed.
Approach: To understand this problem, we must think of the classical harmonic
oscillator and what kind of phenomena it can describe. We must also consider the wave
functions of the quantum harmonic oscillator and understand what kind of calculation
exists that allows us to determine a possible transition.
Solution: One may think of the classical harmonic oscillator as a particle with mass
2
m rolling back and forth in a parabolic (y = kx2 ) potential. The “rolling frequency” is
q
k
then determined to be ω = m
. Or we may also think of the same particle bouncing up
and down with frequency ω in a spring with constant k. Imagine now this particle to be
charged. What will happen? From electrodynamics, we know that accelerated charges
generate radiation and so will happen here as well. The radiation will be waves with
frequency ω. If we now use the photon model, we can say that the harmonic oscillator
radiates photons, each with energy E = ~ω.
And if now study the energy levels of the quantum harmonic oscillator, we get the
result that
1
En = ~ω(n + ).
2
And if we then think of the difference between two levels, ∆En,m ≡ En − Em =
~ω(n−m) and equates this to one kind of photon with energy ~ω we get that (m−n) = ±1
(depending on which of m and n is the largest.)
Next, we shall show that a certain transition is forbidden, while another is not. How
does one calculate that? We should know that the first order transition amplitude n → m
is given by1
Z ∞
hm| x̂ |ni =
dx ψn∗ (x) · x · ψn (x).
−∞
One could go on here and plug in expressions for the wave functions, but this is
tedious and it is much more useful to apply the ladder operator formalism. We define
operators
√
↠ψn (x) = n + 1ψn+1 (x)
√
âψn (x) = nψn−1 (x).
Remember also that x̂ =
immediately see that
q
~
2mω (â
+ ↠) in the ladder formalism. From this we can
1 The electric field comes into play because even vacuum there are vacuum fluctuations in this field
dV (x)
driving transitions. In one dimension E ∝ dx so that for a constant electric field (as the vacuum has
on average), the potential is V (x) ∝ x, meaning that the operator x̂ comes into play.
8
Z
∞
r
∗
ψm
(x)
~
2mω
Z
∞
∗
hm| x̂ |ni =
dx
dx ψm
· x · ψn (x) =
(x) · (â + ↠) · ψn (x) =
−∞
−∞
r
Z ∞
√
√
~
∗
dx ψm
(x) · ( n + 1ψn+1 (x) + nψn−1 (x)) =
2mω −∞
r
√
~ √
( n + 1δm,n+1 + nδm,n−1 ),
2mω
where we have used the orthogonality of the wave functions. From this result we can
calculate
r
√
~ √
( 1 + 1) · 0 + 1 · 0) = 0
2mω
r
r
√
~ √
3~
h2| x̂ |3i =
( 3 + 1) · 0 + 3 · 1) =
2mω
2mω
r
r
√
~ √
~
h1| x̂ |2i =
( 2 + 1) · 0 + 2 · 1) =
.
2mω
mω
h3| x̂ |1i =
And we see, at least for these transitions, that only m = n ± 1 transitions are allowed.
This can be seen also from the Kroenecker delta expressions above.
Discussion: Of course, as mentioned, one may use the explicit form of the Hermite
polynomials to calculate the integrals, but I have preferred to use the ladder operators
instead to demonstrate the power of this formalism. It can not be stated enough how
important the quantum harmonic oscillator is in modern physics. Applications include
vibrational and rotational molecular spectra, formulation of quantum field theory and
many particle physics in for example condensed matter systems. Everybody who wants
to move on to more advanced courses in quantum theory benefits greatly from studying
the quantum harmonic oscillator thoroughly.
7.22 Spin-Orbit Interaction in the Hydrogen Atom
Problem formulation: Why is the ground state of the hydrogen atom not split into
two sub levels by spin-orbit coupling?
Approach: We should understand the reason for spin-orbit coupling, and figure out
what is so special about the hydrogen ground state.
Solution/Discussion: The hydrogen ground state is a 1s-state. That means n = 0,
l = 0, ml = 0. Since the electron in an s-state doesn’t have any orbital angular momentum,
it won’t see any magnetic field (“produced by the nucleus”) and therefore no spin orbit
coupling occurs. Note that the spin-orbit interaction is a relativistic effect, since an
electric field may transform into a magnetic field under Lorentz transformation, thus
producing an “internal Zeeman effect”.
7.26 Angular Momentum Calculations
Problem formulation: (a) What are the possible values of L for a system of two
electrons whose orbital quantum numbers are l1 = 1 and l2 = 3? (b) What are the
possible values of S? (c) What are the possible values of J?
9
Approach: These kind of problems are a bit “Sudoku-like”. One has to learn the
allowed rules. See section 7.8 in Beiser for more information. The rules are:
Quantity
L
S
J
Allowed values
|l1 − l2 |, |l1 − l2 | + 1,...,l1 + l2
|s1 − s2 |, |s1 − s2 | + 1,...,s1 + s2
|L − S|, |L − S| + 1,..., L + S
Solution: Using these rules we get
Particle 1
l1 = 1
s1 = 1/2
j1 = 3/2,1/2
Particle. 2
l2 = 3
s2 = 1/2
j2 = 5/2,7/2
Allowed values
L = 2,3,4
S = 0,1
J = 1,2,3,4,5
Discussion: You will learn more about the angular momentum theory in Kvantmekanik
III where chapter 3 in “Sakurai” is studied. And also how a group-theoretical approach
more rigorously defines the rules.
7.34 More Angular Momentum
Problem formulation: (a) What values can the quantum number j have for a d
electron in an atom whose total angular momentum is provided by this electron? (b)
What are the magnitudes of the corresponding angular momenta of electron? (c) What
are the angles between the directions of L and S in each case? (d) What are the term
symbols for this atom?
Approach: Here, we shall understand the letter notation for the angular momentum
quantum number. We must also understand the rules for angular momentum magnitudes
and angles. Usually angles between vectors can be calculated with scalar products and
we should remind ourselves on how that is done. The last part requires us to find the
rules for term symbols and apply them to the electron described.
Solution: The letter notation for angular momentum goes as follows: l = 0,1,2,3 ⇔
s,p,d,f . Thus we know that a d- electron has angular momentum number l = 2. That,
in turn, yields, by the rule j = l ± s where s = 12 for an electron, that j = 32 and j = 52 .
Moving on to the magnitude, it is given by
p
J = ~ j(j + 1)
√
√
which gives J = 215 ~ or J = 235 ~ for the values above.
Next, we should calculate the angle between L and S. Here we can use a nice trick.
Since we know that J = L + S and we have the magnitude J = |J| = |L + S|, we can
write
J 2 = L2 + S 2 + 2LS cos(θ),
p
p
√
√
where L = |L| = ~ l(l + 1) = 6~ (since l = 2), L = |S| = ~ s(s + 1) = 3/2~ (since
s = 1/2) and θ is the angle between L and S. Remember the formula for the scalar
product! We can rewrite and solve for this angle:
J 2 − L2 − S 2
J 2 − 6~2 − 3/4~2
√ p
cos(θ) =
=
≈
2LS
2 6/ 3/4~2
10
(
−0.707
0.471
√
if J = 15/2~
√
if J = 35/2~
which gives the angles θ3/2 ≈ 2.35 = 135◦ and θ5/2 ≈ 1.08 ≈ 62◦ .
Lastly, the term symbols are to be read as 2S+1 LJ which gives the states 2 D3/2
and 2 D5/2 respectively. One uses, as you might have guessed, capital letters for “total”
quantities.
Discussion: Problems of this kind might be a bit confusing, but after some practice
they should not be too hard. The trick with the scalar product relating different
magnitudes of the angular momentum quantities is extremely useful in many situations
and should be remembered.
7.38 Magnetic Moment of Chlorine
Problem formulation: The ground state of of chlorine is 2 P3/2 . Find its magnetic
moment. Into how many substates will the ground state split in a weak magnetic field?
Approach: From the previous problem, the term symbol should allow us to extract
the quantities S, L and J. From these we must figure out what the magnetic moment
is and how it relates to these quantities. Further, we should identify which mechanism
causes magnetic splitting and how it relates to the magnetic moment.
Solution: We get from deciphering the term symbol that S = 1/2, L = 1 and J = 3/2.
The magnetic moment µJ under LS-coupling (see problem 7.37 in Beiser ) has the
magnitude 2
p
µJ = µB gJ J(J + 1)
where µB =
given by
e~
2m
≈ 5.79 · 10−5 eV /T is the Bohr magneton and gJ is the Landé g-factor,
gJ = 1 +
J(J + 1) − L(L + 1) + S(S + 1)
.
2J(J + 1)
p
Inserting
our
numbers
gives
us
g
=
4/3
and
thus
µ
=
4/3
·
µ
3/2(3/2 + 1) =
J
J
B
p
2µB 5/3.
The meaning of a magnetic moment is that it describes the relation between angular
momentum (actually the torque, which is the change in angular momentum) and magnetic
field.
A weak magnetic field will reveal the values of MJ = −J,J + 1, ..., J ⇒ MJ =
±3/2, ±1/2 so that we get a total of 4 substates. This is the Zeeman effect.
Discussion: Spin orbit coupling reveals the “j-degeneracy”, while the Zeeman effect
reveals the “mj -degeneracy”.
7.40 The Kα Line in a X-ray Spectrum
Problem formulation:
What element has a Kα x-ray line of wavelength 0.144 nm.
Approach: For this problem, the relevant information is found in section 7.9 in Beiser.
We must understand the reason for X-ray emission and also remember what is meant by
the Kα line.
2 There
is a misprint in the book
11
Solution: One can produce x-rays by bombarding atoms with electrons. It may then
be so that an inner shell electron gets knocked away and an outer electron takes its place,
thereby emitting a photon with energy corresponding to the difference between the two
levels. The energies of the photons are high since the inner shell electrons see an almost
unshielded nucleus. The energy levels of a hydrogen-like atom are given by
Z2
,
n2
where E1 = −13.6 eV.We remind ourselves that the Kα -line is the n = 2 → n = 1
transition so that
En = E1
1
1
3
− 2 ) = −E1 · (Z − 1)2 ,
12
2
4
where the formula contains Z − 1 instead of Z due to screening from the remaining
K-shell electron. The energy of a photon is given by Eλ = hc/λ and with λ = 0.144 · 10−9
m we get
EKα = E1 (Z − 1)2 (
4.136 · 10−15 eV s · 2.998 · 108 m/s
≈ 8.6 · 103 eV.
0.144 · 10−9 m
Thus, if these energies are to be equal we shall have
Eλ =
(Z − 1)2 =
4 Eλ
≈ 292 ⇒ Z = 30.
3 −E1
We have that Z = 30 which corresponds to the element Zinc.
Discussion: An atom may instead of x-rays eject an electron instead. This is called
the Auger effect and can be exploited for chemical analysis. See the end of chapter 7 in
Beiser.
7.43 Singlet and Triplet States
Problem formulation:
two outer electrons.
Distinguish between singlet and triplet states in atoms with
Approach: This is again more of a reasoning problem rather than a computation
problem.
