SOLUTIONS HL TOPIC 12 New
Quantum Physics
2. The Photoelectric Effect Tsokos pp. 389 - 392
IB Example
In the photoelectric effect, the maximum speed – kinetic energy of the electrons
emitted by a metal surface when it is illuminated by light depends on which of the
following?
I. Intensity of the light
II. Frequency of the light
III. Nature of the photoelectric surface
(A) I only (B) III only (C) I and II only (D) II and III only
(E) I, II, and III
Answer = D
EXPLANATION : Einstein’s photon theory of light predicts: Intensity vs.
Frequency
1. An increase in intensity means more photons, so more electrons are
ejected; but since the energy of individual photons is not changed, the
maximum kinetic energy of electrons is not affected.
2. Energy depends on the frequency of light E = hf. Thus, if the frequency
is increased, the maximum kinetic energy of electrons will increase.
3. If the frequency is less than the threshold ( critical) frequency f0, then
no electrons will be ejected, no matter how great the intensity.
4. Electrons emitted and critical frequency depend on the type of
photosurface
1
Tsokos p. 396 : 7
From the graph below of electron kinetic energy Ek versus frequency of incoming
radiation, deduce:
a) the critical frequency of the photosurface
b) the work function
c) What is the kinetic energy of an electon ejected when light of frequency =
8.0 x 1014 Hz falls on the surface?
2
d) Another photosurface has a critical frequency of 6.0 x 10 14Hz. Sketch on
the graph the variation with frequency of the emitted electron’s kinetic
energy.
Examples : Q1 and Q 3 p. 392 – 393
Note Q1 : since h = Planck’s constant = 6.63 x 10-34 J.s and is in JOULES
work function of 1.50 eV was converted to joules from 1 eV = 1.6 x 10-19 J
Φ = 1.5eV X 1.6 x 10-19 J/ eV = 2.4 x 10-19 J ( see numerator in solution
below)
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4
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IB EXAM Nov. 2009 Paper 3 B1
Option B — Quantum physics and nuclear physics
B1. This question is about quantum physics and electrons.
(a) Photons of frequency 2.1× 1015 Hz strike the surface of uranium and
electrons are emitted from the surface. The work function of uranium is 3.6
eV.
(i)
Show that the maximum kinetic energy of the emitted electrons is
about 5.0 eV.
(ii) Explain what change to this energy would occur if the light intensity was
doubled
NO CHANGE remember Einstein’s findings:
Einstein’s photon theory of light predicts: Intensity vs. Frequency
1. An increase in intensity means more photons, so more electrons are
ejected; but since the energy of individual photons is not changed, the
maximum kinetic energy of electrons is not affected.
2. Energy depends on the frequency of light E = hf. Thus, if the frequency
is increased, the maximum kinetic energy of electrons will increase.
3. If the frequency is less than the threshold frequency f0, then no
electrons will be ejected, no matter how great the intensity.
4. Electrons emitted and critical frequency depend on the type of
photosurface
6
IB Example
This question is about the photoelectric effect.
In an experiment to investigate the photoelectric effect, light of frequency f is
incident on the metal surface A, shown in the diagram below. A potential difference
is applied between A and B. The photoelectric current is measured by a sensitive
ammeter. (Note: the complete electrical circuit is not shown.)
When the frequency of the light is reduced to a certain value, the current measured
by the ammeter becomes zero. Explain how Einstein’s photoelectric theory accounts
for this observation.
Answer:




light consists of photons/quanta/packets of energy
each photon has energy E= hf / photon energy depends on frequency
a certain amount of energy is required to eject an electron from the metal
if photon energy is less than this energy, no electrons are emitted
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Tsokos pp. 395-396 1 , 2 , 4b
1. a) Explain what is meant by the photoelectric effect
The Photoelectric Effect
The photoelectric effect describes the emission of electrons when light strikes
a metal surface. It is not difficult to imagine that electrons will be emitted,
because the electromagnetic radiation contains energy that can be
transferred to electrons of the atoms of the photo surface, thus enabling them
to pull themselves away from the attraction of the nuclei and leave the surface
altogether. The liberated electrons, called photo-electrons, absorb enough
energy from the incident radiation to overcome a potential-energy barrier,
created by the attraction of positive charges that normally confine the
electrons within the material.
b) A photosurface has a work function of 3.00 eV. What is the critical
frequency?
  hf 0
convert 3.00 eV to J : 3 x 1.6 x 10-19 = 4.8 x 10-19 J
f0 = Φ
h
= 3.00 eV
6.63 x 10-34 J.s
f0 = Φ
h
= 4.8 x 10-19 J
= 7.24 x 1014 Hz
-34
6.63 x 10 J.s
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2. a) What evidence is there for the existence of photons?
Photo electric effect
b) A photosurface has a critical frequency of 2.25 x 1014 Hz. What is the voltage
required to stop electrons emitted from a photo surface when radiation of
frequency 3.87 x 1014 Hz falls on this surface?
