Download (a) Calculate the speed of electrons which have a de Broglie

1
(a)
Calculate the speed of electrons which have a de Broglie wavelength of 1.5 × 10–10 m.
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(2)
(b)
Would you expect the electrons in part (a) to be diffracted by crystals in which the atom
spacing is 0.10 nm? Explain your answer.
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(2)
(Total 4 marks)
2
(a)
The diagram below shows schematically an arrangement for producing interference fringes
using a double slit.
A dark fringe (minimum intensity) is observed at the point labelled P.
(i)
Show clearly on the diagram the distance that is equal to the path difference between
the light rays from the two slits to the point P.
(1)
Page 1 of 67
(ii)
Explain how the path difference determines that the light intensity at point P is a
minimum.
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(3)
(iii)
Explain briefly the role of diffraction in producing the interference patterns (You may
draw a sketch to support your explanation if you wish.)
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(2)
(b)
In one experiment the separation of the slits is 4.0 × 10–4 m. The distance from the slits to
the screen is 0.60 m.
Calculate the distance between the centres of two adjacent dark fringes when light of
wavelength 5.5 × 10–7 m is used.
(2)
(c)
A student has learned that electrons behave like waves and decides to try demonstrate this
using the arrangement in the diagram above. The lamp is replaced by a source of electrons
and the system is evacuated.
The student accelerates the electrons to a velocity of 1.4 × 106 m s–1. The beam of
electrons is then incident on the double slits. The electrons produce light when incident on
the screen.
mass of an electron
Planck constant
(i)
= 9.1 × 10–31 kg
= 6.6 × 10–34 J s
Calculate the de Broglie wavelength associated with the electrons.
(3)
Page 2 of 67
(ii)
Explain briefly, with an appropriate calculation, why the student would be
unsuccessful in demonstrating observable interference using the slit separation of 4.0
× 10–4 m.
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(2)
(Total 13 marks)
3
(a)
Uranium-238 decays by alpha emission to thorium-234. The table shows the masses in
atomic mass units, u, of the nuclei of uranium-238 (
particle (helium-4).
Element
Nuclear mass/u
Uranium-238
238.0002
Thorium-234
233.9941
Helium-4, alpha particle
(i)
), thorium-234, and an alpha
4.0015
1 atomic mass unit, u
= 1.7 × 10–27 kg
speed of electromagnetic radiation, c
= 3.0 × 108 m s–1
the Planck constant, h
= 6.6 × 10–34 J s
How many neutrons are there in a uranium-238 nucleus?
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(1)
(ii)
How many protons are there in a nucleus of thorium?
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(1)
(b)
(i)
Determine the mass change in kg when a nucleus of uranium-238 decays by alpha
emission to thorium-234.
(2)
(ii)
Determine the increase in kinetic energy of the system when a uranium-238 nucleus
decays by alpha emission to thorium-234.
(2)
(c)
Wave particle duality suggests that a moving alpha particle (mass 6.8 × 10–27 kg) has a
wavelength associated with it. One alpha particle has an energy of 7.0 × 10–13 J.
Calculate:
(i)
the momentum of the alpha particle;
(2)
Page 3 of 67
(ii)
the wavelength associated with the alpha particle.
(2)
(Total 10 marks)
4
An electron initially at rest is accelerated through a potential difference. It is then brought to rest
in a collision, and all of its kinetic energy is converted into a single photon of electromagnetic
radiation. Which one of the following quantities is not required to find a value for the wavelength
of the photon?
A
The mass of the electron
B
The charge on the electron
C
The velocity of electromagnetic waves
D
The value of the potential difference
(Total 1 mark)
5
Which one of the graphs best represents the relationship between the energy W of a photon and
the frequency f of the radiation?
(Total 1 mark)
6
The photoelectric effect is one piece of evidence that suggests that light behaves like a stream of
particles or photons.
(a)
State what is meant by the threshold frequency in an experiment to investigate the
photoelectric effect.
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(2)
(b)
State and explain the effect of increasing the intensity of light on the rate at which electrons
are emitted.
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(2)
Page 4 of 67
(c)
In an experiment to investigate the photoelectric effect the radiation incident on the surface
caused the emission of electrons of energy 1.5 × 10−19 J. The work function of the surface
was known to be 3.2 × 10−19 J.
The Planck constant h is 6.6 × 10−34 J s.
The speed of electromagnetic radiation is 3.0 × 108 m s−1.
The mass of an electron is 9.1 × 10−31 kg.
(i)
Calculate the wavelength of the incident radiation.
(2)
(ii)
Calculate the de Broglie wavelength of the emitted electrons.
(3)
(Total 9 marks)
7
For which of the following relationships is the quantity y related to the quantity x by the
relationship
x
y
A
energy stored in a spring
extension of the spring
B
gravitational field strength
distance from a point mass
C
de Broglie wavelength of an electron
momentum of the electron
D
period of a mass-spring system
spring constant (stiffness) of the spring
(Total 1 mark)
8
(a)
Calculate the wavelength of a γ-ray photon which has an energy of 1.6 × 10−15 J.
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(2)
(b)
An X-ray photon is generated which has the same energy as the γ-ray photon described in
part (a).
(i)
How do the speeds in a vacuum of these two photons compare?
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(ii)
How do their abilities to penetrate a given material compare?
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(2)
(Total 4 marks)
Page 5 of 67
9
The diagram below shows electrons being fired at a polycrystalline graphite target in a vacuum.
The electrons are emitted from a heated cathode and pass through an accelerating p.d. The
inside surface on the far side of the chamber is coated with fluorescent material that emits light
when the electrons release their energy to it.
Mass of electron me = 9.1 × 10–31 kg
Planck constant h = 6.6 × 10–34 J s
(a)
The electrons travel at a speed of 4.0 × 107 m s–1. Calculate their de Broglie wavelength.
(1)
(b)
Sketch on the front view of the fluorescent screen shown in the diagram the pattern of light
you would expect to see emitted by the fluorescent material.
Explain why this pattern suggests that electrons have wave-like properties.
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(2)
Page 6 of 67
(c)
Explain one aspect of the experiment that suggests that electrons have particle-like
properties.
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(2)
(Total 5 marks)
10
(a)
Discovery of the photoelectric effect was largely responsible for the development of the
theory that electromagnetic radiation can behave as a particle or as a wave under different
circumstances. The diagram below shows an experimental arrangement used to
demonstrate aspects of the photoelectric effect. When photoelectrons are emitted the
ammeter registers a current.
(i)
The metal plate is illuminated with radiation but does not emit photoelectrons. The
intensity of the radiation is increased. State and explain what effect this increase in
intensity has.
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(2)
Page 7 of 67
(ii)
The metal plate is illuminated with radiation such that photoelectrons are emitted. The
intensity of the radiation is increased. State and explain what effect this increase in
intensity has.
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(2)
(iii)
The metal plate is illuminated with radiation such that photoelectrons are emitted. Air
is now allowed to enter the enclosure. State and explain what effect allowing air into
the enclosure has.
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(2)
(b)
(i)
Show that the de Broglie wavelength of an electron travelling at 0.15c should be
approximately 1.6 × 10–11 m.
the Plank constant, h = 6.6 × 10–34 J s
the speed of electromagnetic waves in a vacuum, c = 3.0 × 108 m s–1
the mass of an electron, me = 9.1 × 10–31 kg
(2)
(ii)
Suggest a suitable material to give an observable diffraction pattern with electrons.
Explain your choice.
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(2)
(Total 10 marks)
Page 8 of 67
11
(a)
Figure 1 shows the electron gun that accelerates electrons in an electron microscope.
Figure 1
(i)
Draw, on Figure 1, electric field lines and lines of equipotential in the region between
the anode and cathode. Assume that there are no edge effects and that the holes in
the plates do not affect the field.
Clearly label your diagram.