Solution/Discussion: A singlet state is a state where the two electron spins are
anti-parallel, |↑↓i, and “cancel out”. Thus S = 0 and ms = 0. Only one value of the
z-component is possible, hence the name singlet.
Since S 2 |↑i = 34 ~2 |↑i and S 2 |↓i = 34 ~2 |↓i, we can construct this state in bracketnotation as
1
|singleti = √ (|↑↓i − |↓↑i),
2
so that the total spin-operator S 2 = (S1 + S2 )2 has eigenvalue zero.
A triplet state is on the other hand when the spins are parallel, |↑↑i, and gives then
S = 1 and ms = −1,0,1. Three possible states gives the name triplet. We can construct
the triple state in three different ways with bracket notation as
1
|tripleti = √ (|↑↓i + |↓↑i) or |↑↑i or |↓↓i .
2
12
All of these give < S 2 >= 1.
The transition rule ∆S = 0 forbids transitions between singlet and triplet states.
13
3
Problem session 3: Molecules
8.5 Molecule Rotation inertia
Problem formulation: When a molecule rotates, inertia causes its bonds to stretch.
(This is why the earth bulges at the equator.) What effect does this stretching have on
the rotational spectrum of the molecule?
Approach: For this problem, we should understand the rotational spectra for diatomic
molecules (the simplest ones). We can find that in section 8.6 in Beiser.
Solution: We have that the rotational energies are given by:
1 L2
J(J + 1)~2
,
=
2 I
2m0 R2
where J is the rotational quantum number (J = 0,1,2,3...), m0 is the reduced mass
and R2 is the distance between the mass centra of the two atoms. Assuming now that
R → R+ > R due to stretching, we immediately see that EJ → EJ− < EJ because of the
inverse proportionality to R2 . So the spectrum is slightly shifted to lower energies.
EJ =
Discussion: It interesting to note that rotations along the molecular symmetry axis
(i.e. “along the bond”) can be neglected. This is due to two factors. Since nuclear mass
can be assumed to lie on this axis it gives zero moment of inertia about this axis. The
electrons however contribute to this rotation energy since they are concentrated in a
1
region whose radius is approximately half of the bond length. But their mass is ≈ 4000
1
of the total molecular mass. And since E ∼ I , the electron contribution to rotational
energies are ∼ 104 times higher than the end over end rotations. This corresponds to
several electron volts, comparable to the bond energies. So the molecule would break in
any environment allowing this kind of transition.
8.7 Rotational Spectrum As an Isotope Identifier
Problem formulation: The J = 0 → J = 1 rotational spectrum line occurs at
1.153 · 1011 Hz in 12 C 16 O and at 1.102 · 1011 Hz in ? C 16 O. Find the mass number of the
unknown carbon isotope.
Approach: Here, we must find a way to relate mass to rotational energies. That is
done through the reduced mass and we should connect the relation of the known isotope
to the unknown one.
Solution: The rotational spectrum is again given by (see the previous problem for
information about the terms, although the reduced mass is now denoted by m):
EJ =
1 L2
J(J + 1)~2
=
.
2 I
2mR2
Looking at the J = 0 → J = 1 energy we get ∆E ≡ E1 − E0 = m0~R2 . The frequency
corresponding to this transition is given by ν = ∆E/h = 2πm10 R2 . We note that if ν 0 < ν
then m0 > m. This means that the unknown mass number should be larger than that
for 12 C.
Next we set mC = 12mp and mO = 16mp where mp is the proton mass (we assume
12·16
that the neutron and proton mass are the same). Thus we have m = 12+16
mp = 48
7 mp .
16M
0
We further denote the unknown mass by M mp so that m = 16+M mp .
14
0
Now we can extract the unknown reduced mass m0 since we have that νν0 = m
m.
Note that we can cancel the factors of R2 since we are dealing with different isotopes.
Additional neutrons don’t add any charge which can alter the bond to that the bond
length stays the same. Simplifying, we can write
1.153 ! m0
ν
7 M
16 · 0.448
=
=
=
⇒M =
≈ 13.
ν0
1.102
m
3 16 + M
1 − 0.448
13
Thus, the unknown isotope is
C.
Discussion: You should notice that this kind of problem is quite common. A nice
thing to remember is that one usually saves a lot of work by canceling different constants.
For example, I never used the value of the proton mass. As have been said before, the
general rule of thumb is; never put in numbers until the end.
8.9 Rotational Spectrum for HCl
Problem formulation:
lengths:
The rotational spectrum of HCl contains the following waveλ1 = 12.03 · 10−5 m
λ2 = 9.60 · 10−5 m
λ3 = 8.04 · 10−5 m
λ4 = 6.89 · 10−5 m
λ5 = 6.04 · 10−5 m.
If the isotopes involved are 1 H and 35 Cl, find the distance between the hydrogen and
chlorine molecule in an HCl molecule.
Approach: Once again, we should use the rotational spectrum and connect it to
various transitions. The bond distance should be extractable from the moment of inertia
I = m0 R2 . We must find which transitions the given wavelengths correspond to.
Solution: The transition energy ∆EJ 0 →J = EJ 0 − EJ is given by
∆EJ 0 →J =
~2 0 0
~2
(J (J + 1) − J(J + 1)) =
n = αn,
2I
I
2
∆E
where α = ~I and n is an integer. The wavelength is as usual given by hc
λ = h ⇒λ=
β 1
hc
∆E = α n , where β = hc.
Next, we can relate different transitions to the number n in the following table:
J 0 \J
1
2
3
4
5
6
7
8
0
1
3
6
10
15
21
28
36
1
x
2
5
9
14
20
27
35
2
x
x
3
7
12
17
24
33
3
x
x
x
4
9
15
22
30
15
4
x
x
x
x
5
11
18
26
5
x
x
x
x
x
6
13
21
6
x
x
x
x
x
x
7
15
7
x
x
x
x
x
x
x
8
8
x
x
x
x
x
x
x
x
This table should be interpreted as taking a J 0 and a J gives the corresponding
n
n according to the definition above. Now we use the fact that λλji = nji (once again,
constants cancel out) so we can relate two wavelengths to their ratio of ns. We use this
for the given wavelengths and find:
λ1
λ2
=
12.03
9.60
≈
5
4
λ1
λ3
λ1
λ4
λ1
λ5
=
=
=
12.03
8.04
12.03
6.89
12.03
6.04
≈
≈
≈
6
4
7
4
8
4
⇒ n1 = 4, n2 = 5: λ1 ∼ J 0 = 4 → J = 3
⇒ n1 = 4, n2 = 5:λ2 ∼ J 0 = 3 → J = 1 or J 0 = 5 → J = 4
⇒ n3 = 6: λ3 ∼ J 0 = 3 → J = 0 or J 0 = 6 → J = 5
⇒ n4 = 7: λ4 ∼ J 0 = 4 → J = 2 or J 0 = 7 → J = 6
⇒ n5 = 8: λ2 ∼ J 0 = 3 → J = 1 or J 0 = 8 → J = 7
Note that a specific n corresponds to one or more possible transitions. We see that
everything is consistent if assume n1 = 4 and thus nk = 3 + k, where k = 1,2,3,4,5.
Finally we can write:
r
~2 n
~ nk λ k
~2 nk λk
0 2
mR =I
.
=
⇒R=
∆Enk
hc
2πc m0
1·35
The reduced mass can be obtained as m0 = 35+1
mp = 35
36 mp . Inserting this for n1 = 4
−5
and λ1 = 12.03 · 10 m, we get:
s
1.055 · 10−34 J · s 4 · 12.03 · 10−5 m
R=
≈ 0.129 nm.
−24 kg
2π · 2.998 · 108 m/s 35
35 · 1.67 · 10
Physics handbook gives the value 0.127 nm so we are quite close.
Discussion: This problem is a bit Sudoku-like when trying to find the transitions.
The important thing to check now and then is the consistency. Since we have several
possibilities for the higher transitions, one has to assume a value and check that it works.
From the table above, we see that n1 = 4 is the lowest possible value (since 5/4 is the
least fraction possible for n2 /n1 ) and trying that is a good start.
8.13 Correspondence Principle
Problem formulation: In section 4.6 it was shown that, for large quantum numbers,
the frequency of the radiation from a hydrogen atom that drops from an initial state of
quantum number n to a final state of quantum number n − 1 is equal to the classical
frequency of revolution of an electron in the nth Bohr orbit. This is an example of
Bohr’s correspondence principle. Show that a similar correspondence holds for a diatomic
molecule rotating about its center of mass.
Approach: This problem should be straight forward given the information above. We
should relate a transition frequency to some quantized quantity related to the rotation.
Solution: Once again we have that
J(J + 1)~2
~2 J
⇒ ∆EJ→J−1 =
.
2I
I
The corresponding photon angular frequency for this transition is then
EJ =
ω = ∆EJ→J−1 /~ =
16
~J
.
I
We also have that
p
LJ = IωJ ⇒ ωJ =
J(J + 1)~
~J
≈
, when J 1.
I
I
Thus, we see that for large J, that is J 1, ω ≈ ωJ , which manifests the correspondence
principle.
Discussion: Generally, the correspondence principle states that quantum theory for
very high quantum numbers should reproduce classical results. One can also say that
the classical theory is a restricted domain of quantum theory,
8.15 Zero Point Energy of Hydrogen Isotopes
Problem formulation: The hydrogen isotope deuterium as an atomic mass approximately twice that of ordinary hydrogen. Does H2 or HD have the greater zero point
energy? How does this affect the binding energies of the two molecules?
Approach: We should by know understand that vibrational spectra for diatomic
molecules are described by the harmonic oscillator. Its zero point energy is defined as
the ground state energy, n = 0.
Solution: We have that E0 =
~
2ω
=
~
2
q
k
m0 .
m0HD
The reduced mass for the hydrogen
2mH mH
= 2m
molecule is
that
= 2m3 H .
H +mH
Since HD is
one neutron added to one of the atoms we may
assume that the potential well is unchanged since there is no change in electrical charge
and we can write that kH2 = k√HD .
√
Then we see that E0HD = 23 E0H2 ( 3/2 ≈ 0.87). The zero point energy for HD is
thus smaller than that for H2 . This means that the HD-molecule vibrational spectra
lies deeper in the potential well and has thus a larger binding energy!
m0H2
= m2H which implies
a H2 -molecule with
Discussion: The parabolic potential well describing the harmonic oscillator is a very
nice approximation to the bottom of a more realistic potential, for example the Morse
potential.
8.16 Zero Vibrational and Rotational Energy
Problem formulation:
tional energy?
Can a molecule have a zero vibrational energy? Zero rota-
Approach: Here we shall try to identify the differences or similarities between the
vibrational spectrum and the rotational spectrum.
Solution: The vibrational spectrum is given by En = ~ω(n + 12 ) with n = 0,1,2,3...
so that the lowest possible energy is E0 = ~ω
2 . We see that there must be at least this
vibrational energy in the molecule system.
2
The rotational energy spectrum is given by EJ = J(J+1)~
with J = 0,1,2,3... Simply
2I
putting J = 0 we have that E0 = 0 and there can be zero rotational energy in the
molecule system.
