1
Observations of the H-α line in the spectrum of a star indicate the presence of hydrogen. The H-α
line has a wavelength of 656 nm and is produced by a transition of electrons into the −3.4 eV
energy level.
Calculate the energy level that the electron moves from when emitting a photon corresponding to
a wavelength of 656 nm. Give your answer in J.
energy level ................................................... J
(Total 4 marks)
2
The diagram shows some energy levels of an atom.
The transition E3 to E1 corresponds to the emission of visible light.
A transition corresponding to the emission of infrared radiation could be
A
E1 to E0
B
E4 to E1
C
E1 to E2
D
E3 to E2
(Total 1 mark)
Page 1 of 50
3
The diagram shows some of the energy levels for a hydrogen atom.
___________________________
first excited state
ground state
0
___________________________
–5.4 × 10-19 J
___________________________
–21.8 × 10-19 J
A free electron of kinetic energy 20.0 × 10–19 J collides with a hydrogen atom in its ground state.
The hydrogen atom is excited from its ground state to the first excited state. The kinetic energy of
the free electron after the collision is
A
1.8 × 10–19 J
B
3.6 × 10–19 J
C
5.4 × 10–19 J
D
16.4 × 10–19 J
(Total 1 mark)
4
The diagram represents some of the energy levels of an isolated atom. An electron with a kinetic
energy of 2.0 × 10–18 J makes an inelastic collision with an atom in the ground state.
(a)
Calculate the speed of the electron just before the collision.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 2 of 50
(b)
(i)
Show that the electron can excite the atom to level 2.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the wavelength of the radiation that will result when an atom in level 2 falls
to level 1 and state the region of the spectrum to which this radiation belongs.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(c)
Calculate the minimum potential difference through which an electron must be accelerated
from rest in order to be able to ionise an atom in its ground state with the above energy
level structure.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 10 marks)
5
The diagram drawn to scale shows some of the energy levels of an atom. Transition P results in
the emission of a photon of wavelength 4 × 10–7 m.
Which one of the transitions A, B, C, or D could result in the emission of a photon of wavelength
8 × 10–7 m?
(Total 1 mark)
Page 3 of 50
6
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.
........................................................................................................................
........................................................................................................................
(1)
(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)
Page 4 of 50
(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.
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)
Page 5 of 50
7
(a)
Describe how the concept of energy levels is useful in the explanation of line spectra.
........................................................................................................................
........................................................................................................................
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........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
The diagram represents some energy levels of the mercury atom.
__________________________________ 0
__________________________________ –1.6 eV
__________________________________ –3.7 eV
__________________________________ –5.5 eV
__________________________________ –10.4 eV ground state
charge of electron = 1.6 × 10–19 C
the Planck constant = 6.6 × 10–34 J s
speed of light in vacuo = 3.0 × 108 m s–1
(i)
What is the ionisation energy, in J, of the mercury atom?
...............................................................................................................
...............................................................................................................
(ii)
Determine which transition corresponds to the emission of radiation of wavelength
141 nm.
...............................................................................................................
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Page 6 of 50
(iii)
State the region of the spectrum in which you would expect to find radiation of this
wavelength.
...............................................................................................................
(7)
(Total 10 marks)
8
The diagram below shows some of the energy levels of the hydrogen atom.
energy/10–19J
0
___________________________ n = ∞
–2.4 ___________________________ n = 3
–5.4 ___________________________ n = 2
–22 ___________________________ n = 1 (ground state)
(a)
Explain how changes of electron energies can produce a line emission spectrum.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(b)
(i)
What is meant by ionisation?
...............................................................................................................
...............................................................................................................
(ii)
State the energy, in J, required to ionise a hydrogen atom from its ground state.
...............................................................................................................
(iii)
Calculate the minimum frequency of radiation that can ionise a hydrogen atom from
its ground state.