(3)
(ii)
Calculate the kinetic energy, speed and momentum of an electron as it passes
through the hole in the anode.
mass of an electron
=
9.1 × 10–31 kg
charge of an electron
=
–1.6 × 10–19 C
(4)
Page 9 of 67
(b)
By calculating the de Broglie wavelength of electrons coming through the anode of this
device, state and explain whether or not they will be suitable for the investigation of the
crystal structure of a metal.
Planck constant
=
6.6 × 10–34 J s
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(4)
(Total 11 marks)
Page 10 of 67
12
The diagram below shows a simple model of a hydrogen atom in which a single electron
stationary wave fits the radius of an atom. This model suggests that the electron cannot exist
outside this atomic sphere. The atomic radius is 3.0 ×10–10 m.
(a)
Explain what the electron stationary wave represents and state what you can infer about
the location of the electron.
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(3)
(b)
(i)
Show that the kinetic energy of the electron can be written as
where p is the momentum and me is the mass of the electron.
(1)
Page 11 of 67
(ii)
Show that the kinetic energy of this electron can be written as
where h is the Planck constant and λ is the wavelength of the electron stationary
wave.
(2)
(c)
(i)
Calculate the kinetic energy of the electron shown in the diagram above.
mass of electron
= 9.1 × 10–31 kg
Planck constant
= 6.6 × 10–34 J s
(3)
(ii)
Calculate the potential energy of the electron in a hydrogen atom when it is at a
distance of 1.5 × 10–10 m from the proton that forms the nucleus of the atom.
permittivity of free space
= 8.9 × 10–12 F m–1
(4)
Page 12 of 67
(iii)
Hence, calculate the total energy of the electron.
(2)
(iv)
State and explain whether this model leads to a stable atom or not.
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(2)
(Total 17 marks)
13
The Bohr model of a hydrogen atom assumes that an electron e is in a circular orbit around a
proton P. The model is shown schematically in Figure 1.
Figure 1
In the ground state the orbit has a radius of 5.3 × 10–11 m. At this separation the electron is
attracted to the proton by a force of 8.1 × 10–8 N.
(a)
State what is meant by the ground state.
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(1)
Page 13 of 67
(b)
(i)
Show that the speed of the electron in this orbit is about 2.2 × 106 m s–1.
mass of an electron = 9.1 × 10–31 k g
(ii)
Calculate the de Broglie wavelength of an electron travelling at this speed.
Planck constant = 6.6 × 10–34 J s
(iii)
How many waves of this wavelength fit the circumference of the electron orbit? Show
your reasoning.
(7)
(c)
The quantum theory suggests that the electron in a hydrogen atom can only exist in certain
well-defined energy states. Some of these are shown in Figure 2.
Figure 2
An electron E of energy 2.5 × 10–18 J collides with a hydrogen atom that is in its ground
state and excites the electron in the hydrogen atom to the n = 3 level.
Page 14 of 67
Calculate
(i)
the energy that is needed to excite an electron in the hydrogen atom from the ground
state to the n = 3 level,
(ii)
the kinetic energy of the incident electron E after the collision,
(iii)
the wavelength of the lowest energy photon that could be emitted as the excited
electron returns to the ground state.
speed of electromagnetic radiation = 3.0 × 108 m s–1
(5)
(Total 13 marks)
14
(a)
(i)
Explain what is meant by the duality of electrons.
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(ii)
State the relation between the electron mass, electron velocity and the wavelength for
a monoenergetic beam of electrons.
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(3)
Page 15 of 67
(b)
The spacing of atoms in a crystal is 1.0 × 10–10 m.
mass of the electron
the Planck constant
=
=
9.1 × 10–31 kg
6.6 × 10–34 J s
Estimate the speed of electrons which would give detectable diffraction effects with such
crystals.
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(4)
(c)
Give one piece of evidence to demonstrate that electrons have particle properties.
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(1)
(Total 8 marks)
15
(a)
(i)
Explain what is meant by duality of electrons.
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(ii)
Give one example of each type of behaviour of electrons.
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(3)
Page 16 of 67
(b)
Electrons, of mass m and charge e, are accelerated from rest through a potential
difference, V, and acquire a kinetic energy of ½mυ2, where υ is the final velocity of the
electrons.
Show that
where h is the Planck constant and λ is the electron wavelength.
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(3)
(Total 6 marks)
16
In an electron diffraction tube, high speed electrons are produced by an electron gun at one end
of the tube. The electrons are incident on a thin slice of a polycrystalline material.
The diagram shows the pattern of bright rings that is formed on the fluorescent screen at the
other end of the tube.
(a)
Explain how the production of bright rings suggests that the electrons behave like waves.
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(1)
Page 17 of 67
(b)
The electrons in the tube have a velocity of 3.5 × 107 m s–1.
Calculate the de Broglie wavelength of the electrons.
de Broglie wavelength ................................................. m
(2)
(Total 3 marks)
17
The diagram shows some of the electron energy levels of an atom.
An incident electron of kinetic energy 4.1 × 10–18 J and speed 3.0 × 106 m s–1 collides with the
atom represented in the diagram and excites an electron in the atom from level B to level D.
(a)
For the incident electron, calculate
(i)
the kinetic energy in eV,
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Page 18 of 67
(ii)
the de Broglie wavelength.
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(4)
(b)
When the excited electron returns directly from level D to level B it emits a photon.
Calculate the wavelength of this photon.
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(3)
(Total 7 marks)
18
(a)
Electrons and electromagnetic waves exhibit properties of both waves and particles.
Suggest evidence which indicates that
(i)
electrons have wave properties,
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(ii)
electromagnetic radiation has particle properties,
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(iii)
electromagnetic radiation has wave properties.
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(3)
(b)
Calculate the de Broglie wavelength of an electron travelling at 5.0 × 106 m s–1. You should
ignore relativistic effects.
Page 19 of 67
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(3)
(Total 6 marks)
19
(a)
(i)
State what is meant by the wave-particle duality of electromagnetic radiation.
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(ii)
Which aspect of the dual nature of electromagnetic radiation is demonstrated by the
photoelectric effect?
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(2)
(b)
A metal plate is illuminated with ultra violet radiation of frequency 1.67 × 1015 Hz. The
maximum kinetic energy of the liberated electrons is 3.0 × 10–19 J.
(i)
Calculate the work function of the metal.
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(ii)
The radiation is maintained at the same frequency but the intensity is doubled. State
what changes, if any, occur to the number of electrons released per second and to
the maximum kinetic energy of these electrons.
number per second ...............................................................................
maximum kinetic energy ........................................................................
Page 20 of 67
(iii)
The metal plate is replaced by another metal plate of different material. When
illuminated by radiation of the same frequency no electrons are liberated. Explain
why this happens and what can be deduced about the work function of the new
metal.
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(8)
(Total 10 marks)
20
(a)
Calculate the energy, in J, of a photon of wavelength 4.50 × 10–7 m.
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(b)
Calculate the speed of an electron which has the same wavelength as the photon in part
(a).
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(Total 5 marks)
21
Electrons travelling at a speed of 5.00 × 105 m s–1 exhibit wave properties.
(a)
What phenomenon can be used to demonstrate the wave properties of electrons? Details
of any apparatus used are not required.
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(1)
Page 21 of 67
(b)
Calculate the wavelength of these electrons.
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(2)
(c)
Calculate the speed of muons with the same wavelength as these electrons.
Mass of muon = 207 × mass of electron
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(3)
(d)
Both electrons and muons were accelerated from rest by the same potential difference.
Explain why they have different wavelengths.
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(2)
(Total 8 marks)
Page 22 of 67
22
(a)
Calculate the de Broglie wavelength of an electron travelling at 2.00% of the speed of light.
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(b)
Determine the frequency of the electromagnetic radiation that would have the same
wavelength as this electron.
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(Total 5 marks)
23
(a)
Electrons behave in two distinct ways. This is referred to as the duality of electrons.