17
Discussion: The zero point energy of the harmonic oscillator has some very interesting
consequences. In quantum field theory, is means that the vacuum state is not actually
empty but contains some energy. One can remedy this by stating that it is only relevant
to talk about energy differences anyway, but this does not work in general relativity.
There, absolute energy is in fact important.
8.18 Return of the Harmonic Oscillator
Problem formulation: Assume that the H2 molecule behaves exactly like a harmonic
oscillator with a force constant of 573 N/m. (a) Find the energy (in eV) of its ground
and first excited vibrational states. (b) Find the vibrational quantum number that
approximately corresponds to its 4.5-eV dissociation energy.
Approach: This problem should be straight forward having some experience of the
harmonic oscillator.
Solution: Again we write En = ~ω(n + 12 ). Further, ω =
0
m
eV
= m2H . Then we get that ~ω
and E1 = 32 ~ω ≈ 0.82 eV.
q
k
m0
and we use that
≈ 0.54 eV. We proceed by calculating E0 = 12 ~ω ≈ 0.27
For the second part, we can write ED = En = ~ω(n + 12 ) = 4.5 eV. Re-arranging
1
gives n = 4.5eV
~ω − 2 = 4.5/0.54 − 0.5 ≈ 7.83 ≈ 8. Note that by the Boltzmann relation
(see the discussion of the next problem) 4.5 eV ∼ 54000 K which is the temperature need
to “split” the hydrogen molecule by vibrating it.
Discussion: Being handy with the harmonic oscillator both in theory and “practice”
is very handy and makes it possible to do many calculations fast. Again, it is good to
use proper units.
8.21 Room Temperature Vibrations
Problem formulation: The bond between the hydrogen and chlorine atoms in a
H 35 Cl molecule has a force constant of 516 N/m. Is it likely that an HCl molecule will
be vibrating in its first excited vibrational state at room temperature?
1
Approach: We should try to figure out the energy required for the mentioned transition
and also how room temperature comes into play.
Solution: We note that the vibrational spectrum is given by
1
En = ~ω(n + ),
2
q
where ω = mk0 , k being the force constant constant and m0 the reduced mass. The
energy between the first excited state and the ground state can then be determined to be
r
1
1
k
∆E = E1 − E0 = ~ω(1 + − 0 − ) = ~ω = ~
.
2
2
m0
q
1·35
36·k
We now calculate the reduced mass as m0 = 1+35
mp which gives ∆E = ~ 35m
=
p
q
36·516N/m
6.582 · 10−16 eV s 35·1.67·10
−27 kg ≈ 0.37eV .
18
The famous Boltzmann relation E = kB T , with the Boltzmann constant kB =
8.617 · 10−5 eV/K, states that room temperature T = 300 K corresponds to an energy
of approximately 26 meV. We can thus deduce that at room temperature, the energy
content is by no means near the energy required to excite the HCl-molecule. It is
therefore not likely to find it in an excited vibrational state.
Discussion: The relation E = kB T is used very often and it is useful to memorize
some values for this relation. For example that TRoom ∼ 26 meV and also that 1eV
∼ 12000 K. You will learn more about this in chapter 9 in Beiser.
19
4
Problem session 4: Time Dependent Perturbation
Theory
Preliminary Theory
The material for first order time dependent perturbation theory is found in chapter 9 in
Beiser, but a short resume is given here:
If a system initially is in the state |ai at time t = t0 and a time dependent
perturbation,H 0 (t), is turned on, the state will at time t be in the state
X (1)
|ψ(t)i = |ai e−iEa /~ +
cb (t) |bi e−iEb /~ ,
b6=a
(1)
where cb (t) = − ~i
Rt
t0
0
Eb −Ea
. The probability of finding
~
(1)
= |cb (t)|2 . Remember that this is
0
Hba
(t0 )eiωba t dt0 and ωba =
the system in the state |bi is then given by Pa→b (t)
valid for first order perturbation theory.
Problem 1, Hydrogen Atoms in an Electrical Field
A system of hydrogen atoms in their respective ground states are situated between the
electrodes in a capacitor. The capacitor consists of two parallel plates in vacuum. A
voltage pulse is applied over the capacitor, which yields a homogeneous electrical field:
E(t) = E0 e−t/τ , for t > 0 and E(t) = 0 otherwise.
(a) Show that after long time, the proportion of atoms in the 2p(m = 0) state is, in
first order approximation, given by
215 a20 e2 E02
,
310 ~2 (ω02 + τ12 )
where a0 is the Bohr radius and ~ω is the energy difference between the 2p− and the
ground state.
(b) What is the proportion of atoms in the 2s state?
Solution suggestion
The electrical field perturbation is added to the hydrogen Hamiltonian by the term
H 0 (t) = ezE0 e−t/τ , for t > 0 and H 0 (t) = 0 otherwise.
This is motivated by the electrical potential energy which has the form −eV = − dE
dz in
one dimension, where we have chose our coordinate system such that the electrical field
points in the z-direction. The electron charge is −e.
Now we can calculate the matrix elements of H 0 (t). They are given by
0
Hba
(t) = hb| ezE0 e−t/τ |ai = eE0 e−t/τ hb| z |ai ≡ eE0 e−t/τ Tba .
The first order approximation for the amplitude of finding the atom in the 2p-state
(m = 0 is suppressed for now) starting of in the 1s-state is given by
20
(1)
c2p (t)
i
=−
~
Z
t
0
0
H2p,1s
(t0 )eiω0 t dt0
0
i
= − eE0 T2p,1s
~
t
Z
0
1
et (iω0 − τ ) dt0 =
0
1
i
et(iω0 − τ ) − 1 t→∞ i
1
− eE0 T2p,1s (
) →
eE0 T2p,1s
1
~
~
iω0 − τ
iω0 −
1
τ
.
Then we have after long time that the transition probability is given by
(1)
P1s→2p = |c2p (∞)|2 =
1
1
|T2p,1s |2 e2 E02 2
~2
ω0 +
1
τ2
.
We move on by calculating the matrix element
Z ∞
T2p,1s = h2p| z |1si =
d3 x ψ ∗ (x)2p · z · ψ(x)1s .
−∞
We switch to spherical coordinates and use the analytical forms of the hydrogen wave
functions. They can be found for example in Physics Handbook and the relevant ones
are given by
ψ2s (r,θ,φ) = ψ100 (r,θ,φ) = R10 (r)Y00 (θ,φ) =
2
3/2
a0
1
e−r/a0 · √
4π
r
1
ψ2p,m=0 (r,θ,φ) = ψ200 (r,θ,φ) = R20 (r)Y10 (θ,φ) = √ 3/2 e
2 6a0
−r/(2a0 )
3
cos(θ),
4π
where a0 is the Bohr radius. Inserting this into the expression above and using that
d3 x = r2 drdΩ and z = r cos(θ) we get
Z ∞
Z
3
∗
T2p,1s =
drr R10 (r)R20 (r) · dΩY10
(θ,φ)Y00 (θ,φ) ≡ Ir · IY ,
0
where we have exploited the separation of variables to get two integrals we can calculate
separately. We get
Ir = √
1
6a40
∞
Z
0
3 r
1 2
drr4 e− 2 a0 = {r = 2/3a0 x} = √ ( )5 a0
6 3
Z
0
∞
a0 28
dxx4 e−x = √ 4 ,
63
where we have used that the x-integral is actually Γ(4) = 24. Next, we move on to the
other integral:
√ Z
√ Z +1
3
3
1
2
IY =
dΩ cos (θ) =
d(cos(θ)) cos2 (θ) = √ .
4π
2 −1
3
Multiplying the two integrals we then get that T2p,1s =
probability is finally given by
P1s→2p =
1
1
|T2p,1s |2 e2 E02 2
2
~
ω0 +
21
1
τ2
=
a0 28
√
235
and the transition
215 a20 e2 E02
310 ~2 (ω02 + τ12 )
If we instead calculate the P1s→2s , we can proceed just as above. But when we
calculate the IY -integral this time we get
Z
Z
1
1
1 +1
d(cos(θ)) cos(θ) = 0,
IY = dΩ √ cos(θ) √
=
2 −1
4π
4π
so that the total probability is actually zero. The physical reason for this is the transition
rule ∆l = ±1 which originates from the fact that photons carry angular momentum
(spin) which is 1 which must be conserved in a transition. The transition 1s → 2s would
violate this law and must therefore be zero.
Problem 2, Perturbed Harmonic Oscillator
Calculate in the lowest non trivial approximation the probability that an electron in the
ground state in a simple harmonic oscillator at t = −∞ will be in an excited state at
t = ∞ if it is perturbed by an electrical field
t −2
E0 e−( τ )
√
E(t) =
πτ
The electron remains bound by the simple harmonic potential. (In this problem, it
is useful to represent theqelectron position, say x, in terms of step operators for the
~
2mω (a
harmonic oscillator, x =
+ a† ).
Solution suggestion
In this problem we can apply the theory pretty much in the same way as above and we
write the perturbative term as
2
2
exE0
H 0 (t) = √ e−t /τ .
πτ
Here x denotes our spatial direction (in this problem we are dealing with a one dimensional
system in contrast to the problem above).
eE0 −t2 /τ 2
0
With this perturbation, the matrix elements are given as Hba
(t) = √
e
hb| x |ai
πτ
≡
eE0 −t2 /τ 2
√
e
Tba .
πτ
The first order approximation for the amplitude of finding the atom in the b-state
(we start as stated from the state |ai = |0i) is given by
(1)
Z
Z t
0
t02
i t 0 0 iω0 t0 0
i eE0
Hb0 (t )e
dt = − √ Tb0
e(ibωt − τ 2 ) dt0 =
~ 0
~ πτ
−∞
i eE0
≡ − √ Tb0 · I(t).
~ πτ
cb = −
We move on and calculate I(t):
Z
t
I(t) =
02
(ibωt0 − tτ 2 )
e
−∞
t0
dt = {x = } = τ
τ
0
Z
t
e−x
2
+ibωτ x
dx.
−∞
Now we let t → ∞ and denote α ≡ ibτ ω. Then we get
Z
∞
I(∞) = τ
e−x
2
+αx
√
√
2
2 2 2
dx = τ πeα /4 = τ πe−ω b τ /4 ,
−∞
22
by using the standard formulas for Gaussian integrals. Note that i doesn’t matter that α
is a complex number. This can be shown by extending the integral to the complex plane,
choosing an appropriate contour integral and using that the function to be integrated is
actually analytic on the whole complex plane.
Next, we calculate Tb0 by using the ladder representation of the x-operator:
r
Tb0 = hb| x |0i =
~
hb| (a + a† ) |0i =
2mω
r
~
,
2mω
where only b = 1 yields a non-zero result.
Combining our results we get
(1)
c1 (∞)
2 2
i
= − e−ω τ /4
~
r
~
eE0 ,
2mω
and then
P0→1 (∞) =
(1)
|c1 (∞)|2
=
e−ω
2 2
τ /2 2
e E02
.
2m~ω
Note here that in first order perturbation theory, only transitions to the first excited
state is possible. Other transitions become manifest in higher orders.
