School of Physics & Astronomy
PHS1022 Quantum Physics - Sample Problem Set 1
QUESTION 1
(a)
Monochromatic light of wavelength 300 nm is incident normally on a sodium (Na)
surface of area 4 cm2. If the intensity of the light is 0.15 Wm-2, determine the rate at
which photons strike the Na surface.
Solution
The energy per photon is:
The power (or energy flux) is given by:
.
Hence, the rate at which photons strike the Na surface is:
(b)
What is the range of kinetic energies of photoelectrons emitted from the Na surface
when illuminated by light of wavelength 300 nm?
Hint: The work function of Na is 2.28 eV.
Solution
The maximum kinetic energy is determined from Einstein’s photoelectric equation:
In the photoelectric effect an electron with the highest amount of energy in the metal
absorbs a photon of energy hf. The kinetic energy of these “surface” electrons is given
by the Einstein equation, where the work function, W, represents the minimum
amount of energy necessary to remove an electron from the surface of the metal.
However, it is possible for electrons to be emitted with kinetic energies less than Kmax.
These electrons come from lower energy electron states in the metal, where they are
bound more tightly. Therefore photoelectrons can emerge from the surface with 0 eV this corresponds to the minimum kinetic energy.
QUESTION 2
Consider an antihydrogen atom, which consists of an antielectron (i.e., a positron,
)
orbiting an antiproton
The masses of the positron and the antiproton are identical to
that of an electron and proton, respectively. However, the positron has a charge +e and the
antiproton has a charge -e.
(a)
Calculate the radius of the first Bohr orbit in the antihydrogen atom.
Solution
Note that since we have simply interchanged a positron for an electron and an
antiproton for a proton, the quantum mechanics of an antihydrogen atom is expected
to be identical to that of a hydrogen atom. We can therefore utilise the results of the
Bohr model of a H-atom directly.
The radius of the first Bohr orbit (n = 1) is given by:
where Z = 1 (the “nucleus” consists of a single negative charge), and the reduced mass
for antihydrogen is:
(b)
Calculate the ionization energy in eV for an antihydrogen atom in the ground state.
Solution
Using the reduced mass M and Z = 1, we have:
The ionization energy is then given by:
(c)
Calculate the wavelength of the photon emitted when an antihydrogen atom makes a
transition from the n = 2 state to the n = 1 state. To what part of the electromagnetic
spectrum does this radiation correspond?
Solution
Using the expression for the energy obtained in part (b) we write:
The difference in energy between the two states is:
The wavelength is therefore given by:
This lies within the ultraviolet (UV) part (< 400 nm) of the electromagnetic spectrum.
QUESTION 3
Figure 1 shows a schematic representation of a Young’s double slit experiment that uses
electrons. The source produces monoenergetic electrons with a fixed energy. A moveable
detector D is used to record the number of electrons arriving at a point x on a distant screen.
Screen
Double Slit
Path 2
D
Slit 2
x
Electron Source
Path 1
d
0
Slit 1
L
Figure 1: Young’s double slit experiment with electrons.
(a)
Discuss briefly the experimental phenomenon of electron diffraction and its
interpretation in terms of quantum physics.
Solution
A beam of collimated electrons is sent through a double slit (see Figure 1). An
interference pattern consisting of maxima and minima is observed when a large
number of electrons are recorded (e.g., 70,000 electrons were recorded in the
Tonomura experiment, see e.g., http://www.hqrd.hitachi.co.jp/global/doubleslit.cfm).
The fringes exhibit a
interference pattern that is characteristic of the
Young’s two slit experiment with light. Here
, where
is the de
Broglie wavelength of the monoenergetic electrons. The arrival of the electrons at the
detector (screen) is random (non-deterministic), and the electrons are localised on
detection as one would expect from a particle; however, the fringe pattern is only
determined correctly if one attributes a de Broglie wavelength to the electrons, in
which case constructive or destructive interference between the electron
wavefunctions occurs depending on their phase (path) difference. The probability of
detection is determined by superposing the wavefunctions as indicated in part (b)
below. A full discussion of electron diffraction can be found in the PHS1022
Quantum Physics Workshops that can be accessed via Moodle.
(b)
Write an expression for the probability of detecting an electron at position x, between
x and x+ dx, on the distant screen. Take care to define all symbols.
Solution
According to quantum physics the probability of detecting an electron at a point x
between x and x + dx, is given by:
Here
denotes the wavefunction that describes an electron passing through slit 1,
travelling over path 1 and being detected at D. Similarly,
is the wavefunction
describing an electron passing through slit 2, travelling over path 2 and being detected
at D. Note that if the electron wavefunctions are in phase (i.e.,
the
probability of detection is a maximum. However, if the wavefunctions are out of
phase (i.e.,
the probability of detection is a minimum. The quantum
expression for the probability distribution exhibits interference (maxima and minima)
in the distribution of electrons.
The interference terms become apparent if we expand the expression for the
probability, i.e.,
Here φ denotes the phase difference between the electron wavefunctions
(c)
and
In what way does the probability obtained in (b) differ from the predictions of
classical physics?