(i)
State what is meant by the duality of electrons.
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(ii)
Give one example of each type of behaviour of electrons.
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(3)
(b)
Calculate the speed of electrons that have a de Broglie wavelength of 1.70 × 10–10 m.
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(2)
(Total 5 marks)
Page 23 of 67
24
(a)
(i)
Name a force which acts between an up quark, u, and an electron. Explain, with
reference to an exchange particle, how this force operates.
You may be awarded marks for the quality of written communication in your answer.
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(ii)
With what particle must a proton collide to be annihilated?
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(4)
(b)
A sigma plus particle, Σ+, is a baryon.
(i)
How many quarks does the Σ+ contain?
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(ii)
If one of these quarks is an s quark, by what interaction will it decay?
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(iii)
Which baryon will the Σ+ eventually decay into?
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(3)
(Total 7marks)
25
(a)
Experiments based on the photoelectric effect support the particle nature of light. In such
experiments light is directed at a metal surface.
(i)
State what is meant by the threshold frequency of the incident light.
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(1)
Page 24 of 67
(ii)
Explain why the photoelectric effect is not observed below the threshold frequency.
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(2)
(b)
Monochromatic light of wavelength 5.40 × 10–7 m is incident on a metal surface which has
a work function of 1.40 × 10–19 J.
(i)
Calculate the energy of a single photon of this light.
answer = ...................................... J
(2)
(ii)
Calculate the maximum kinetic energy of an electron emitted from the surface.
answer = ...................................... J
(2)
(iii)
Calculate the maximum speed of the emitted electron.
answer = ...................................... m s–1
(2)
Page 25 of 67
(iv)
Calculate the de Broglie wavelength of the fastest electrons.
answer = ...................................... m
(2)
(Total 11 marks)
26
(a)
State what is meant by the wave-particle duality of electrons.
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(1)
(b)
Electrons of wavelength 1.2 × 10–10 m are required to investigate the spacing between
planes of atoms in a crystal.
(i)
Calculate the momentum of an electron of this wavelength stating an appropriate unit.
momentum of electron = ................................................
(3)
(ii)
Calculate the speed of such an electron.
speed of electron = .............................................. m s–1
(2)
Page 26 of 67
(iii)
Calculate the kinetic energy of such an electron.
kinetic energy of electron = .................................................... J
(2)
(Total 8 marks)
27
Electrons exhibit wave properties.
(a)
What phenomenon can be used to demonstrate the wave properties of electrons? Details
of any apparatus used are not required.
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(1)
(b)
Calculate the de Broglie wavelength of electrons travelling at a speed of 4.50 × 105 m s–1.
answer = ..................................... m
(2)
Page 27 of 67
(c)
The muon has a mass equal to 207 times the mass of an electron.
Calculate the speed of muons with the same de Broglie wavelength as the electrons in part
(b).
answer = ........................................ m s–1
(3)
(Total 6 marks)
28
When a clean metal surface in a vacuum is irradiated with ultraviolet radiation of a certain
frequency, electrons are emitted from the metal.
(a)
(i)
Explain why the kinetic energy of the emitted electrons has a maximum value.
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(2)
(ii)
Explain with reference to the work function why, if the frequency of the radiation is
below a certain value, electrons are not emitted.
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(2)
(iii)
State a unit for work function.
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(1)
Page 28 of 67
(b)
Light energy is incident on each square millimetre of the surface at a rate of
3.0 × 10–10 J s–1. The frequency of the light is 1.5 × 1015 Hz.
(i)
Calculate the energy of an incident photon.
answer = ....................................... J
(2)
(ii)
Calculate the number of photons incident per second on each square millimetre of the
metal surface.
answer = .........................................
(2)
(c)
In the wave theory model of light, electrons on the surface of a metal absorb energy from a
small area of the surface.
(i)
The light striking the surface delivers energy to this small area at a rate of
3.0 × 10–22 J s–1.
The minimum energy required to liberate the electron is 6.8 × 10–19 J.
Calculate the minimum time it would take an electron to absorb this amount of
energy.
answer = ....................................... s
(1)
Page 29 of 67
(ii)
In practice the time delay calculated in part c (i) does not occur. Explain how this
experimental evidence was used to develop the particle model for the behaviour of
light.
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(2)
(Total 12 marks)
29
(a)
When free electrons collide with atoms in their ground state, the atoms can be excited or
ionised.
(i)
State what is meant by ground state.
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(1)
(ii)
Explain the difference between excitation and ionisation.
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(3)
Page 30 of 67
(b)
An atom can also become excited by the absorption of photons. Explain why only photons
of certain frequencies cause excitation in a particular atom.
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(4)
(c)
The ionisation energy of hydrogen is 13.6 eV. Calculate the minimum frequency necessary
for a photon to cause the ionisation of a hydrogen atom. Give your answer to an
appropriate number of significant figures.
answer ..........................................Hz
(4)
(Total 12 marks)
Page 31 of 67
30
An electron has a speed of 8.4 × 105 m s–1.
Calculate the de Broglie wavelength of this electron.
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de Broglie wavelength .......................................................... m
(Total 2 marks)
31
(a)
When monochromatic light is shone on a clean cadmium surface, electrons with a range of
kinetic energies up to a maximum of 3.51 × 10–20 J are released. The work function of
cadmium is 4.07 eV.
(i)
State what is meant by work function.
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(2)
(ii)
Explain why the emitted electrons have a range of kinetic energies up to a maximum
value.
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(4)
Page 32 of 67
(iii)
Calculate the frequency of the light. Give your answer to an appropriate number of
significant figures.
answer = ................................ Hz
(4)
(b)
In order to explain the photoelectric effect the wave model of electromagnetic radiation was
replaced by the photon model. Explain what must happen in order for an existing scientific
theory to be modified or replaced with a new theory.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 12 marks)
32
(a)
J.J. Thompson investigated the nature of cathode rays in discharge tubes.
Suggest how he could have demonstrated that the cathode rays were negatively charged
particles.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 33 of 67
(b)
In an experiment, electrons are incident on a thin piece of graphite. The electrons emerging
from the graphite strike a fluorescent screen and produce the pattern shown in the figure
below.
State and explain the evidence this provides about the nature of moving electrons.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(c)
High energy electrons may be used to investigate the nature of protons of diameter
2.4 × 10–15 m.
(i)
Calculate the lowest value of the momentum of the high energy electrons that would
be suitable for this investigation.
State an appropriate unit for your answer.
momentum ................................
unit ....................................
(3)
Page 34 of 67
(ii)
Calculate the kinetic energy of the electrons.
kinetic energy ............................................... J
(2)
(Total 9 marks)
33
(a)
What phenomenon can be used to demonstrate the wave properties of electrons?
........................................................................................................................
(1)
(b)
Calculate the wavelength of electrons travelling at a speed of 2.5 × 105 ms–1.
Give your answer to an appropriate number of significant figures.
wavelength .......................................... m
(3)
(c)
Calculate the speed of muons with the same wavelength as these electrons.
mass of muon = 207 × mass of electron
speed ..................................... ms–1
(2)
(Total 6 marks)
Page 35 of 67
34
An electron has a kinetic energy E and a de Broglie wavelength λ. The kinetic energy is
increased to 4E. What is the new de Broglie wavelength?
A
B
C
λ
D
4λ
(Total 1 mark)
35
Electrons and protons in two beams are travelling at the same speed. The beams are diffracted
by objects of the same size.
Which correctly compares the de Broglie wavelength λe of the electrons with the de Broglie
wavelength λp of the protons and the width of the diffraction patterns that are produced by these
beams?
comparison of de
Broglie
wavelength
diffraction pattern
A
λe > λp
electron beam width > proton beam width
B
λe < λp
electron beam width > proton beam width
C
λe > λp
electron beam width < proton beam width
D
λe < λp
electron beam width < proton beam width
(Total 1 mark)
36
When comparing X-rays with UV radiation, which statement is correct?