Problem 3, Quantum Atom-Photon Interaction
Spontaneous emission of light from excited atoms can not be explained by a classical
model of light, but can in a quantum mechanical description. A light quantum, that
is a photon, is then described by its wave vector and polarization state, collectively
called a mode and denoted here by k. The modes are well described by harmonic
oscillators where the excitation level for each mode corresponds to the number of photons
nk and the corresponding states are denoted |nk i. The number of photons in a mode
is not limited, since the photons are bosons with spin S = 1. Interaction (electrical
dipole approximation) with an atom (two level model) can then be represented by the
Hamiltonian operator:
r
X
~ωk
0
H (t) = ie
(d21 ak e−iωk t |2i h1| − d12 a†k e+iωk t |1i h2|),
20 V
k
q
~ωk
where 20 V is the so called electrical field per photon and dij |ii hj| is the dipole operator
between the atomic states i and j.
The combined atom-photon states can be written as |nk i |ii ≡ |nk , ii, where i = 1,2
on which dij and ak , a†k acts separately on the atomic and photonic parts respectively.
(a) Calculate the matrix elements for absorption and emission for a photon in a given
mode k. Classically, light must always be present for interaction. What happens in
quantum mechanics?
(b) Assume that the atom initially is in the state |2i. Show by first order perturbation
theory that the Einstein coefficient for spontaneous emission is given by
A21 =
πe2 X
ωk |d12 |2 δ(ωk − ω12 ).
~0 V
k
23
The following representation of Dirac’s delta function
sin2 ( t (ω0 −ω))
δ(ω0 − ω) = (2/π)limt→∞ t(ω20 −ω)
might be useful here.
2
Solution suggestion
Our perturbation is in this problem given by
r
X
~ωk
H 0 (t) = ie
(d21 ak e−iωk t |2i h1| − d12 a†k e+iωk t |1i h2|).
20 V
k
We calculate the matrix elements for a given k by multiplying with combined bras
and kets for that k. Then only terms for that particular k give contributions:
r
√
~ωk
(d21 e−iωk t hi|2i h1|ji hmk |nk − 1i nk
20 V
√
− d12 e+iωk t hi|1i h2|ji hmk |nk + 1i nk + 1) =
r
√
~ωk
ie
(d21 e−iωk t nk δi,2 δj,1 δmk ,nk −1 −
20 V
√
d12 e+iωk t nk + 1δi,1 δj,2 δmk ,nk +1 ).
0
Hm
(t)
k ,i;nk ,j
= ie
We see now that we need nk − mk = ±1 for a transition between the atomic states
1 ↔ 2. This can be interpreted as transitions involves absorption or emission of a photon.
Specific matrix elements can now be calculated as
r
√
~ωk
d21 e−iωk t nk
20 V
r
√
~ωk
d12 e+iωk t nk + 1
Hn0 k +1,1;nk ,2 (t) = −ie
20 V
Hn0 k −1,2;nk ,1 (t)
= ie
So if there are no photons present, nk = 0, there can be no transition from 1 → 2 but
it is still possible for 2 → 1 to occur and it is accompanied by the creation of a photon.
The element for emission is also slightly larger than that for absorption and they both
grow with nk .
Next, we use our perturbation formula again and we assume that we initially, at t = 0
have the atom in state |2i with zero photons present. We have then
Z
0
i t 0
(1)
cnk +1,1 (t) = −
H
(t0 )e−iω12 t dt0 =
~ 0 1,1;0,2
r
Z t
√
0
~
i · ie
d12 0 + 1
eit (ωk −ω12 ) dt0 ≡
−
~
20 V
0
r
i · ie
~
−
d12 I(t).
~
20 V
Looking at the integral part we have
#t
" 0
Z t
2 sin( 2t (ωk − ω12 ))
t
eit (ωk −ω12 )
it0 (ωk −ω12 ) 0
e
dt =
= ei 2 (ωk −ω12 )
.
i(ωk − ω12 )
(ωk − ω12 )
0
0
Combining the integral with our previous terms we then get
24
(1)
cnk +1,1 (t)
e
=
~
r
2 sin( 2t (ωk − ω12 ))
t
~
d12 · ei 2 (ωk −ω12 )
,
20 V
(ωk − ω12 )
and further
(1)
Pk (t) = |cnk +1,1 (t)]2 =
2 sin2 ( 2t (ωk − ω12 ))
e2 ωk
,
|d12 |2
~0 V
(ωk − ω12 )2
The total probability for a transition is then a sum over all values of k and we write
the Einstein coefficient as the mean transition rate over a time t. Hence,
A21 =
2 sin2 ( 2t (ωk − ω12 ))
P (t) X Pk (t)
e2 X
.
=
=
|d12 |2 ωk
t
t
~0 V
t(ωk − ω12 )2
k
k
Next, we let t → ∞ and use the given identity (2/π)limt→∞
ω12 ). This gives us the final result
A21 =
sin2 ( 2t (ωk −ω12 ))
t(ωk −ω12 )2
= δ(ωk −
πe2 X
ωk |d12 |2 δ(ωk − ω12 ).
~0 V
k
The identity can be understood qualitatively as follows:
For ω 6= 0 we have that sin2 (ωt/2)/(tω 2 ) has a bounded nominator. When t → ∞
the function goes to zero because of the denominator is blowing up. When ω ≈ 0 we
2 2
Taylor expand the sine function and get sin2 (ωt/2)/(tω 2 ) ≈ 14 ωω2tt = t/4 which grows
without bound as t → ∞. This exactly mimics the behavior of the delta function. But
beware, the identity should be thought of as valid inside integrals just as delta functions
should. The π factor is needed to normalize the function.
25
5
Problem session 5: Quantum Statistics
9.5 Population of Hydrogen Rotational States
Problem formulation: The moment of inertia of the H2 molecule is 4.64 · 10−48
kg · m2 . (a) Find the relative populations of the J = 0,1,2,3 and 4 rotational states at
300 K. (b) Can the populations of the J = 2 and J = 3 states ever be equal? If so, at
what temperature does this occur?
Approach: Here, we should understand and apply the introduction section of chapter
9 in Beiser. If we understand this, the rest should be straight forward.
2
Solution: The energies of the rotational states are give by EJ = J(J+1)~
with a
2I
degeneracy of 2J + 1 for each level. (Remember that mJ = −J, − J + 1,...,J − 1,J).
Since H2 is a molecule, we apply the Maxwell-Boltzmann (MB) distribution. It says that
the number of molecules in the energy state is given by
n() = A · g() · e−/(kB T ) ,
where A is a normalization constant, g() is the degeneracy of the energy , kB is
Boltzmann’s constant and T is the temperature. Using this on our H2 molecules, we get
n(J ) = A · (2J + 1) · e−
J(J+1)~2
2I
/(kB T )
.
We see immediately that n(0 ) = A. Inserting T = 300 K, I = 4.64 · 10−48 kg · m2 and
J = 1,2,3,4 we get
n(1 )
n(0 )
n(2 )
n(0 )
n(3 )
n(0 )
n(4 )
n(0 )
= 1.68
= 0.88
= 0.21
= 0.028
To see which temperature that gives an equal ratio between n(2 ) and n(3 ), we write
1=
2
n(3 )
2 · +1
− ~
(2(3+1)−2(2+1))
=
· e 2IkB T
,
n(2 )
2·2+1
and solve for T . We get
1=
7 − Ik3~2T
1 3~2
·e B ⇒T =
≈ 1548K.
5
log( 75 ) IkB
Heating the H2 molecules to this temperature, gives equal populations for the J = 2 and
J = 3 states.
Discussion: We apply the MB-distribution because we treat the hydrogen molecules
as non interacting classical particles. This works fine for a diatomic gas.
26
9.14 Neutron Beam Density
Problem formulation: A flux of 1012 neutrons/m2 emerges each second from a port
in a nuclear reactor. If these neutrons have a Maxwell-Boltzmann energy distribution
corresponding to T = 300 K, calculate the density of neutrons in the beam.
Approach: We must relate flux to density in some way and understand what the
MB-distribution tells us about this relation.
Solution: Particle flux, φ is defined as the number of particles crossing a unit area per
N
mN
mN
second. We can write φ = A·t
. The density however, is given by ρ = mN
V = A·L = A·v·t .
m
So we see that ρ = φ v . Here m is the neutron mass, V = A · L is the volume and v is the
particle velocity. Here we have assumed that all particles have the same velocity v = L/t.
But which velocity? We choose to take the
q mean velocity from the MB-distribution that
the neutrons follows. It is given by v =
we get:
ρ=φ
8kB T
πm
. Inserting this into the density formula
m
m
1.6736 · 10−27 kg
= φq
≈ 7 · 10−19 kg/m3 .
≈ 1012 m−2 s−1
v
2510m/s
8kB T
πm
Discussion: This density is not very high. The number of particles per cubic meter
is ρ/m = φ/v ≈ 4 · 108 m−3 . It is in fact lower than the number of particles per cubic
meter in the atmosphere of the moon 4 · 1011 m−3 . 3
9.15 Pressures Of Distributions
Problem formulation: At the same temperature, will a gas of classical molecules, a
gas of bosons, or a gas of fermions exert the greatest pressure. The least pressure? Why?
Approach: This is one of these discussion questions again. The information can be
found in section 9.4 in Beiser (surprise!).
Solution/Discussion: Since the fermions follow the Fermi-Dirac (FD) distribution
they will have a high amount of high energy particles since the exclusion principle forbids
double occupancy. This is most easily seen at low temperatures where the fermions
occupy states to the Fermi level. Bosons, on the other hand, tend to fill the same states
(can be seen from the Bose Einstein (BE) distribution) and at low temperatures they
will preferably fill low energy states. Classical molecules follow the MB distribution and
should fall in between. Since pressure is directly related to particle energy. The energy
∂E
“rankings” imply the same “pressure ranking”. P = ∂V
|T =const .
9.20 Sunspots
Problem formulation: Sunspots appear dark, although their temperature are typically 5000 K, because the rest of the sun’s surface is even hotter, about 5800 K. Compare
the radiation rates of surfaces of the same emissivity whose temperatures are respectively
5000 and 5800 K.
3 See
for example: en.wikipedia.org/wiki/vacuum
27
Approach: This problem should be straight forward if one knows the Stefan-Boltzmann
law.
Solution: The Stefan-Boltzmann law states that the energy radiated per unit area per
second, denoted R, is proportional to the emissivity, (which depends on the material)
times the temperature to the power 4 times a constant σ = 5.670 · 10−8 W/(m2 · K 4 ). We
4
R5800
58004
write R = σT 4 . We then immediately get R
= σ5800
σ50004 = 50004 ≈ 1.8.
5000
Discussion: The emissivity takes values from 0, a perfect reflector which doesn’t
radiate at all, to 1, for a blackbody.
9.33 Gas Cloud Properties
Problem formulation: A gas cloud in our galaxy emits radiation at a rate of 1.0 · 1027
W. The radiation has its maximum intensity at a wavelength of 10µm. If the cloud is
spherical and radiates like a blackbody, find its surface temperature and its diameter.
Approach: We need to relate the maximum intensity wavelength to the temperature
and also find some relation between the area/radius of a sphere to the radiated power.
Wien’s displacement law and the Stefan Boltzmann law should do the trick.