Solution
According to classical physics there is no interference between the electrons. The
probability of detecting an electron at the distant screen is given classically by:
where the wavefunctions are defined in (b). The classical expression represents the
sum of the probabilities, without the interference terms. Hence in classical physics
there is no possibility of obtaining a Young’s interference pattern with a double slit!
_____________________
School of Physics & Astronomy
PHS1022 Quantum Physics - Sample Problem Set 2
QUESTION 1
A beam of X-rays from a synchrotron is incident on a vessel containing gaseous krypton (Kr).
Figure 1 shows the X-ray absorption spectrum near the K-edge of Kr.
Figure 1: X-ray absorption spectrum for gaseous Kr.
(a)
Calculate the wavelength (in nm) of X-rays at the absorption K-edge of Kr.
Solution
From Figure 1 the absorption K-edge of krypton occurs at approximately
Equating the K-edge energy with the photon energy we have:
Thus the wavelength of X-rays corresponding to the K-edge in krypton is 0.113 nm.
(b)
The power in the synchrotron beam at the energy corresponding to the K-edge is 7.3W.
Calculate the number of X-ray photons per second incident on the vessel containing
Kr.
Solution
For monochromatic radiation the power is equal to the number of photons per second
multiplied by the energy of each photon. Hence the number of X-ray photons per
second is calculated from:
QUESTION 2
An electroscope consists of a very thin piece of gold foil (known as gold leaf) fixed at the top
to a rod of copper. A zinc plate is attached to the copper conductor, as shown in Figure 2.
Figure 2: Schematic of a gold leaf electroscope.
Charge can be transferred to the electroscope by bringing the zinc plate in contact with a
charged object. The charge flows over the conducting copper and gold, and the gold leaf
moves away from the copper rod as it is repelled by the charge on the copper conductor.
(a)
It is observed that when an initially negatively charged zinc plate is exposed to
ultraviolet (UV) radiation the gold leaf collapses. Briefly explain this observation.
Solution
Ultraviolet (UV) radiation incident on the zinc plate ejects photoelectrons from the
metal surface, which are repelled from the zinc plate by its negative charge. When
negatively charged electrons are removed from the zinc plate, the remaining zinc atoms
are positively charged. The zinc atoms are neutralised when electrons flow from the
copper rod. As a consequence the repulsive force between the gold leaf and the copper
rod decreases and the leaf collapses. This situation is illustrated schematically below.
(b)
If the zinc plate is charged in the same manner as in part (a), and the intensity of the
UV light is increased what do you expect to happen?
Solution
The intensity of the UV light depends on the number of UV photons per second.
Increasing the intensity of the UV light causes more photoelectrons per second to be
ejected from the zinc plate, and consequently the gold leaf collapses more quickly.
(c)
X-ray photoelectron spectroscopy (XPS) uses photo-ionisation of electrons to analyse
the surface composition of a material. Figure 3 shows a simplified energy level
diagram for a sodium atom in NaCl.
Figure 3: Energy level diagram for a sodium atom.
If X-rays with energy 1.253 keV are used, calculate the kinetic energy of the sodium 2s
photoelectron peak expected in an XPS spectrum.
Solution
The Einstein photoelectron equation is used to calculate the kinetic energy of the 2s
photoelectron peak in the XPS spectrum, i.e.,
where hf is the energy of the incident X-ray photon (1.253 keV) and W is the photoionization energy (i.e., the energy required to remove an electron from the 2s level in
sodium). From Figure 3 we note that for the 2s level in sodium,
hence
QUESTION 3
When white light between 400 nm and 700 nm is shone on atomic hydrogen a spectrum
consisting of dark lines is observed (see Figure 4 ). Briefly discuss the origin of these lines in
the spectrum of hydrogen. What determines the wavelengths of these spectral lines?
656nm
486nm
434nm 410nm
Figure 4: Dark lines in the spectrum of atomic hydrogen between 400 nm and 700 nm.
Solution
Figure 4 shows the absorption spectrum of atomic hydrogen in the range 400 nm to 700 nm.
When white light (having this range of wavelengths) is shone onto a gas of atomic hydrogen,
photons with energy (wavelength) corresponding to differences in the energy levels of the
hydrogen atom are absorbed, causing an electron to make an “upward” transition between
quantized (discrete) energy levels. The wavelengths of the absorption lines are determined
from:
(1)
where
and
Upon substituting
This corresponds to the Balmer series of the hydrogen atom.
and
into Eq. (1) we obtain the wavelengths
shown in Figure 4.
_____________________
School of Physics & Astronomy
PHS1022 Quantum Physics - Sample Problem Set 3
QUESTION 1
(a)
When black and white photographic film is exposed to light, molecules of silver
bromide (AgBr) contained in the photosensitive emulsion are dissociated (i.e., broken
apart). The minimum energy required to dissociate a AgBr molecule is 0.68 eV.
Calculate the maximum wavelength (in nm) beyond which this film will not record
light.