A
X-rays have a lower frequency.
B
X-rays travel faster in a vacuum.
C
X-rays do not show diffraction and interference effects.
D
Using the same element, photoelectrons emitted using X-rays
have the greater maximum kinetic energy.
(Total 1 mark)
Page 36 of 67
37
The intensity of a monochromatic light source is increased. Which of the following is correct?
Energy of an emitted
photon
Number of photons
emitted per second
A
increases
increases
B
increases
unchanged
C
unchanged
increases
D
unchanged
unchanged
(Total 1 mark)
Page 37 of 67
Mark schemes
1
(a)
υ=
(1)
= 4.86 × 106 ms–1 (1)
2
(b)
yes (1)
same order as λ (1)
2
[4]
2
(a)
(i)
path difference clearly indicated correctly
B1
(1)
(ii)
path difference must be (n + ½)λ not just ½λ
(accept odd number of wavelengths) (allow diagram)
.
waves arrive at P antiphase / totally out of phase / exactly out of phase / out of
phase by by π or 180° (allow diagram)
B1
interfere destructively / cancel / “crests and troughs” at same time
B1
(3)
(iii)
idea of waves spreading out at a slit / slits
(allow diagram for this mark)
B1
production of overlapping beams from the double slit or illuminating both slits in
double slit
B1
(2)
(b)
fringe spacing = λD / d
or correct substitution of data
C1
0.83 mm
A1
(2)
Page 38 of 67
(c)
(i)
momentum of electrons = 1.27 × 10−24 (kg m s−1)
or m × v and correct substitution
C1
wavelength = h / p
C1
5.2 × 10−10 m
A1
or
wavelength = h / p and p = mv; or h / mv
C1
correct substitution of data
C1
correct answer
A1
(3)
(ii)
calculation of fringe spacing ( = 7.8 × 10−7 m)
allow e.c.f. for d / D confusion in (b)
or ratio λ / d for each
B1
the fringe spacing is too small (to enable separate fringes to be seen)
B1
or
calculation of D or of λ / d or speed for λ ≈ d
B1
speed in apparatus much larger than
that needed to make λ ≈ d
B1
(2)
[13]
3
(a)
(i)
146
B1
(1)
(ii)
90
B1
(1)
(b)
(i)
0.0046 u or 4.0061 u
B1
their mass change in u × 1.7 × 10–27 or
7.8 × 10–30 or 6.8 × 10–27 kg
B1
(2)
Page 39 of 67
(ii)
E = mc2 (or recalls 1 u = 931 MeV)
C1
their (i) × 9 × 1016
6.9 – 7.0 × 10–13 or 4.82 MeV
A1
(2)
(c)
(i)
speed determined correctly from their (ii)
(1.43 × 107 m s–1)
or
p2 / 2m = E or E = ½mv2
and momentum (p) = mv
C1
9.5 – 9.8 × 10–20 kg ms–1
A1
(2)
(ii)
wavelength = h / mv
C1
their value of h / their (i)
6.6 – 6.9 × 10–15 m
A1
(2)
[10]
4
5
6
A
[1]
B
[1]
(a)
the frequency needed to liberate an electron (electrons) from the surface of a material
or
minimum frequency to cause photoelectric effect
C1
the minimum frequency of the radiation / light / photon needed to liberate an electron
(electrons) from (the surface of) a material or from the surface
A1
(2)
(b)
the rate increases or more electrons per second
M1
there are more photons striking the surface each second
A1
no change in rate if frequency is below threshold frequency – allow 1
(2)
Page 40 of 67
(c)
(i)
Calculation using hc / E for (4.7 or 1.5 or 3.2) × 10–19 J
use of 1.5 leads to 1.32 × 10–6;
use of 3.2 leads to 6.2 × 10–7
C1
4.2 ×
10–7
m
A1
(2)
(ii)
use of 1.5 × 10–19 J
B1
p = √(2mE) and λ = h / p or E = ½ mv and λ = h / mv
2
C1
correct answer for their energy
1.26 × 10–9 m for 1.5 × 10–19 J
1.2 × 10–9 m for 1.7 × 10–19 J
0.86 × 10–9 m for 3.2 × 10–19 J
0.71 × 10–9 m for 4.7 × 10–19 J
A1
(3)
[9]
7
8
C
[1]
= 1.2(4) × 10–10 m (1)
(a)
(2)
(b)
(i)
same (1)
(ii)
same (1)
(2)
[4]
9
(a)
1.8/1.81 × 10–11 m
B1
1
(b)
circular bands of light on diagram
B1
diffraction/interference effect or electron λ ≈ atomic
spacing in graphite
B1
2
Page 41 of 67
(c)
state correct and appropriate particulate aspect
B1
quote evidence from this expt that shows electrons possess
aspect
B1
2
e.g.
electrons carry momentum/kinetic energy to screen
excite other electrons in atoms/cause emission of energy/light
or
electrons carry charge
can be accelerated by electric field/p.d.
etc
[5]
10
(a)
(i)
no electrons will be released / no current / no effect
B1
work function energy not being exceeded / insufficient
photon energy to exceed work function / photon
frequency below threshold frequency
B1
(ii)
more electrons released per second / current increases
B1
more photons (of sufficient energy) striking metal surface per second
B1
(iii)
answers must be in corresponding pairs below
cause – B1
electrons collide with air
molecules
photons absorbed by air
consequence – B1
less electrons reach anode
(s–1)
less photons reach plate so
fewer electrons emitted (s–1)
air contaminates plate
(work function ↑) so fewer
photons sufficiently energetic
to release electrons
cause must be
everything in one
pair of boxes above
ammeter reading or current
falls
Page 42 of 67
(b)
(i)
λ=
or correctly substituted values irrespective of powers of 10
B1
1.61 × 10–11 m
N.B. “show that”
B1
(ii)
crystal (or named crystalline material) / graphite
B1
atomic spacing (condone atomic diameter or distance
between nuclei) ≈ λ electrons
B1
[10]
11
(a)
(i)
Lines of equipotential parallel to the plates
B1
Field lines perpendicular to plates, evenly spaced
and with arrows upwards
B1
Lack of clear labelling of at least one of the types
of line loses 1 mark
Either field shown to be uniform
B1
3
(ii)
KE = 8.8 × 10–17 J
B1
Use of ½ mv2
C1
Speed = 1.4 × 107 m s–1
ecf
A1
Momentum =1.27 × 10–23 kg m s–1
ecf
B1
4
Page 43 of 67
(b)
Use of de Broglie wavelength = h/mv
C1
5.2 × 10–11 m
ecf
A1
diffraction of electrons necessary
M1
will work because wavelength is of same order as atomic
separation (not just wavelength is too small)/argument
consistent with their (a) (ii).