Solution: We have from Wien that λmax T = 2.898 · 10−3 m · K which immediately
givesqus T ≈ 290K. Stefan and Boltzmann tell us that R = I/A = I/(4πr2 ) = σT 4 ⇒
r=
I
4πσT 4 ,
where we have used that the cloud is a blackbody implying = 1. Inserting
q
1027 W
11
numbers gives us r = 4π5.670·10−8 W/(m
m. Then the diameter
2 ·K 4 )·2904 K 4 ≈ 4.45 · 10
is simply the double of this value: d = 2r ≈ 8.9 · 1011 m.
Discussion: The radius is very big and in comparison, the sun diameter is ≈ 1.4 · 109 m.
9.37 Free Electron Gas
Problem formulation: Show that the median energy in a free-electron gas at T = 0
is equal to F /22/3 = 0.630F .
Approach: Here, we should figure out what is meant by the median energy. And also
how we can find it from the FD-distribution, which governs the behavior of electrons
which are fermions.
Solution: One should figure out that for a system of N electrons, the median energy
is the energy of the N/2:th electron state. For T = 0, we know that the number of
electrons in the energy range to d is given by
3N −3/2 √
d for < F and 0 for > F .
2 F
The median energy, m can then be obtained by finding the energy that corresponds
to N/2 electrons. We can write
n()d =
N
=
2
Z
0
m
3N −3/2 √
3N −3/2 2 3/2
F
d =
⇒ m = 2/3 .
2 F
2 F 3 m
2
28
There is a simpler way of obtaining this result. We have that the Fermi energy is
~2 3N 2/3
given by F = 2m
( 8πV ) . Writing m as “the fermi energy for N/2 electrons” we get
~2 3N/2 2/3
F
immediately m = 2m
( 8πV )
= 22/3
.
Discussion: The FD-distribution explains a wide variety of phenomena, for example
why not all electrons in a metal contribute to the conductivity.
9.49 Classical Limit of Bose Einstein Distribution
Problem formulation:
The formulation is very long. See Beiser, page 333.
Approach: The approach is quite straight forward from the hints and descriptions in
the problem formulation.
Solution: The number of particles in the
where

1


 eα e kB T −1
1
f () =
α kB T


 e e − k +1
A · e BT
energy range is given by n()d = g()f ()d,
for bosons
for fermions
for classical particles
We have that the Helium atoms have spin 0 so they are bosons. And further, we notice
that for eα e kB T 1, both the FD and BE distributions become the MB distribution
with A = e−α . Now, in the energy range ≈ kB T , we see that the identification becomes
eα 1. Let us use the MB- distribution and check whether A = e−α 1 or not under
the given circumstances.
We write
Z √
Z
Z
4 2πV m3/2 √ − k T
− k T
e B d =
N = g()fBE ()d ≈ A g()e B d = A
h3
√
√
√
Z
√ − k T
4 2πV m3/2
4 2πV m3/2 π
AV
B d = A
=A
e
(kB T )3/2 = 3 (2πkB T m)3/2 .
3
3
h
h
2
h
−3/2
3
Then we solve for A and get A = N
. We have from the problem
V h (2πkB T m)
23
3
formulation that N/V = 1000 · 6.022 · 10 /(22.4m ) and m = 4u = 4 · 1.66 · 10−27 kg.
This gives us A ≈ 3.6 · 10−6 1 so the assumption valid!
Discussion: For really low temperatures, He has some remarkable properties, such as
zero viscosity, extremely effective heat transport and zero entropy.
29
6
Problem session 6: Solid State Physics
10.2 Madelung Constant of NaCl
Problem formulation: Show that the five first terms in the series of the Madelung
12
24
constant of NaCl are α = 6 − √
+ √83 − 62 + √
− ...
2
5
Approach: We must understand the structure of the NaCl crystal which is a
Solution: The structure of NaCl is the so called face centered cubic structure. Each
atom sits in the corner of a cube of distance r0 with six nearest neighbors of the other
kind. One may draw a picture to visualize the structure. Now, the Madelung constant
can be viewed as the sum of nearest neighbor hierarchies divided by their respective
distance (in units of r) to a specific atom. The charge of the atom kind must be accounted
for:
−α = ±
# nearest neighbours
# next nearest neighbours
±
± ...
distance to nearest neighbors distance to next nearest neighbors
We can do this in a systematic way. Let us put a Na+ atom at coordinate (0,0,0).
The nearest neighbors are then Cl− atoms at positions (0,0, ± 1), (0, ± 1,0) and (±1,0,0),
i.e. at distance 1. So we have 6 nearest neighbors with charge −1 (in units of e). The
first term is thus −6/1 = 6. The next term in the series is then Na+ atoms at positions
(±1, ± 1,0), √
(±1,0, ± 1) and (0, ± 1,√± 1). That gives us in total 12 atoms with charge +1
at distance 2. The term is +12/ 2.
The next next nearest neighbors are Cl− at distances
(±1, ± 1, ± 1) which are 8 atoms
√
in total, with charge −1. We get the term −8/ 3.
Third order neighbors are Na+ atoms at positions (±2,0,0), (0, ± 2,0) and (0,0 ± 2).
We get the term +6/2.
Finally, fourth order is given by Cl− at (±1,±2,0), (±2,±1,0), (0,±1,±2),
√ (0,±2,±1),
√
(±1,0, ± 2) and (±2,0, ± 1)√giving 4 · 6 = 24 atoms in total with distance 22 + 12 = 5.
We obtain the term −24/ 5.
Adding up all terms and swapping the minus sign from the definition, we get:
12
8
6
24
α = 6 − √ + √ − + √ ...
2
3 2
5
Discussion: There is a serious problem with the method used above. The sum does
not converge! The problem arises since we add larger and larger terms with alternating
signs. Another way of thinking of it is that we add atoms with the same sign from spheres
which radii grows as r while the number of atoms on the sphere grows approximately as
r2 . So the sum grows as r.
A method to make this sum convergent (which is possible but the book cheats a bit
by omitting this issue), is to sum over cubes with a specific side, containing atoms of
both signs. It is possible to show that the sum will converge in this case to the value
1.748. See for example http://en.wikipedia.org/wiki/Madelung_constant.
10.3 Cohesive Energies
Problem formulation: (a) The ionization energy of potassium is 4.34 eV and the
electron affinity of chlorine is 3.61 eV. The Madelung constant for the KCl structure is
1.748 and the distance between ions of opposite sign is 0.314 nm. On the basis of these
30
data only, compute the cohesive energy of KCl. (b) The observed cohesive energy of KCl
is 6.42 eV per ion pair. On the assumption that the difference between this figure and
that obtained in a is due to the exclusion-principle repulsion, find the exponent n in the
formula Br−n for the potential energy arising from this source.
Approach: We should be clear what the different energies mean and be careful about
energy costs and energy gains for certain processes.
Solution: The cohesive energy, U0 , is the sum of the Coulomb energy between the ions,
UC and the electron transfer energy, Ue . We can write U0 = UC + Ue . The Coulomb
2
energy is as usual given by UC = −α 4πe 0 r0 which can readily be calculated since we have
α and r0 . We get
UC =
−1.748
· 1.44 eV nm ≈ −8.016 eV,
0.314 nm
where we have calculated e2 /(4π0 ) directly in suitable units. The Coulomb energy
means that it costs −UC = 8.016eV to break a pair KCl → K + + Cl− .
Next we treat the electron transfer energy. We have that for K → K + + e− the
energy required is EK = 4.34 eV while for Cl + e− → Cl− an energy amount of
ECl = 3.61 eV is gained. The net energy gain for K + + Cl− → K + Cl is then
EK + ECl = 4.34eV − 3.61eV = 0.73eV . Adding the two contributions, breaking the
pair and moving an electron, requires then in total 8.016eV − 0.73eV ≈ 7.29eV for each
pair. The cohesive energy is then (7.29/2)eV ≈ 3.64eV per ion.
The true value is given as 6.42eV per ion pair. Assuming that the difference comes
from exclusion repulsion we can write
αe2
1
(1 − ) ⇒ −8.016eV + (7.29 − 6.42)eV =
4π0 r0
n
1
1
− 8.016eV (1 − ) ⇒ (7.29 − 6.42)eV = 8.016eV ⇒ n ≈ 9.26.
n
n
UTot. = U0 + UExcl. = −
Discussion: We see that the exclusion energy vary strongly with r (from the derivation
in the book, UExcl. ∼ r− n. It significantly alters the bonding energy and is required
for accurate results. It might be confusing how to add energies with signs. A good
convention is that positive energies are costs and negative energies are gains.
10.5 The Joule-Thomson Effect
Problem formulation: The Joule-Thomson effect refers to the drop in temperature
a gas undergoes when it passes slowly from a full container to an empty one through
a porous slug. Since the expansion is into a rigid container, no mechanical work is
done. Explain the Joule-Thomson effect in terms of the van der Waal attraction between
molecules.
Approach: Understanding the van der Waal interaction and the circumstances for the
system, one should be able to deduce the mechanism from the effect.
Solution/Discussion: What happens is that since no work is done on the gas, and
neither is any heat exchanged with the environment, the gas energy is conserved. The
release of the gas into the empty container will expand it, increasing the average distance
between the molecules. Since the van der Waal interaction depends on the inverse
31
distance between the molecules, the potential energy will then increase. By energy
conservation, the kinetic energy must then decrease which lowers the temperature.
10.9 Return of the Free Electron Gas
Problem formulation: Does the “gas” of freely moving electrons in a metal include
all the electrons present? If not, which electrons are members of the “gas”?
Approach: This problem is more of reasoning type, and information can for example
be found in Beiser section 10.5.
Solution/Discussion: It is only the outer (valence) electrons that contribute to the
metallic electron gas. The inner shell electrons are very localized (d- and f-orbitals) and
move in the vicinity of their respective atom. The electron gas is responsible for the
characteristic properties of metals, such as electrical conductivity and heat transfer.
10.14 Band Structure of Metals
Problem formulation: What are the two combinations of band structure and occupancy by electrons that can cause a solid to be a metal?
Approach: Here, we must understand the connection between band structure and
metal-insulator-semiconductor structure.
Solution/Discussion: When atoms are brought together, forming a solid, the electrons
form energy bands. These bands are separated by gaps of varying sizes. From the FermiDirac distribution, which electrons obey, there is a maximum energy that electrons fill up
these bands to. It is called the Fermi energy. Now, if one band is perfectly full, there is a
gap to the next band which an electron has to overcome to be able to move to free states
(the other states are Pauli blocked!), thereby conducting heat or electrical current for
example.. If this gap is not too small large we have an insulator, because no states are
available unless a large amount of energy is supplied. If the gap is very small, we have a
semiconductor, were just a little stimuli is required for electrons to acquire free states. If
a band is partially filled, electrons have access to free states and we have a metal.
There is however another case. If one filled band and one empty band happen to
overlap, free states become available to the electrons by jumping from one band to the
other.
A very interesting field of research is that of topological insulators, which is an
exception to the rules mentioned above. Certain materials allow for conduction on edges
and surfaces while at the same time being insulators in the bulk. This is a consequence
of certain symmetries and interaction types that puts constraints on the energy bands.