Find work function first then stopping voltage :
  hf 0
note f0 = 2.25 x 1014 Hz
Vs ( stopping voltage) =
  hf 0
Ek = hf – Φ
NOTE : f = 3.87 x 1014 Hz
NOT f0
note f0 = 2.25 x 1014 Hz
Φ = (6.63 x 10-34 J.s ) x (2.25 x 1014 Hz ) = 1.49175 x 10-19 J
Vs ( stopping voltage) =
Ek = hf – Φ
= (6.63 x 10-34 J.s ) x (3.87 x 1014 Hz) - 1.49175 x 10-19 J
= 2.5658 x 10-19 J - 1.49175 x 10-19 J
= 1.07406 x 10-19 J
convert to eV : 1 eV = 1.6 x 10-19 J
Vs ( stopping voltage) = 0.671 eV
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4 b) When light of wavelength 2.08 x 10-7 m falls on a photo surface, a voltage of
1.40 V is required to stop the emitted electrons from reaching the anode. What is
the largest wavelength of light that will result in emission of electrons from this
photosurface? TRICKY
STRATEGY:
Need to find largest wavelength from critical frequency ( f0) by using :
f0 = c
λ
c = constant = speed of light = 3.0 x 108 ms-1
But to find f0 need work function (Φ ) from : Vs = hf – Φ
NOTE : f NOT f0
Φ = h f0
STEPS:
1) Calculate f from light of wavelength 2.08 x 10-7 m :
f=c
λ
c = constant = speed of light = 3.0 x 108 ms-1
f = 1.4423 x 1015 Hz
2) Calculate Φ from :
Φ = hf - Vs
Vs = hf – Φ
NOTE : f NOT f0
h = Planck’s constant = 6.63 x 10-34 J.s
= (6.63 x 10-34 J.s ) x (1.4423 x 1015 Hz) - 1.40 eV
= 9.562 x 10-19 - 1.40 eV ( need to convert to J because Plancks constant)
1.40 eV = 2.24 x 10-19 J
= 9.562 x 10-19 - 2.24 x 10-19 J = 7.322 x 10-19 J
3) Calculate f0 from : Φ = h f0
f0 = Φ
h
= 7.322 x 10-19 J = 1.104 x 1015 Hz
6.63 x 10-34 J.s
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4) FINALLY calculate wavelength from
f0 = c
λ
c = constant = speed of light = 3.0 x 108 ms-1
λ =c =
f0
2.717 x 10-7 m
Miscellaneous Questions
1. In the photoelectric effect, the maximum speed – kinetic energy of the electrons
emitted by a metal surface when it is illuminated by light depends on which of the
following? D
I. Intensity of the light
II. Frequency of the light
III. Nature of the photoelectric surface
(A) I only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II, and III only
Explanation:
Einstein’s photon theory of light predicts: Intensity vs. Frequency
1. An increase in intensity means more photons, so more electrons are
ejected; but since the energy of individual photons is not changed, the
maximum kinetic energy of electrons is not affected.
2. Energy depends on the frequency of light E = hf. Thus, if the
frequency is increased, the maximum kinetic energy of electrons will
increase.
3. If the frequency is less than the threshold ( critical) frequency f0, then
no electrons will be ejected, no matter how great the intensity.
4. Electrons emitted and critical frequency depend on the type of photo
surface
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Questons 2-3
2. Which graph shows the maximum kinetic energy of the emitted electrons versus
the frequency of the light? A
(A) (B)
(C)
(D)
(E)
3. Which graph shows the total photoelectric current versus the intensity of the
light for a fixed frequency above the cut off frequency? D
(A) (B) (C) (D) (E)
4.
The photoelectric effect involves the emission of electrons from B
A.
the surface of a metal when the metal is heated.
B.
the surface of a metal when it is illuminated with electromagnetic radiation.
C.
an atom of a material when the material is negatively charged.
D. an atom of a material when the material is heated.
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3. Wave- Particle Duality of Matter and Light / de Broglie
wavelength
Tsokos p. 396 : 8 and 9
λ=h
p
a)
( p = momentum = mv)
λ=h
mv
λ = 6.63 x 10-34 J.s
(0.25) (10)
= 2.652 x 10-34 m
b) No
Since the value of h is so small , ordinary macroscopic objects do not display
wave-like behavior. For example, a baseball ( m = 0.15kg) thrown at a speed of
40 ms-1 has a de Broglie wavelength of
λ=h = h =
p
mv
6.63 x 10-34 J*s
=
-1
(0.15kg) (40ms )
1.1 x 10-34 m
This is much too small to measure. However, with subatomic particles, the wave
nature is clearly evident.
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a)
The electron as a wave and the Electron Microscope
The question is , given an electron, when doe we treat it as a particle or as a
wave? Remember that if we call something a wave then it must show wave-like
properties – in particular diffraction. A wave of wavelength λ will diffract around
an obstacle of size d if , and only if, the λ is comparable to or bigger than d. To
find a typical electron wavelength, consider an electron moving at a speed of 10 5
ms-1 . It has a momentum p = 9.1 x 10-26 kgms-1 and therefore λ = 7.2 x 10-9 m.