A1
4
[11]
12
(a)
e– likely where amplitude is max [or at r = 1.5 × 10–10 m]
B1
or impossible where amplitude is zero/probability of
finding electron amplitude of stationary wave is related
to probability
C1
probability is proportional to amplitude squared
A1
(b)
(i)
shows ½ mv2 = ½ (mv)2/m and states p = mv
B1
(ii)
States mv = h/λ
B1
½ mv2 = ½ (mv)2/m (i.e. re-writes k.e. in momentum
terms)
[k.e. = ½ h2/λ2/m]
B1
Page 44 of 67
(c)
(i)
λ = 6 × 10–10
B1
k.e. = ½ × (6.6 × 10–34)2 /(6 × 10–10 )2 × 9.1 × 10–31 [ecf]
C1
= 6.65 × 10–19 J
A1
(ii)
Uses kQ/r or variant
C1
p.e. = kQ1Q2/r1 = 8.98 × 109 × (1.6 × 10–19)2/1.5 × 10–10
C1
= (–)15.3 × 10–19 J
A1
makes it explicit that this is a negative
quantity relative to k.e.
allow use of symbol/stated value for electron/proton
charge
A1
(iii)
adds ((c)(i)) and ((c)(ii)) calcs correctly without regard
to sign
C1
quotes answer as –ve, addition is correct
(cao range –8 → –9 –8.57 × 10–19 J)
A1
(iv)
this is stable [ecf, must be consistent with (c)(iii)]
B1
negative total energy so energy must be supplied
to break up
B1
[17]
Page 45 of 67
13
(a)
lowest energy state/level that the electron can occupy
or state in which electron needs most energy to be released
B1
1
or the level of an unexcited electron (not lowest orbit)
(b)
(i)
force = mv2/r or mrω2 and v = rω
B1
8.1 × 10−8 = 9.1 × 10−31 × v2/5.3 × 10−11
or (v2 =) 4.72 × 1012 seen
B1
2.17 × 106 (m s−1)
B1
(ii)
λ = h/mv or 6.6 × 10−34/9.1 × 10−31 × 2.2 × 106
C1
7
3.3 × 10−10 m
A1
(iii)
circumference = 2π5.3 × 10−11 = 3.3 × 10−10 m
M1
1 (allow e.c.f. from (ii))
A1
(c)
(i)
1.9(4) × 10−18 J
B1
(ii)
5.6 × 10−19 J (e.c.f. 2.5 × 10−18 − their (i))
B1
Page 46 of 67
(iii)
energy difference E = 3 × 10−19 J
(condone any difference)
C1
E = hc/λ or E = hf and c=fλ
or their E = 6.6 × 10−34 × 3.0 × 108/λ
C1
6.6 or 6.7 × 10−7 m
A1
5
[13]
14
(a)
(i)
electrons behave sometimes as particles (1)
and sometimes as waves (1)
(ii)
mυ ∝ 1 / λ (or mυ = h / λ) (1)
(3)
(b)
For (crystal) diffraction, electron wavelength must be of order of atom spacing (1)
hence λ ≈ 10–10 m (1)
(1)
4
(c)
deflection in E-field
(or deflection in B-field, or any other correct evidence) (1)
(1)
[8]
15
(a)
(i)
(ii)
electrons behave sometimes as particles, sometimes as waves (1)
particle example:
deflection in electric [or magnetic] field (1)
wave example:
diffraction (1)
(3)
Page 47 of 67
(b)
(kinetic energy)
mυ2 = eV (1)
p (= mυ ) =
λ=
, gives result (1)
(3)
[6]
16
(a)
(Constructive) interference / superposition occurs
or
Waves arrive in phase so produce maximum intensity
Diffraction alone is not enough
B1
1
(b)
Correct substitution of numerical value in h / mv irrespective of powers of 10
C1
2.1 × 10−11 (m)
A1
2
[3]
17
(a)
(i)
k.e. =
(1)
= 26 (eV) (1) (25.6 eV)
(ii)
(use of λdB =
gives) λdB =
(1)
= 2.4 × 10–10 m (1) (2.42 × 10–10 m)
4
Page 48 of 67
(b)
(use of hf = E1 – E2 gives) f =
(1)
(= 1.05 × 1015 (Hz))
(use of λ =
gives) λ =
(1)
= 2.9 × 10–7 m (1) (2.86 × 10–7 m)
3
[7]
18
(a)
(i)
(wave property)
(electron) diffraction (1)
(ii)
(particle-property)
photoelectric effect (1)
(iii)
(wave property)
interference / diffraction / refraction (1)
(3)
(b)
(momentum of electron =) mυ = 9.11 × 10–31 × 5.0 × 106 (1)
(= 4.56 × 10–24 (kg m s–1)) (1)
[(λ = h/mυ gives) λ = 6.6(3) × 10–34 / 4.56 × 10–24 (1)
(allow e.c.f. for value of mυ)
= 1.5 × 10–10 m (1)
(1.45 × 10–10 m)
(3)
[6]
19
(a)
(i)
electromagnetic radiation behaves either as a particle or as a wave (1)
(ii)
(electromagnetic radiation) behaves as a particle (1)
(2)
(b)
(i)
hf = φ + Ek (1)
φ = (6.63 × 10–34 × 1.67 × 1015) – (3.0 × 10–19) (1)
= 8.1 × 10–19 (1) J (1) (8.07 × 10–19)
(ii)
(number per second) doubled (1)
(maximum kinetic energy) remains constant (1)
Page 49 of 67
(iii)
(all) electrons have insufficient energy to leave the (new) metal (1)
the work function of the (new) metal is greater than hf
[or the work function of the (new) metal is greater than
that of the original metal] (1)
The Quality of Written Communication marks were awarded primarily for the
quality of answers to this part.
(8)
[10]
20
(a)
(use of f =
gives) f =
(1) (= 6.6.7 × 1014Hz)
(use of E = hf gives) E = 6.63 ×10–34 × 6.67 × 1014 (1)
= 4.42 × 10–19(J) (1)
(b)
(use of λ =
gives) v (=
)=
(1)
= 1.62 × 103 m s–1 (1)
[5]
21
(a)
electron diffraction or interference (1)
1
(b)
(use of λ =
gives) λ =
(1)
= 1.46 × 10–9 m (1)
2
(c)
mm = 207 × 9.11 × 10–31 (kg) (1) ( = 1.89 × 10–28 (kg))
(use of meve = mμvμ , = when λ is constant, gives)
vμ =
(1)
= 2.4 × 103 m s–1(2.41 × 103 m s–1)
[or recalculate using v =
]
3
Page 50 of 67
(d)
gain in energy or work done on particle is the same for both (1)
wavelength is inversely proportional to momentum (1)
gain in momentum is different for both (1)
the smallest mass has the largest acceleration/gain in speed (1)
[or wavelength proportional to m–½ with constant k.e.]