10.26 The SQUID
Problem formulation: A SQUID magnetometer that uses a superconducting ring
2.0 mm in diameter indicates a change in the magnetic flux through it of 5 flux quanta.
What is the corresponding magnetic field change?
Approach: This problem should be straight forward knowing how a SQUID works.
Relevant information is found in section 10.10 in Beiser.
32
Solution: Since a SQUID quantizes the flux through it, we may directly write
φ = AB = nφ0 ⇒ ∆B =
where φ0 =
yields
h
2e
∆nφ0
∆φ
=
,
A
A
≈ 2.068 · 10−15 T·m2 is the magnetic flux quantum. Inserting numbers
∆B =
5 · 2.068 · 10−15 T · m2
≈ 3.3 · 10−9 T.
π(1.0 · 10−3 m)2
Discussion: Earth’s magnetic field is roughly 25−65µT so a SQUID can easily measure
magnetic field changes superimposed on it. In fact, the SQUID can measure magnetic
field changes in biological systems and are thus used for example in medicine.
33
7
Problem session 7: The Atomic Nucleus
11.2 Boron Mixtures
11
Problem formulation: Ordinary boron is a mixture of the 10
5 B and 5 B isotopes and
has a composite mass of 10.82 u. What percentage of each isotope is present in ordinary
boron?
Approach: This problem should be straight forward, knowing the concept of weighted
mean value.
Solution: The composite mass is just the weighted mass mean value which is given by
m = x · mA + (1 − x) · mB ⇒ x =
m − mB
.
mB − mA
Putting mA = 10 u, mB = 11 u and m = 11.82 u gives x ≈ 0.18. So there is 82%
and 18% 10
5 B in ordinary Boron.
11
5 B
Discussion: The result is reasonable since the composite mass tends more towards 11
u than 10 u.
11.14 Oxygen Beta Decay
19
Problem formulation: Both 14
8 O and 8 O undergo a beta decay. Which would you
expect to emit a positron and which an electron? Why?
Approach: We must understand the properties negative and positive beta decay and
relate that to the difference between the isotopes.
Solution/Discussion: As stated in Beiser section 11.3 negative beta decay decreases
the neutron:proton proportions by turning a neutron into a proton and an electron
(actually, an anti-neutrino is released as well) while positive beta decay increases it by
turning a proton into a neutron plus a positron (and a neutrino). The difference between
19
the isotopes is that 14
8 O has more protons than neutrons and the opposite is true for 8 O.
For small atoms,Z < 20 or so, it is often energetically preferable to have equal numbers
19
−
of neutrons and protons. Following this rule, one might guess that 19
8 O →9 F + e and
14
14
+
8 O →7 N + e . It turns out that in the energy spectrum the nuclei, these processes in
fact lowers the energy as expected.
11.17 Mass-Energy Relation
Problem formulation: Find the energies needed to remove a neutron for 42 He, then
to remove a proton, and finally to separate the remaining neutron and proton. Compare
the total with the binding energy of 42 He.
Approach: Here it is useful to know that 1 u = 931.49 MeV/c2 .
34
Solution: For an atom,A
Z X ,the following relation is satisfied by energy conservation
and Einstein’s relation:
Z · (mp + me ) + (A − Z) · mn = mA
+ EBinding /c2 ,
ZX
where the m:s are the proton, electron, neutron and atom mass in order from left to
right.
Then we have that the energy to remove a certain particle from the system is “equal”
to the mass difference between that particle and the total atom. Removing a neutron,
we get
∆E = (m32 He + mn − m42 He ) = (3.016029 + 1.008665 − 4.002603) · 931.49M eV
≈ 20.58M eV
Then removing a proton gives
∆E = c2 (m21 H + m11 H − m32 He ) = (1.007825 + 2.014102 − 3.00160293) · 931.49M eV
≈ 5.49M eV,
where we have used m11 H to include the electrons.
And then separating the neutron and proton gives
∆E = c2 (mn + m11 H − m21 He ) = (1.008665 + 1.007825 − 2.014102) · 931.49M eV
≈ 2.2M eV
The sum of all these is 20.58 + 5.49 + 2.22eV = 28.29eV . We can calculate the binding
energy directly also as:
EBinding = (2 · 1.008665 + 2 · 5.486 · 10−4 + 2 · 1.08665 − 4.002603) · 931.49M eV
≈ 29.30M eV.
So we see that they are pretty much the same.
Discussion: Sometimes one uses c = 1 to make the units of energy and mass equal.
In this way one can directly omit any factors of c2 . You will probably hear physicists
talk about elementary particle masses in units of M eV or GeV (the Higgs boson for
example).
11.18 Magnesium Atomic Mass
Problem formulation:
mass.
The binding energy of
24
12 M g
is 198.25 MeV. Find its atomic
Approach: The Einstein relation should once again do the trick.
+ EBinding /c2 so
Solution: We again have that Z · (mp + me ) + (A − Z) · mn = mA
ZX
we can directly write mA
= Z · (mp + me ) + (A − Z) · mn − EBinding /c2 . Inserting
ZX
proper values Z = 12, A − Z = 12 gives then
m24
= 12 · (1.008665 + 5.486 · 10−4 + 1.08665)u − 198.25/931.49u
12 M g
≈ 23.985u.
35
Discussion: The result in the tables in Beiser page 508 gives m24
≈ 23.985045u so
12 M g
the theory is consistent.
11.22 Mirror Isobars
Problem formulation: Two nuclei with the same mass for which Z1 = N2 and
Z2 = N1 , so that their atomic numbers differ by 1, are called mirror isobars; for
15
example, 15
7 N and 8 O. The constant a3 in the coulomb energy of Eq. (11.18) can be
evaluated from the mass difference between two mirror isobars, one of which is odd-even
and the other even-odd (so that the pairing energies are zero). (a) Derive a formula
for a3 in terms of the mass difference between two such nuclei, their mass number A,
the smaller atomic number Z of the pair, and the masses of the hydrogen atom and the
neutron (Hint: First show that (A − 2Z)2 = 1 for both nuclei.) (b) Evaluate a3 for the
15
case of the mirror isobars 15
7 N and 8 O.
Approach: Here we are given some hints on what to do. We should start with
investigating what makes the mirror isobars so special. Then we must find a way to
relate these properties to nuclear binding energies. Both in the Liquid drop model and
the Einstein relation.
Solution: For the mirror isobars, we have that |Z1 − Z2 | = 1, N2 = Z1 and N1 = Z2 .
From this we can derive that A1 = Z1 + N1 = N2 + Z2 = A2 ≡ A. The number of protons
+ neutrons are the same for both nuclei. Further we get that A − 2Zi = Zi + Ni − 2Zi =
Ni − Zi = Ni − (Zj ± 1) = Ni − (Ni ± 1) = ∓1. Then it is clear that (A − 2Zi )2 = 1.
Next, the binding energy of a nucleus is in the Liquid Drop model given by
EB = a1 A − a2 A2/3 − a3
a5
Z(Z − 1)
− a4 (A − 2Z)2 + (±,0) · 3/4 ,
A1/3
A
where the terms are in order: volume energy, surface energy, coulomb energy, asymmetry
energy and paring energy. For our mirror isobars, A is the same so the difference of
their volume energy and surface energy immediately vanishes. We have also shown that
(A − 2Z)2 = 1 for both nuclei so the energy difference of these terms also vanishes. The
symmetry term will finally also vanish, since if one of the isobars is “even-odd’, ’the other
must automatically be “odd-even” so both terms vanish since one of them must have one
of these possibilities because of the mirror symmetry.
We finally then get, with Z2 = Z1 + 1, Z1 the smaller one, that the difference in their
binding energies is given by:
∆EB ≡ EB,Z2 − EB,Z1 = −a3 (
− a3
Z2 (Z2 − 1)
1/3
A2
−
Z1 (Z1 − 1)
1/3
)=
A1
1
2a3 Z1
((Z1 + 1)Z1 − Z1 (Z1 − 1)) = − 1/3 .
A1/3
A
Now we can compare this with the binding energy difference from the mass differences.
Here we have that
∆EB /c2 = (mp + me ) · (Z2 − Z1 ) + mn · (A2 − Z2 − (A1 − Z1 )) − (mA2 X2 − mA1 X1 ) =
Z2
m11 H · (Z2 − Z1 ) − mn · (Z2 − Z1 ) − ∆m = m11 H − mn − ∆m.
36
Z1
We have again used that the mirror isobars have the same A and Z2 − Z1 = 1. ∆m is
the difference in nuclei mass between the isobars.
Combining our two expressions, we get
A1/3
(2Z1 + 1)1/3
∆EB c2 = −
(m11 H − mn − ∆m)c2 .
2Z1
2Z1
15
For the two mirror isobars 15
7 N and 8 O, we have Z1 = ZN = 7. We then get
a3 = −
(2 · 7 + 1)1/3
(1.007825 − 1.008665 − (15.003065 − 15.000109)) · 931.49M eV ≈
2·7
0.62M eV.
a3 = −
Discussion: From Beiser section 11.5, we see that a common value that gives good
results is a3 = 0.595 MeV. So we are quite close.
11.23 More Mirror Isobars
Problem formulation: The coulomb energy of Z protons uniformly distributed
throughout a spherical nucleus of radius R is given by
3 Z(Z − 1)e2
.
5 4π0 R
(a) On the assumption that the mass difference ∆M between a pair of mirror isobars
is entirely due to the difference ∆m between the 11 H and neutron masses and to the
difference between their coulomb energies, derive a formula for R in terms of ∆M , ∆m
and Z, where Z is the atomic number of the nucleus with the smaller number of protons.
15
(b) Use this formula to find the radii of the mirror isobars 15
7 N and 8 O.
EC =
Approach: With the previous problem in mind, we know the properties of N , Z and
A for the mirror isobars. Then we follow the instructions. The steps should be straight
forward.
Solution: From the Einstein relation we have that for the mirror isobars that
∆EB /c2 = m11 H − mn − ∆M = ∆m − ∆M
where ∆m ≡ m11 H − mn and ∆M is the larger atom mass minus the smaller one. If
we now, as instructed, assume that ∆M = −∆EB /c2 + ∆m = ∆EC /c2 + ∆m, we can
write
3 e2
(Z2 (Z2 − 1) − Z1 (Z1 − 1)) = {Z2 = Z1 + 1} =
5 4π0 R
3 2Z1 e2
3Z1 e2
3Z1 e2
⇒R=
=
.
5 4π0 R
10π0 ∆EC
10π0 c2 (∆M − ∆m)
∆EC = EZ2 − EZ1 =
Finally, to calculate actual radii of the two isotopes we just insert proper numbers:
∆m = (1.007852−1.008665)·931.49M eV /c2 ≈ −0.7825M eV /c2 and ∆M = (15.003065−
15.000109) · 931.49M eV /c2 ≈ 2.753M eV /c2 .
Then we get
R=
3Z1 e2
6 · 7 · (1.44M eV · f m)
≈
≈ 3.42f m.
10π0 c2 (∆M − ∆m)
5(2.753 + 0.7825)M eV
37
Discussion: Once again wee see that mirror isobars allow us to extract nice information
from their nucleon symmetry.