Solution
Equating the dissociation energy of a AgBr molecule with the energy of the incident photon:
Thus the maximum wavelength for dissociation is given by:
(b)
A 0.1 kg mass falls through a height of H = 1 m. If all the energy acquired in the fall
were converted to optical photons (λ = 680 nm), how many photons would be
emitted?
Solution
Equate the gravitational potential energy with the energy of the photons, according to:
where N is the number of photons, each having energy hf. Hence we have:
(c)
Light form a laser diode (λ = 680 nm) is incident on an antimony oxide photocathode,
whose work function is 2.0 eV. Clearly stating your reasons indicate if photoelectrons
would be emitted from the surface of the photocathode.
Solution
The energy of a photon from the laser diode is:
Since the energy of the incident photon is less than the work function for antimony
oxide, no photoelectrons are emitted from the surface of the photocathode.
QUESTION 2
A diffraction pattern is observed on a fluorescent screen when electrons are accelerated
through a potential difference of 5 kV, and subsequently interact with a carbon thin film.
Figure 1 shows the resulting diffraction pattern.
Figure 1: Diffraction pattern observed when electrons interact with a thin film of carbon.
(a)
Calculate the de Broglie wavelength of electrons that give rise to the diffraction pattern
in Figure 1.
Solution
The de Broglie wavelength is given by:
Thus
(b)
Discuss briefly what happens to the diffraction pattern when the accelerating potential
is reduced to 1 kV.
Solution
Since
, if we reduce the accelerating potential from 5 kV to 1 kV, the de
Broglie wavelength will increase by a factor of
The separation of maxima in the
diffraction pattern depends on wavelength according to:
therefore
is
increased and the circular diffraction pattern expands.
(c)
The separation of the fringes in Figure 1 is approximately 1 cm. Estimate the distance
between carbon atoms in the thin film. Clearly show how you arrived at your answer.
Take the distance between the carbon film and the fluorescent screen to be D ~ 12 cm.
Solution
Assume that interference arises when two electron waves are scattered off adjacent
carbon atoms in the thin film. We can model this on the basis of a Young’s double slit
arrangement as shown in Figure 2.
Screen
xm ~ 1 cm
Atom 2
Distance between
d
C-atoms
Atom 1
D ~ 12 cm
Figure 2: Interference between two electron waves scattering off adjacent carbon atoms.
The locations of the maxima are given by:
,
where
is the de Broglie wavelength calculated in Question 2 (i). Since
we can use the following approximation:
Hence we can write:
This gives:
is small
A value of
is in agreement with the expected interatomic distance for
carbon atoms in a thin film.
_________________________
School of Physics & Astronomy
PHS1022 Quantum Physics - Sample Problem Set 4
QUESTION 1
(a)
A laser diode produces optical radiation of wavelength 690 nm. Calculate the energy
(in eV) of a single photon of light from the laser diode.
Solution
The energy of a single photon of light is given by:
Converting to eV, where
(b)
gives:
A 0.1 gram mass falls through a height of 1 cm. If all the energy acquired in the fall
were converted to optical photons (8 = 690 nm), how many photons would be
emitted?
Solution
Assuming all the gravitational potential energy is converted into photons, we have:
where N is the number of photons and l is the height through which the mass falls.
The number of photons is determined from:
(c)
Calculate the momentum of a single photon of light from the laser diode (8 =690 nm).
Solution
The momentum of the photon is given by:
(d)
Light form the laser diode is incident on an antimony oxide photocathode, whose work
function is 2.0 eV. Clearly stating your reasons indicate if photoelectrons would be
emitted from the surface of the photocathode.
Solution
The Einstein photoelectric equation determines the maximum kinetic energy of
photoelectrons emitted from the surface of the photocathode, i.e.,
where W = 2.0 eV is the work function of antimony oxide. We note that
Since the incident photon energy is less that the minimum amount of energy (i.e., the
work function) needed to free an electron from the surface of the photocathode, there
can be no photoemission.
QUESTION 2
Bucky balls, or more correctly buckministerfullerene (
), are large molecules consisting of
sixty tightly bound carbon atoms in the shape of a soccer ball. Under certain circumstances a
double slit interference pattern can be observed using bucky balls.
(a)
Hot
molecules from an oven are collimated into a fine beam, which is incident on
a double slit of separation, d, made from silicon nitride (SiN). A detector is used to
record the number of molecules arriving at a point beyond the slit. The average speed
of the
molecules is 210
Calculate the de Broglie wavelength of the
molecules.
[Assume the mass of a bucky ball is:
Solution
The de Broglie wavelength of a
molecule is given by:
(b)
In order to produce an observable interference pattern with bucky balls, the SiN double
slit is chosen to have a slit separation of d = 150 nm. Calculate the spacing of the
maxima in the interference pattern. Take the distance between the SiN double slit and
the detector to be L = 1.0 m.
Solution
The maxima in the interference pattern are determined from
where
is the de Broglie wavelength of a bucky ball and m = 0,1,2,... is the order of the
maximum.
Since
where
we can use the approximation:
is the distance to the mth maximum. Hence:
and the spacing of the maxima (fringe spacing) in the interference pattern is given by:
whence
________________________