max 2
[8]
22
(a)
(1) (=6.00 × 106(m s-1))
speed of electron =
(use of λ =
gives) λ =
(1)
= 1.21 × 10-10 m (1)
(b)
(use of c = fλ gives) f =
(1)
(allow C.E. for value of λ from (i))
= 2.48 × 1018Hz (1)
5
[5]
23
(a)
(i)
electrons behave as both particles and waves (1)
(ii)
particle: deflection in an electromagnetic field
or other suitable examples (1)
wave: electron diffraction (1)
3
(b)
(use of λ =
gives)
=
(1)
= 4.28 × 106 m s–1 (1)
2
[5]
Page 51 of 67
24
(a)
(i)
(named force) from weak (nuclear), electromagnetic or gravity (1)
uses a mediating/exchange particle, named particle from W(±) (boson),
(g) photon or graviton (1)
to transfer energy/momentum (1)
when electron emits/receives exchange particle,
disappearance/creation of new particle occurs (1)
QWC 1
(ii)
anti proton (1)
max 4
(b)
(i)
3 (quarks) (1)
(ii)
weak (nuclear) (1)
(iii)
proton (1)
3
[7]
25
(a)
(i)
below a certain frequency (called the threshold frequency)
no electrons emitted (1)
or minimum frequency for electrons to overcome work function
1
(ii)
(light travels as photons) energy of a photon depends on
frequency (1)
below threshold frequency (photon) does not have enough
energy to liberate an electron (1)
or reference to work function eg a photon does not have enough
energy (to allow the electron) to overcome the work function
2
(b)
(i)
(use of E = hc/λ)
E = 6.63 × 10–34 × 3.00 × 108/5.40 × 10–7 (1)
E = 3.68 × 10–19 (J) (1)
2
(ii)
(use of hf = Ek +
)
3.68 × 10–19 = Ek + 1.40 × 10–19 (1)
Ek = 2.28 × 10–19 (J) (1)
2
Page 52 of 67
(iii)
(use of Ek = mv2/2)
2.28 × 10–19 = 1/2 × 9.11 × 10–31 × v2 (1)
v2 = 2 × 2.28 × 10–19/9.11 × 10–31 = 5.0 × 1011
v = 7.1 × 105 (m s–1) (1)
2
(iv)
(use of λ = h/mv)
λ = 6.63 × 10–34(9.11 × 10–31 × 7.1 × 105) (1)
λ = 1.03 × 10–9 (m) (1)
2
[11]
26
(a)
electrons can have wavelike properties and particle like properties (1)
1
(b)
(i)
(use of λ = h/mv)
mv = 6.63 × 10–34/1.2 × 10–10 (1)
mv = 5.5 × 10–24 (1) kg m s–1 (1) (or Ns)
(ii)
v = 5.5 × 10–24/9.11 × 10–31 (1)
v = 6.1 × 106 m s–1 (1)
(iii)
(use of E = ½mv2)
E = ½ × 9.11 × 10–31 × (6.1 × 106)2 (1) (must see working
or equation)
E = 1.6(9) × 10–17 J (1) (no working max 1)
7
[8]
Page 53 of 67
27
(a)
(electron) diffraction/interference/superposition (1)
1
(b)
(use of λ = h/mv)
λ = 6.63 × 10–34/(9.11 × 10–31 × 4.50 × 105) (1)
λ = 1.6 × 10–9 (m) (1)
2
(c)
207 × 9.11 × 10–31 (1) × v = 6.63 × 10–34/1.6 × 10–7 (1)
v = 2200 (2170) (m s–1) (1)
3
[6]
28
(a)
(i)
hf is energy available/received or same energy from photons (1)
energy required to remove the electron varies (hence kinetic
energy of electrons will vary) (1)
2
(ii)
(work function is the) minimum energy needed to release
an electron (1)
(or not enough energy to release electron)
below a certain frequency energy of photon is less than
work function or energy of photon correctly related to f (1)
2
(iii)
joule (1) (accept eV)
1
(b)
(i)
(use of E = hf)
energy = 6.63 × 10–34 × 1.5 × 1015 (1)
energy = 9.9 × 10–19 (J) (1)
2
(ii)
number of photons per second = 3.0 × 10–10/9.9 × 10–19 (1)
number of photons per second = 3.0 × 108 (1)
2
(c)
(i)
(time taken = 6.8 × 10–19/3 × 10–22)
time taken = 2.3 × 103 s (1)
1
Page 54 of 67
(ii)
light travels as particles/ photons (1)
(or has a particle(like) nature)
(which transfer) energy in discrete packets (1)
or 1 to 1 interaction
or theory rejected/modified (in light of validated evidence)
2
[12]
29
(a)
(i)
when electrons/atoms are in their lowest/minimum energy (state) or
most stable (state) they (are in their ground state)
1
(ii)
in either case an electron receives (exactly the right amount of) energy
excitation promotes an (orbital) electron to a higher energy/up a level
ionisation occurs (when an electron receives enough energy) to leave
the atom
3
(b)
electrons occupy discrete energy levels
and need to absorb an exact amount of/enough energy to move to a higher level
photons need to have certain frequency to provide this energy or e = hf
energy required is the same for a particular atom or have different energy levels
all energy of photon absorbed
in 1 to 1 interaction or clear a/the photon and an/the electrons
4
(c)
energy = 13.6 × 1.60 × 10−19 = 2.176 × 10−18 (J)
hf = 2.176 × 10−18
f = 2.176 × 10−18 ÷ 6.63 × 10−34 = 3.28 × 1015 Hz
3 sfs
4
[12]
Page 55 of 67
30
correct substitution into formula, condone power of ten error
C1
8.7 × 10–10 (m)
A1
[2]
31
(a)
(i)
minimum energy required
to remove electron from metal (surface) OR cadmium OR the material
2
(ii)
photons have energy dependent on frequency OR energy of photons constant
one to one interaction between photon and electron
Max KE = photon energy – work function in words or symbols
more energy required to remove deeper electrons
4
(iii)
(use of hf = Ø + Ek(max))
6.63 × 10–34 × f = 4.07 × 1.60 × 10–19
f = 1.04 × 1015 (Hz) OR 1.03 × 1015 (Hz)
+ 3.51 × 10–20
(3 sig figs)
4
(b)
by repeatable/checked by other
theory makes predictions tested
scientists/peer reviewed (experiments) OR new evidence that is repeatable/
checked by other scientists/peer reviewed
2
[12]
Page 56 of 67
32
(a)
passed them between charged plates / near charged object
or
use magnetic field
M1
correct deviation
or
circular path in direction indicating negative charge
A1
2
(b)
diffraction
B1
electron is behaving as a wave
B1
2
(c)
(i)
p = h/λ or substitution of wavelength into λ = h/p or λ = h/mv
C1
2.76 or 2.8 × 10–19
A1
kg m s–1 / N s / J s m–1 / J Hz–1 m–1
B1
3
(ii)
EK = p2/2m or quotes p = mv and Ek = ½ mv2
(symbols or numbers)
C1
4.1 or 4.2 × 10–8 (J)
A1
2
[9]
Page 57 of 67
33
(a)
(electron) diffraction / interference / superposition ✓
Accept derfraction
1
(b)
(use of λ = h / mv)
λ = 6.63 × 10-34 / (9.11 × 10-31×2.5×105) ✓
λ = 2.9 × 10-9m ✓ ✓ (2 sig figs.)
3
(c)
v = 2.5 × 105 / 207 ✓
v = 1200 m s-1 ✓
OR use v = h / mλ with CE from part (b)
Answer alone gets 2 marks
2
[6]
34
35
36
37
B
[1]
A
[1]
D
[1]
C
[1]
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Examiner reports
1
Most candidates found part (a) to be straightforward and did it well.
Many candidates showed that they had clear, correct ideas in part (b), but a few talked vaguely of
electrons which were too small or going too fast.
2
3
(a)
(i)
Surprisingly few candidates were able to show the path difference clearly. This was
considered to be an easy beginning to the question considering that knowledge of
this distance is the first basic step in determining whether a signal is maximum or
minimum at a given point.
(ii)
There were many good thorough answers to this part. Even candidates who could not
identify what the path difference was in (i) were able to produce a clear answer in
many instances. Some candidates spoiled their answer by being specific and stating
that the path difference had to be a half wavelength. The phase difference resulting
from the different path lengths was the most commonly overlooked point.
(iii)
Candidates usually gained credit for showing the spreading of waves at a slit but
fewer explained clearly that this diffraction was necessary to produce two coherent
sources or overlapping beams, which then interfered. Many ignored the question and
proceeded to describe the conditions for maxima and minima.
(b)
This part was usually well done. Some candidates spoiled their attempt by doubling the
answer obtained using the correct formula.
(c)
(i)
This was completed successfully by the majority of the candidates although there was
a reluctance to give the unit as m. Many tried to deduce a composite formula and
many of these attempts were unsuccessful.
(ii)
Many good answers were given in this part. Candidates were able to undertake a
variety of relevant calculations but the most important feature of an explanation was
that the fringe spacing would be too small to be visible. Some candidates were
clearly confused between slit width and slit separation. These candidates often
deduced the ratio of λ / d and then wrote about the need to have a slit width of the
same order as the wavelength to produce diffraction. Candidates need to be more
sceptical about such a statement as in a typical Young's slits experiment fringes are
visible although this condition is clearly not met. A few candidates appreciated that
for the same slit width less diffraction would take place and therefore there would be
less likelihood of the beams overlapping within 0.6 m.
(i)
This was usually correct.
(ii)
There were many correct answers to this part but not as many as for part (i) as many
misread the question and presumably thought the question was still referring to the
uranium nucleus.
(i)
There were two reasonable interpretations of this question, both of which were
equally rewarded. Some candidates determined the difference in mass between the
uranium and the thorium nucleus and others the difference in mass between the
parent nucleus and the products of the decay. There were a large number of errors in
adding and subtracting the numbers but the majority knew how to convert u to kg.