11.29 Deuteron Model
Problem formulation: A simplified model of the deuteron consists of a neutron and
a proton in a square potential well 2 fm in radius and 35 MeV deep. Is this model
consistent with the uncertainty principle?
Approach: We should remember the “good ol’ ” uncertainty principle from quantum
mechanics and apply it to our system. Is it fulfilled or not with the given values?
Solution/Discussion: The uncertainty principle for momentum and position is given
by
∆x · ∆p ≥
~
~
⇔ ∆p ≥
2
2∆x
The energy uncertainty can then, by the relation
write the uncertainty for a nucleon’s energy as
∆E =
p2
2m
= E, be estimated. So we can
~2
(~c)2
(197M eV · f m)2
=
=
≈ 1.3M eV
2
2
2
8m(∆x)
8mc (∆x)
8 · 940M eV (2f m)2
Since the energy 1.3 MeV < 35 MeV, the nucleon can without problem be said to be
confined in the potential. The model is consistent!
38
8
Problem session 8: Nuclear Reactions
12.19 Upper Bound on the Age of Our Solar System
Problem formulation: As discussed in this chapter, the heaviest nuclides are probably
created in supernova explosions and become distributed in the galactic matter from
which later stars (and their planets) form. Under the assumption that equal amounts
of the 235 U and 238 U now in the earth were created in this way in the same supernova,
calculate how long ago this occurred from their respective observed relative abundances
of 0.7 and 99.3 percent and respective half-lives of 7.0 · 108 y and 4.5 · 109 y.
Approach: Understanding the radioactive decay law, this problem should be straight
forward given the assumption above. We should apply it to ratios instead of actual
particle numbers though.
Solution: For any nuclear decay we have that the number of nuclei at time t, denoted
N (t), decays exponentially by the decay law:
N (t) = N0 e−λt ,
where N0 is the number of particles at time t = 0 and λ is the decay constant which
is related to the half-live by T1/2 = log 2/λ. Now we assume that at time t = 0, the
abundance of the two uranium isotopes are equal, N0,1 = N0,2 ≡ N0 , so we can write
their abundance ratio, Q, at any other time as
Q(t) ≡
N0,1 e−λ1 t
= e−(λ1 −λ2 )t .
N0,2 e−λ2 t
Solving for t to see which time is needed to produce the ratio we have today, we get
t=
log(Q(t))
1
log(Q(t))
=
λ2 − λ1
log(2) ( T 1 −
1/2,1
Now, we can insert numbers, calling
get:
t=
log(0.7/99.3)
log(2)
10−9
4.5
235
1
−
1
T1/2,2
.
U nucleus number 1 and
10−8
7.0
238
U number 2. We
y ≈ 5.92 · 109 y ≡ 5.92Ga
Discussion: What we have calculated is an upper bound of the solar system age.
More accurate experiments and calculations give approximately 4.5 Ga. The age of the
universe is somewhere around 13.7 Ga. Note that the calculation we made is based on the
assumption that equal amounts of the isotopes were created a the supernova explosion.
Assumptions like this are always required and different independent experiments and
calculations are needed to give accurate estimations.
12.21 Uranium Decay Chain
Problem formulation: The radionuclide 238
92 U decays into a lead isotope through the
successive emissions of eight alpha particles and six electrons. What is the symbol of the
lead isotope? What is the total energy released?
39
Approach: Understanding how a nucleus changes under alpha decay and electron
emission (electron or negative beta decay) is all we need for the first part this problem.
The by now well used Einstein relation should be handy for the second part.
Solution: For alpha decay we have that
A
ZX
A−4
→Z−2
Y +42 He,
while for electron beta decay, we have
A
ZX
−
→A
Z+1 Y + e + ν.
If 238
92 U decays eight times by α-emission then A → A − 4 · 8 = 238 − 32 = 206 and
Z → Z −8·2 = 92−16 = 76. Six successive beta decay increases Z → Z +6 = 76+6 = 82.
We then end up with 206
82 P b, a lead isotope.
The energy released in the decay is as usual the mass difference (up to a factor of c2 )
of the initial nucleus and the particles remaining afterwards. We can write (assuming
the anti-neutrino is massless)
mU = 8 · mα + 6 · me + mP b + ERelease /c2 ⇒ ERelease = c2 (·238.050786 − 8 · 4.002603
− 205.974455 − 6 · 5.486 · 10−4 ) · 931.49M eV /c2 ≈ 48.64M eV.
Discussion: Experiments have shown that neutrinos and anti-neutrinos in fact have
masses but they are incredibly small, on the order of 1eV! Hence, their mass should be
negligible for our purposes.
12.22 More Uranium Decay
Problem formulation: The radionuclide 232 U alpha-decays into 228 T h. (a) Find the
energy released in the decay. (b) Is it possible for 232 U to decay into 231 U by emitting
a neutron? (c) Is it possible for 232 U to decay into 231 P a by emitting a proton? The
atomic masses of 231 U and 231 P a are respectively 231.036270 u and 231.035880 u.
Approach: The Einstein relation should tell us which reactions that may occur spontaneously. Those are the reactions that release energy since they then the system energy
is lowered.
Solution: The decays we are interested are
232
U →228 T h +42 He
232
U →231 U +1 n
232
U →231 P a +1 p,
were we have drawn the conclusion that it is alpha decay for the first process since the
mass number decreases by 4.
The Einstein relation together with energy conservation allow us to write:
Ebind. = c2 (mi − mf − mX ),
where Ebind. is the binding energy, mi is the initial nuclear mass, mf the final nuclear
mass and mX is the emitted particle mass. For our three cases, we get
40
Ebind. /c2 = (232.037238 − 228.028750 − 4.00260323) · 931.49M eV /c2 ≈ 5.48M eV /c2
Ebind. /c2 = (232.037238 − 231.036270 − 1.008665) · 931.49M eV /c2 ≈ −7.17M eV /c2
Ebind. /c2 = (232.037238 − 231.035880 − 1.007276) · 931.49M eV /c2 ≈ −5.51M eV /c2 .
We see that only the first process occurs spontaneously since it is the only one that
lowers the system energy.
Discussion: This problem is a standard one, when it comes to nuclear physics. Einstein’s relation is truly useful. The only complication is to get all masses right.
12.24 Radium Alpha-Decay
Problem formulation: The energy liberated in the alpha decay of 226 Ra is 4.87 MeV.
(a) Identify the daughter nuclide. (b) Find the energy of the alpha particle and the recoil
energy of the daughter atom. (c) If the alpha particle has the energy in b within the
nucleus, how many of its de Broglie wavelengths fit inside the nucleus? (d) How many
times per second does the alpha particle strike the nuclear boundary?
Approach: In all particle kinematics, conservation laws are useful to derive relations
between masses, energies and velocities of the involved particles. It should not be too
hard to figure out which equations that should be set up. Next, we should just remember
the de Broglie relation, and finally to find some reasonable way of obtaining the nuclear
radius size. Then the last part should be easy.
Solution: The daughter nuclide is easy to identify. Alpha decays decreases A by 4
and Z by 2 so we get 222 Rn. Next we employ conservation of momentum and energy.
We assume that the Ra-nuclide is stationary so that the initial momentum and kinetic
energy is zero. The momentum is then distributed to the daughter nuclide and the alpha
particle. We get
mRa vRa = 0 = mα vα + mRn vRn ⇒ vRn = −vα
mα
.
mRn
The total kinetic energy after the alpha release is then
2
mRn vRn
mα vα2
+
= {using the relation above} =
2
2
mα vα2
mα vα2
mα vα2
mα
mα
+
=
(1 +
) = Eα (1 +
)≈
2
2mRn
2
mRn
mRn
4
226
4
Eα (1 +
) = Eα (1 +
) = Eα
.
ARn
222
222
E=
222
Thus the energy of the alpha-particle is then Eα ≈ 226
4.87M eV ≈ 4.78M eV , and
the recoil energy of the daughter nuclei is thus ERn = E − Eα ≈ 0.086M eV .
Next, the de Broglie relation states that λ = hp . We use the energy from above to
calculate the momentum of the particle and get.
h
4.136 · 10−15 eV s · c
12.4 · 10−13 M eV m
1
√
≈√
≈
·
≈
M
eV
35620.1
2mE
2 · 4 · 931.49M eV · 4.78M eV
6.6f m.
λ= √
41
We can approximate the nuclear size by the radius approximation formula R ≈
1.2f m · A1/3 = 1.2f m · 2261/3 ≈ 7.2f m. So we see that approximately 2 wavelengths fits
across the nuclear diameter 2R.
The frequency with which the alpha particle strikes the nuclear boundary is approximately the inverse time it takes for it to cross the nuclear diameter. We write:
v
ν=
=
2R
p
2E/m
=
2R
p
2 · 4.78M eV /(4 · 931.49M eV /c2 )
≈ 1.05 · 1021 Hz
2 · 7.2 · 10−15 m
Discussion: Section 12.4 in Beiser contains nice information about the alpha decay.
See also the appendix to chapter 12.
12.51 Nuclear Fusion
Problem formulation: (a) A particle of mass mA and kinetic energy KEA strikes a
stationary nucleus of mass mB to produce a compound nucleus of mass mC . Express the
excitation energy of the compound nucleus in terms of mA , mC , KEA and the Q value
of the reaction. (Note: |Q| mc2 .) (b) An excited state in 16 O occurs at an energy of
16.2 MeV. Find the energy need by a proton to produce a 16 O nucleus in this state by
reaction with a stationary 15 N nucleus.
Approach: Conservation of momentum and energy together with the definition of
Q-value should provide us with all we need.
Solution: Consider a reaction of the type A + B → C. If the B-particle is stationary
we can write the conservation of energy as
mA c2 + mB c2 + KEA = mC c2 + KEC + E ∗ ⇒
E ∗ = c2 (mA + mB − mC ) + KEA − KEC ≡ Q + KEA − KEC ,
where E ∗ is the energy that excites the daughter nucleus.
The conservation of momentum gives that pA = pC and by the relation EK =
we get that
EKA − EKC =
p2
2m
p2A
p2
p2
p2
p2
mA
mA
− C = A − A = A (1 −
) = EKA (1 −
).
2mA
2mC
2mA
2mC
2mA
mC
mC
So in total we can write
E ∗ = Q + KEA (1 −
mA
).
mC
For the second part, we have A = a proton so mA = m11 H = 1.007825 u. B =15 N
so mB = mN = 15.000109 u and C =16 O so mC = mO = 15.994915 u. We solve the
m1 H
equation above for KEA = KE11 H above and get KE11 H = (E ∗ − Q)/(1 − m1O ). The
Q-value is calculated to be Q = 12.1271M eV .
KEp =
16.2M eV − 12.1271M eV
≈ 4.34M eV
1 − m11 H /mO
42
Discussion: This energy corresponds, by the Boltzmann relation 1eV ∼ 11600 K, to
approximately 4.9 · 106 · 11600K ≈ 5.6 · 1010 K. The temperature for this to happen by
heating is kind of high. Note, that we choose the laboratory frame for this calculation.
The center of mass frame could also be used of course.