This is quite small. To see the wavelike nature of this electron we would need an
opening of about this size. This is where the electron microscope comes in. Its
lenses are on the order of 10-8 m which is the de Broglie wavelength of the
electron used.
b) me = 9.11 x 10-31 kg
λ = 680 nm = 680 x 10-9 m h = 6.63 x 10-34
λ=h
mv
680 x 10-9 m =
6.63 x 10-34 J.s
(9.11 x 10-31 kg) (v)
v = 1071 ms-1
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Miscellaneous Questions
1. If the momentum of an electron doubles, its de Broglie wavelength is
multiplied by a factor of _____________________
λ=h
mv
λ=1
2
2. What is the energy of a photon with a wavelength of 2.07 nm?
E  hf
E = 1.3724 J
3. What is the wave length of a proton with a linear momentum of 3.3 x 10-23
kg m s-1 ?
λ=h
mv
=
2.00 x 10-11 m
4. What would happen to the energy of a photon if its wavelength were
reduced by a factor of 2?
E  hf
Frequency is inversely proportional to wavelength. So reducing wavelength by a
factor of 2 INCREASES f by a factor of 2. E is DIRECTLY proportional to f so E
increases by a factor of 2
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5.
Which of the following phenomena provides evidence for de Broglie’s hypothesis? A
A.
Electron diffraction = best answer diffraction related to wave behavior
B.
X-ray production = nuclear decay
C.
Line spectra
D.
Nuclear energy levels
6. Calculate the deBroglie wavelength of a 0.20 kg ball moving with a speed of
15ms-1
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IB Example
The de Broglie wavelength of an electron with energy 5.0 keV is λ.
a) Determine λ.
E = p2
2m
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b) A student makes the statement “The electron wavelength is not real, it is
just a mathematical construction. Electrons are particles and never waves.”
Outline evidence which suggests that the student’s statement is not correct.
Answer :

Diffraction : Experiments show that electrons can be diffracted and
therefore must be waves

Photoelectric effect – frequency and its effect on photoelectrons:
To increase the kinetic energy with which the photoelectrons leave, you must
increase the frequency of light. This is strong evidence that light can also act as
a wave.
IB Example
This question is about the de Broglie hypothesis.
(a)
State the de Broglie hypothesis.
(2)
Since an electromagnetic wave can behave like a particle, can a particle of
matter behave like a wave? In 1923, the French physicist Louis de Broglie said
yes. His conjecture, which has been supported by experiment, is that a particle of
mass m and speed v and thus linear momentum p = mv, has an associated
wavelength. This wavelength is called the de Broglie wavelength.:
λ=h
p
p = momentum = mv
(b)
λ=
Determine the de Broglie wavelength of a proton that has been accelerated
from rest through a potential difference of 1.2 kV.
h
2meV
8.3 × 10–13 m
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4. Quantum Theory and the Uncertainty Principle –
Tsokos pp. 398 –404
no problems to solve
5. Nuclear Structure and Beta Decay II Tsokos pp.
407-411
Calculating energy released during nuclear decay reaction
Example 1 - done for you
Fig. 6.4 below shows the lowest nuclear energy levels of the magnesium nucleus
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Mg
12
Also shown is a gamma decay ( Υ ) as the nucleus undergoes gamma decay
and releases a photon. The nuclear energy levels that are involved are shown.
By looking at the levels you should note that the change in energy if from 5.24
MeV to 1.37 eV.
So the energy released is in the gamma decay reaction is:
5.24 – 1.37 = 3.87 MeV.
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Example 2
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Fig 6.5 shows an energy level of plutonium and a few of the energy levels of
uranium
242
238
Pu
94
U
92
Calculate the energy released during alpha decay for both alpha particles:
Alpha particle 1 ( top graph) :
4.983 – 0.307 = 4.676 MeV
Alpha particle 2 ( bottom graph) : 4.983 – 0.148 = 4.835 MeV
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22
Example
The following are statements concerning radioactive decay.
I.
Alpha particles have discrete energies.
II.
The beta-energy spectrum is a broad continuous distribution of energies.
III.
Gamma rays are emitted with discrete energies.
Which statement(s) is(are) evidence for the existence of nuclear energy levels? D key word
discrete
A.
I only
B.
II only
C.
III only
D.
I and III only
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6. Half Life II – Review
IB Exam May 2010 B3
B3. This question is about radioactive decay.
(a) The decay constant for a particular isotope is λ = 0.048 s–1. A sample of the
isotope initially contains 2.0 x 1012 nuclei of this isotope.
(i) Define decay constant.
(ii) Estimate the number of nuclei that will decay in the first second.
dN
 N
dt
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Tsokos p. 413 : 13 b) Just determine ½ life from graph
½ life = 7 min.
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