(a)
(b)
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6
(ii)
Only the total change in mass was appropriate in this part. Many gained a mark for E
= mc2 but a large number of candidates seemed unaware of the physics involved
here and used ½mv2 or ½: mc2.
(c)
(i)
The use of ½ mc2 to determine the speed of the alpha particle was correct in this part
followed by momentum = my. The common error was to calculate momentum
assuming the alpha particle to travel at 3 × 108 m s−1. Some tried to use p = h / λ.
(a)
A large number of candidates gave a complete answer. However, failure to state that it is
the frequency of the electromagnetic radiation (or light) that is relevant or to state that
electrons were emitted from the surface were causes of many lost marks. A significant
proportion of the candidates did not know what happens in the photoelectric effect and had
the idea that photons were emitted due to the incidence of electrons.
(b)
Many gave loose answers that did not refer to the rate at which electrons were emitted and
stated simply ‘more electrons emitted’. The fact that higher intensity meant more photons
arriving per second per square metre was not well known.
(c)
Many were confused between the equations that they needed to use for electromagnetic
radiation and for particles. Correct answers to the two parts were, therefore, frequently
seen the wrong way round.
(i)
Those who used the correct formula in this part often used the wrong energy (usually
1.5 × 1019 J).
(ii)
Although many quoted h / mv they were clearly confused and 20 to 30% of the
candidates used c as the velocity. Most who knew the correct process used 1.5 ×
1019 J in their attempt to determine the velocity but errors with arithmetic were
common.
9
(a)
This simple calculation of the de Broglie wavelength was well done by many.
(b)
A pleasing number could give an indication of the likely diffraction pattern on the front of the
tube and then go on to offer an explanation of the wave behaviour in terms of diffraction or
interference effects.
(c)
This was less impressive. Only about one-third of candidates reasoned that (for example)
charge is a particulate property and that it is demonstrated by the acceleration of the
electron in the electric field.
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10
(a)
(b)
11
12
(a)
(i)
Answers to this were variable with only better candidates making direct references
relating the photon energy to the threshold frequency or work function.
(ii)
This part was often very poorly answered – very few candidates made reference to
increased rate of arrival of photons and to the rate of emission of photoelectrons.
Many candidates appeared to believe that greater intensity meant more energetic
photons.
(iii)
Again there few totally convincing answers given to this part. Many candidates
recognised that the likely outcome would be that the current would fall but only the
best candidates were able to give satisfactory reasons for this either in terms of
absorption of some photons by the air or the collisions between the photoelectrons
and the air molecules (reducing the rate at which photoelectrons reached the anode).
Many candidates answered in a manner which indicated confusion between the
photons and the photoelectrons.
(i)
Most candidates were able to show that the de Broglie wavelength was
approximately equal to the given value. Weaker candidates were confused by the
difference between c and v in the momentum relationship.
(ii)
Most candidates were able to suggest a crystalline or polycrystalline material. The
majority of these candidates did not make an overt comparison between the atomic
(ionic) spacing and the de Broglie wavelength of the electron.
(i)
Most of the candidates could draw the field using both lines of equipotential and
electric lines. A few omitted to label the lines. A more common mistake was to draft
the diagram carelessly so that it was not clear that the field was apparently uniform.
(ii)
Weaker candidates got very tangled in this calculation, attempting to use
½ mv2
to calculate kinetic energy rather than using it to calculate speed once they had found
the kinetic energy by using the potential difference in the field.
(b)
The calculation in this part was done quite well. Few candidates could go on to explain
whether or not the de Broglie wavelength made the electrons suitable for the investigation
of metallic crystal structures. Some had no idea what the typical values for atomic
separations are in metallic crystals. More surprisingly, those who did know the separations
tended to be unclear about whether the wavelength was too big or too small or broadly
applicable.
(a)
Candidates were usually able to suggest what the diagram showed (in terms of the likely
position of the electron) but only better candidates went on to discuss the relationship
between amplitude and probability.
(b)
(i)
This simple algebraic manipulation was done well by many.
(ii)
Similarly, many were able to identify the relationship required and to manipulate it for
an easy two marks.
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(c)
(i)
Although most were able to spot that they needed the equation from (b)(ii) for this
calculation (surprisingly some could not), a number of common errors crept in. Some
failed to recognise that the wave shown in the diagram is half of a wavelength, some
recognised this but divided the half wavelength by two rather than multiplying by two
and arrived at answers that were considerably adrift of the correct value. The unit
was often missing here.
(ii)
Unfortunately, the value of the charge of the electron was omitted from this question.
However, the value appeared twice elsewhere in the paper and the omission did not
present any problem to the vast majority of candidates. (The mark scheme allowed
full credit for solutions which used an incorrect value for e or which provided an
algebraic solution.) In fact, the major problem in this part was not the absence of the
charge, but a complete misunderstanding of the physics. About half the candidates
used the equation for electric field strength (with the correct values for the charge on
the electron and proton). Others, whilst recognising the correct equation, failed to
include the proton charge. A very common error was to disregard the sign of the
answer, expressing it as a positive value. This had consequences for the remaining
parts of the question.
(iii)
Errors were carried forward to part (iii) but even so there was much poor physics.
Even those who had arrived at both the correct magnitude and sign for part (ii)
dropped the sign as they moved to part (iii), and lost a mark in consequence. Again,
algebraic/incorrect values were accepted here as ecf from (ii)
(iv)
There was little understanding of the concept of atomic stability shown in this part.
Often, candidates made incorrect assertions about stability. There is no general
understanding that a negative value of total energy means a stable situation because
energy will be required from elsewhere to release the electron.
The remaining questions were based on a text passage that dealt with some
elements of the physics of loudspeakers.
13
(a)
This was generally known well.
(b)
(i)
This was done well by the majority of the candidates although setting out of the
working left something to be desired in many instances.
(ii)
This part was usually correct but some omitted a unit.
(iii)
Correct calculation of the circumference was essential to this part. A small but
significant number of candidates did this incorrectly or compared the wavelength with
the radius or diameter and therefore gained no credit.
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(c)
(i)
This was usually correct. A few ignored the factor of 1019 or gave no unit.
(ii)
This was also generally successfully completed.
(iii)
The most common error was use of an incorrect energy difference. A small minority
used λ =
15
Most candidates answered part (a)(i) well, but part (a)(ii) produced many inaccurate or poorly
expressed ideas.
Part (b) produced many answers which earned full marks, but it also showed that many
candidates could not rearrange simple algebraic equations. Common mistakes were to rearrange
½ mʋ2 =eV into ʋ = (2meV)½ , or to use ½ mʋ2 = meV.
16
17
(a)
Candidates needed to explain the production of the bright ring by constructive interference.
The mention of diffraction alone was insufficient.
(b)
Most candidates obtained the correct wavelength. Weaker candidates often quoted the
equation and then left the mass component blank, suggesting that they did not know that
the mass of the electron was on the formula and data sheet. Some substituted the value for
the electron charge for the mass.
Parts (a) and (b) of this question showed errors at different levels of ability. The slightly better
candidates used the correct equations but often used the speed of light in the de Broglie
relationship. The very weak candidates did not know when to apply
or
. In other cases, the wrong energy was used to calculate the wavelength of a
photon and it was not uncommon to see the electron energies at levels D and B being added
together.
Part (a) (i) in particular showed a variety of errors. Multiplying, rather than dividing by the electron
charge was the obvious error. Some of the better candidates used the electron speed to
calculate the kinetic energy, arriving at the energy in joules which had already been given in the
question, and then failing to convert this to eV. About 15% of the candidates incurred a significant
figure error on this question by quoting an answer to five significant figures.
18
There was some evidence that quite a number of candidates were not prepared for this topic.
Candidates who were conversant with it usually gave a completely correct answer, but others
usually failed to score at all. In the calculation in part (b), a number of candidates started with
½mv2, showing that they were unfamiliar with the subject.