43
9
Problem session 9: Elementary Particles
12.55 Nuclear Electrostatics
Problem formulation: Assume that immediately after the fission event shown in Fig.
12.17 the fission fragment nuclei are spherical and in contact. What is the potential
energy of the system?
Approach: This problem requires us to estimate the nuclei as spheres in contact. We
need to obtain the charges and the radii of these spheres. After some simplifications, the
rest should be ordinary electrostatics.
Solution: The reaction we are interested in is
1
236
∗
94
140
1
n +235
92 U →92 U →38 Sr +54 Xe + 2 n
We imagine the 94
38 Sr-nucleus as a sphere in contact with another sphere representing
Xe
the 140
nucleus.
Then the charges of these nuclei are +38e and +54e respectively.
54
Since the spheres are in contact the distance between their centers is the sum of the radii
which can be obtained from the empirical radius formula on the form RA ≈ 1.2 fm · A1/3 .
Then we can write the electrostatic potential energy between the nuclei as
V =
e2
38 · 54
≈ 1.44M eV · f m · 175.6 · f m−1 ≈ 253M eV.
4π0 1.2 fm (941/3 + 1401/3 )
Discussion: The energy released in the reaction is given by the Q-value:
Q = c2 (mn + mU − 2 · mn − mSr − mXe ) =
c2 (235.043925 − 139.92164 − 93.915361 − 1.008665) · 931.49M eV /c2 ≈ 185M eV.
The electrostatic energy is thus larger than the released energy which may be confusing.
The remedy here is that nuclear attractive forces reduce the energy somewhat.
12.61 Proton Energy Barrier
Problem formulation: The initial reaction in the carbon cycle from which stars
13
hotter than the sun obtain their energy is 11 H +12
6 C →7 N + γ. Find the minimum
energy the proton must have to come in contact with the 12
6 C nucleus.
Approach: This should be just application of the coulomb potential. We must find
the radii of the involved particles though.
Solution: If we again view the nuclei and the proton as hard spheres, we can write
the electrostatic energy between them as
e2
6·1
,
4π0 rC + rp
where rC and rp are the carbon nuclei and proton radii respectively. The carbon radius
can be approximated by the usual formula r ≈ 1.2 fm · A1/3 giving, with A = 12,
rC ≈ 2.47 fm. The proton radius has a value which is approximately rp ≈ 0.81 fm.
Inserting this to our formula above yields
E=
E = 1.44MeV · fm
6
≈ 2.61M eV.
3.28fm
44
Discussion: Since 1 eV ∼ 12000 K we have the the energy above corresponds to
T ≈ 31 GK. I might seem unreasonable then that the reaction above may occur in
stars slightly hotter than the core of the sun (T ≈ 15 MK). But the fact that the stars
have a tremendously high density together with the long tail of the Maxwell-Boltzmann
distribution deems this reaction possible after all!
13.1 Virtual Photon Interaction
Problem formulation: The interaction of one photon with another can be understood
by assuming that each photon can temporarily become a “virtual” electron-positron
pair in free space, and the respective pairs can then interact electromagnetically. (a)
How long does the uncertainty principle allow a virtual electron-positron pair to exist if
hν 2mc2 , where m is the electron mass? (b) If hν > 2mc2 , can you use the notion
of virtual electron-positron pairs to explain the role of a nucleus in the production of
an actual pair, apart from its function in ensuring the conservation of both energy and
momentum?
Approach: This problem is an application of the time-energy uncertainty relation
which we should know at this part of the course. We need also to reflect on the role
of a nucleus in the vicinity of the photon. What physical laws come into play in real
electron-positron creation?
Solution: The lower limit of the time-energy uncertainty relation states that ∆E∆t = ~2 .
It may be thought of setting the time scale for an “energy loan” of ∆E, which may on
this time scale be used to create particles. Or in other words: the time allowed to violate
energy conservation.
Thus, if a photon as an energy hν 2mc2 , where m is the electron mass, it may still
become an electron-positron pair by “borrowing” the required energy in the time
~
6.51 · 10−16 eV s
≈ 3.22 · 10−22 s.
=
2
4mc
4 · 5.11keV
A nucleus in the vicinity of the electron-positron pair will, by its strong electric field,
separate the pair (the electron and positron have opposite charge) sufficiently so that
they are too far apart to recombine.
∆t =
Discussion: From energy and momentum conservation, we actually need a nucleus
to create real electron-positron pairs. The same laws determines that electron-positron
annihilation actually must produce two photons. Remember that the photon never can
have zero momentum!
13.5 Electron-Positron Pair Creation
Problem formulation: Show that 4me c2 , where me is the electron mass, is the
minimum energy needed by a photon to to produce an electron-positron pair when it
collides with an electron in the process γ + e− → e− + e+ + e− .
Approach: We must find the process in which the least energy is required and set
up kinematic equations to solve for the photon energy. Some reflection about different
frames might be useful.
45
Solution: Our sought process should be the one where the created particles are standing
still. That is, only the rest mass energy will be produced in the process. This happens if
the process where a photon and an electron collide with momenta of equal magnitude
and opposite direction so that it is in total zero. The momentum afterwards should then
also be zero, by conservation laws, and the energy of the collision is used only to create
particles and not moving them. This is a center of mass frame calculation.
This process is actually equal to the laboratory frame (or equivalently, the electron
rest frame) process where a photon collides with a stationary electron and all three
products continue onwards with the same total momentum as the initial photon.
We write the energies and momentum as
Ein = Eγ + me c2
pin = pγ = Eγ /c
r
pout c 2
Eout = 3 m2e c4 + (
)
3
pout =?
Employing the conservation of energy and momentum, we write Eout = Ein and
pout = pin . We get
r
r
r
p
p
c
Eγ
out
in
)2 = 3 m2e c4 + (
)2 = 3 m2e c4 +
Eγ + me c = 3 m2e c4 + (
3
3
9
E
γ
⇒ Eγ2 + 2Eγ me c2 + m2e c4 = 9(m2e c4 +
) ⇒ Eγ = 4me c2 .
9
2
In this approach the total initial energy has to be at least Eγ +me c2 = 4me c2 +me c2 =
5me c2 .
Consider now the center of mass (CM) approach. Here, we let the photon and electron
smash into each other with equal and opposite momentum. Then
Ein = Eγ + Ee = Eγ +
p
m2e c4 + (pe c)2
pin = pγ + pe = Eγ /c + pe
Eout = 3me c2
pout = 0
Then by conservation laws, Ein = Eout and pin = pout , we get
Ee =
p
Eγ +
q
m2e c4 + (pe c)2 =
q
m2e c4 + Eγ2 ⇒ {Ein = Eout } ⇒
m2e c4 + Eγ2 = 3me c2 .
Solving for Eγ and Ee we get Eγ = 43 me c2 and Ee = 53 me c2 . So the total input
energy required is then Ein = 3me c2 . This is lower than in the laboratory frame! It is
more efficient to smash particles into each other than smashing one particle into another
stationary one.
One might get the impression that energy is created or saved in the CM-calculation
since 3me c2 < 5me c2 . This is not the case! The solution is that the electron will by
moving see, in its rest frame, a Doppler-shifted photon energy given by:
46
Eγ0 = γEγ (1 + ve /c),
where γ = √
1
1−ve2 /c2
is the Lorentz-factor,Eγ is the CM photon energy and ve is the
electron speed.
From above, we have that the electron, in the CM-frame has the energy
Ee =
p
5
25 2 4
4
m2e c4 + (pe c)2 = me c2 ⇒
m c = m2e c4 + (pe c)2 ⇒ pe = mc.
3
9 e
3
So that we now have the electron momentum in the CM-frame. But this momentum is
of course also, by special relativity, given by:
me ve
⇒ p2e (c2 − ve2 ) = m2e c2 ve2 ⇒
pe = γme ve = q
ve2
1 − c2
ve2 =
p 2 c2
pe c2
p2e c2
= 2e 2 ⇒ ve =
=
2
2
2
me c + pe
Ee /c
Ee
Then we obtain the γ-factor as γ =
q 1
2
1− 452
4
2
3 me c
c
5
2
3 me c
=
4
c
5
= 53 .
Inserting this into the shifted photon energy, we obtain
Eγ0 = γEγ (1 + v/c) =
4
20
9
54
me c2 (1 + ) =
me c2 = 4me c2 .
33
5
9
5
So the electron actually sees a 4me c2 -energy photon approaching in the CM-frame as
well.
Discussion: It is a good exercise to also do this problem with with 4-vectors. In
general it is more efficient to have collider accelerators than smashing particles into
stationary targets. One may show that its getting more and more beneficial when the
number of particles created increases.
13.22 Meson Properties
Problem formulation: One kind of D meson consists of a c and a u quark. What is
its spin? Its charge? Its baryon number? Its strangeness? Its charm?
Approach: This problem is pretty much table work at the level of this course.
Solution:/Discussion Both the c-, charm, and the u quark are spin 1/2 fermions. So
they can either exist in a singlet S = 0 or triplet state S = 1. The charge of the quarks
are + 23 for c and − 23 e for u. So the total charge must be 0. The baryon number is zero
since the D meson consists of one quark and one anti-quark having baryon numbers + 13
and − 13 respectively. Since we have no quark of the strange kind but one of the charm
kind, we have strangeness 0 and charm −1 (it is +1 for the anti-charm quark).
13.26 The Weak Interaction
Problem formulation: The “carriers” of the weak interaction are the W ± , whose
mass is 82 GeV/c2 , and the Z 0 , whose mass is 93 GeV/c2 . Use the method of Sec. 11.7
to find an approximate figure for the range of the weak interaction.
47
Approach: We must understand what can be said from the energy-time uncertainty
relation (note that this relation is of another kind that that of non-commuting hilbert
space operators). We need to find some rough estimates for the quantities involved.
Solution/Discussion: The energy uncertainty principle is on the form
~
.
2
It states that energy conservation can be violated in a small amount of time set by a
certain limit. Let us assume that the vector bosons (Z 0 ,W ± ) travel at a speed v ∼ c (of
course v < c) and that emission of a vector boson implies a temporary energy difference
of ∆E ≈ mc2 , where m is a vector boson mass. We have here neglected any kinetic
energy. Then the uncertainty relation ∆E∆t ∼ ~ implies that the distance traveled in
c~
time ∆t can be estimated as r ∼ c∆t ∼ mc
2.
Now we have that ~c ≈ 0.197 GeV·fm. Using a vector boson mass of m ≈ 90 GeV,
we get
∆E∆t ≥
0.197GeV · f m
≈ 2 · 10−3 f m = 2am.
90GeV
This is really a sub-nuclear distance since nuclei usually are of the size 1 − 10 fm. The
weak interaction is responsible for radioactive decays and at sufficiently high energies,
there is actually a unification of weak and electromagnetic interactions: the electroweak
interaction.
r≈
48
10
Epilogue
This document have been the first edition of a solution manual. There are material to
add, more efficient explanations to be made and also pictures to add. If the course is
given again, this would be a nice update to this document. For suggestions, errors and
comments, don’t hesitate to mail me. My mail address is printed on the title page.
Christian Spånslätt, Stockholm, 13/11/27
49