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19
Overall, the candidates had a sound understanding of the photoelectric effect and there was a
good response to part (a). The most common error was for candidates to refer to the
wave-particle duality of electrons rather than of electromagnetic radiation.
The calculation in part (b)(i) was carried out correctly by most candidates but a large number of
answers were presented without units. Good candidates had no difficulty with part (ii) but many
simply stated that “the number of electrons released per second increases”, rather than “the
number doubles”. Answers to part (iii) indicated that candidates seemed to have a much better
understanding of the photoelectric effect than those who sat the examination in January.
Pleasingly few candidates referred to electrons in shells, ionisation or the electronic bond with a
single atom. The weaker candidates often lost marks because they expressed ideas in an
unclear fashion, for example, “the work function needs to be higher in the new metal” was a
statement commonly seen.
20
21
This question provided very good discrimination with the majority of candidates failing to make
any worthwhile attempt at all the calculations. Although a significant number of candidates
completed part (a) correctly, many of the others simply invented formulae to fit the data provided
e.g. energy = hλ appeared quite often. In part (b) the majority of candidates failed to use the de
Broglie relationship, but instead chose to misinterpret the question and tried to find the speed of
an electron having the same energy as the photon.
Most candidates knew that electrons exhibited wave properties when they were diffracted but it
was sad to see that the correct spelling of diffraction was accomplished by only a minority of
candidates. The de Broglie calculation in part (b) was attempted correctly by most candidates but
a significant number gave an answer to only 1 significant figure. This showed that candidates did
not set their calculator to display scientific notation and 0.000000001 was unthinkingly used as
the final answer.
Part (c) was found to be a little more difficult than part (b) due to the given information being fairly
complicated and also the need to rearrange the working equation.
Part (d) proved to be difficult even for the good candidate. Many candidates made links between
variables without any justification; for example, “the wavelength must be different because their
masses are different”, appeared frequently. Also, vague statements rather than specific
statements were made, e.g. candidates stated that the mass of the moon was different rather
than larger than that of the electron.
22
Most candidates showed that they could perform these types of calculations with ease. A number
of candidates however did pick up a significant figure penalty. Part (b) did show more physics
errors than part (a), the most common being using the electron speed rather than the speed of
light in the equation c = fλ
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23
Most candidates were aware of wave particle duality, but they sometimes lost marks through lack
of care. For example, a statement such as “an electron can behave as a wave or as a photon”
was common but did not gain any marks. In identifying the behaviour of electrons, weaker
candidates often gave an example but failed to state which type of behaviour it represented.
In part (b) only the weaker candidates had trouble with the calculation. They either failed to use
the de Broglie equation or could not rearrange the equation to make the speed, v, the subject.
24
In part (a) most candidates referred correctly to the Weak Nuclear Force, giving the W+ or Wboson as the exchange particle. Fewer candidates gave the alternative gravity force or
electromagnetic force as answers, but a few thought a possible interaction was through the
Strong Nuclear Force, even though the electron is a lepton and not subject to the SNF. Even
good candidates found it difficult to obtain full marks in this part because only a few stated that
the exchange particle transfers energy or momentum, and an even smaller number wrote about
particles being created or annihilated. Almost all candidates answered (a) (ii) correctly.
Part (b) turned out to be quite a discriminating section. The better candidates appreciated that
this question could be answered by just knowing that a baryon was involved and the fact it was a
Σ+ particle was purely incidental. Surprisingly, even when this fact was understood, the three
sub-sections proved to be too difficult for candidates lacking the required knowledge.
25
Part (a) was not answered well and there was much confusion as to the processes involved in
the photoelectric effect. As has been the case in the past, a significant number of candidates
confused the effect with excitation and ionisation of atoms. Only a minority of candidates were
able to link the energy of a photon to its frequency and there was much confusion between
threshold frequency and the work function. It was not unusual to see responses that stated the
threshold frequency has to equal the work function. It also seems that some candidates are
under the impression that the photoelectric effect involves the emission of photons – presumably
due to confusion between excitation and line spectra.
Part (b) proved to be a familiar calculation for the majority of candidates and many competent
answers were seen. The only significant misunderstanding occurred in (b) (iii) when candidates
assumed that the kinetic energy of the electron is found by using the equation for the energy of a
photon.
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26
Part (a) was answered reasonably well and candidates generally understood the meaning of
particle wave duality. Some candidates just referred to wave properties, presumably assuming
that particle properties were self-evident.
In part (b) (i) it was noticeable that less able candidates confused momentum with energy. The
unit for momentum also caused significant problems. The penalty for not being able to calculate
momentum was not a major one, as consequential error was allowed for parts b (ii) and b (iii).
27
28
This question was well answered and the majority of candidates appreciated that diffraction is a
wavelike property that electrons exhibit. The calculation in part (b) proved to be quite
straightforward and full marks were obtained by a pleasing number of candidates.
The initial parts of this question caused considerable problems to candidates. They found it very
difficult to explain why the kinetic energy of the emitted electrons had a maximum value and also
fully explain the link between photon energy, work function and maximum kinetic energy. The
idea that some electrons require more energy to be emitted than others did appear to be well
understood. Candidates also had a tendency to confuse the photoelectric effect with excitation
and ionisation. Evidence from this and previous papers suggests that this is a topic candidates
find very difficult and this is particularly true when they are required to explain aspects of the
phenomena.
Parts (b) and (c) proved much more accessible and candidates used the various relevant
equations confidently. Full marks for calculations were quite common. Part (c) (ii), which
assessed How Science Works, did confuse some candidates. When this happened, candidates
tended to explain the significance of validated evidence in general terms, rather than how it was
used to develop the particle model of light.
29
Many students were able to distinguish between excitation and ionisation successfully and also
to define the ground state. They clearly found the structured format of this question helpful.
However, students were not so good at explaining the process of excitation of atoms by the
absorption of photons. It was common to see muddled answers that confused the photoelectric
effect with excitation. The term work function was often used incorrectly in candidate responses
as was threshold frequency. A significant minority focused on the photon released after excitation
rather than the incident photon.
The calculation in part (c) was generally done well and most students gave answers to the
correct number of significant figures. A common error by some students was to fail to convert
electron volts to joules, this mistake limited them to a maximum of two marks.
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30
31
32
Many candidates were able to obtain full marks for this calculation. The two most common errors
made were the lack of substitution for the mass of the electron and a calculator error involving
the order in which the calculation was performed. This calculation error meant that candidates
performed (h ÷ m) × v.
Students have found questions on the photoelectric effect quite challenging in previous series. In
view of this it was pleasing to see more confident answers this time around. The explanation of
work function and the calculation were well answered by a significant proportion of students.
Explanations of the range of kinetic energies were less sound and as has been the case in the
past there was frequent confusion between the photoelectric effect and excitation of electrons
from discrete energy levels. The idea of validated evidence which was required in part (b), seems
now to be well understood.
Strictly correct answers to part (a) were rare but credit was given to answers involving charged
objects rather than parallel plates. Those who chose to use magnets, almost universally
mentioned positive and negative poles of magnets! This was penalised.
Although there were some good answers to part (b) referring to the diffraction pattern and top the
wave–particle duality, some interpreted the diagram as representing the energy levels in an
atom.
In part (c) (i), the correct equation was generally used but students often did not use the correct
substitution for wavelength. The unit for momentum was often correct. A disappointing number
chose to give a version of the unit derived from the equation instead of an appropriate,
remembered unit. A significant minority of students did not attempt (c) (ii). Most had no idea of
how to find the kinetic energy from the momentum.
33
This question was well answered and there and there were no major issues raised by it.
A significant proportion of candidates did however; lose the significant figure mark in part (b).
This was usually because they gave their answers to three significant figures instead of two.
It was quite common for candidates to obtain full marks for part (c) by to using the de Broglie
equation to calculate the speed of the muons rather than simply dividing the speed of the
electrons by 207.
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