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 7 of 50
(iv)
Explain what happens to an electron in the ground state of a hydrogen atom when it
receives radiation of a frequency greater than the minimum frequency obtained in
part (b)(iii).
...............................................................................................................
...............................................................................................................
(5)
(c)
Calculate the wavelength of the radiation emitted when an electron falls from level n = 3 to
level n = 2 in the hydrogen atom.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 11 marks)
9
The lowest energy levels of a hydrogen atom are represented in the diagram below, which is not
to scale.
(a)
Describe what happens when a hydrogen atom is ionised.
........................................................................................................................
........................................................................................................................
(b)
State the minimum amount of energy, in J, required to ionise a hydrogen atom from its
ground state.
........................................................................................................................
Page 8 of 50
(c)
A hydrogen atom excited to the n = 3 energy level may emit either a single photon or two
photons in returning to the ground state.
Describe what happens to the electron in each case.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(d)
Use the diagram above to identify the transition which produces a photon of energy
2.09 × 10–18 J.
(e)
Calculate the frequency of an emitted photon due to a transition from level n = 2 to the
ground state.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(Total 8 marks)
10
(a)
State what happens in an atom when line spectra are produced.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 9 of 50
(b)
The diagram below represents some energy levels of the lithium atom.
(i)
Calculate the ionisation energy, in J, of the lithium atom.
...............................................................................................................
...............................................................................................................
(ii)
An excited lithium atom may emit radiation of wavelength 6.1 × 10–7 m.
Show that the frequency of this radiation is approximately 5.0 × 1014 Hz.
...............................................................................................................
...............................................................................................................
(iii)
Calculate the energy, in J, of each photon of this radiation.
...............................................................................................................
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Page 10 of 50
(iv)
Draw, on the diagram, an arrow between two energy levels which shows the
transition responsible for the emission of a photon of energy 2.0 eV.
(v)
Two transitions emit radiation of similar frequencies. One of them is the transition
between A and C. What is the other?
...............................................................................................................
(vi)
A transition between which two levels would give radiation of the longest possible
wavelength?
...............................................................................................................
(9)
(Total 11 marks)
11
The circuit diagram shows a light emitting diode (LED) connected in series with a resistor, R, and
a 3.0 V battery of negligible internal resistance.
(a)
The LED lights normally when the forward voltage across it is 2.2 V and the current in it is
35 mA.
Calculate
(i)
the resistance of R,
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
the number of electrons that pass through the LED each second.
...............................................................................................................
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...............................................................................................................
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(4)
Page 11 of 50
(b)
The LED emits light at a peak wavelength of 635 nm.
(i)
Calculate the energy of a photon of light of this wavelength.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Estimate the number of photons emitted by the LED each second when the current
through it is 35 mA. Assume all the photons emitted by the LED are of wavelength
635 nm and that all the electrical energy produces light.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(Total 8 marks)
12
(a)
A fluorescent tube is filled with mercury vapour at low pressure. In order to emit light the
mercury atoms must first be excited.
(i)
What is meant by an excited mercury atom?
...............................................................................................................
...............................................................................................................
(ii)
Describe the process by which mercury atoms become excited in a fluorescent tube.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(b)
What is the purpose of the coating on the inside surface of the glass in a fluorescent tube?
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 12 of 50
(c)
The lowest energy levels of a mercury atom are shown below. The diagram is not to scale.
(i)
Calculate the frequency of an emitted photon due to a transition, shown by an arrow,
from level n = 4 to level n = 3.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Draw a line on the diagram to show a transition which emits a photon of a longer
wavelength than that emitted in the transition from level n = 4 to level n = 3.
(3)
(Total 8 marks)
13
A diffraction grating was used to measure the wavelength of a certain line of a line emission
spectrum.
(a)
The grating had 600 lines per millimetre. The angle of diffraction of the second order line
was 35.8°.
(i)
Calculate the wavelength of this line.
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Page 13 of 50
(ii)
Calculate the energy, in eV, of a photon of this wavelength.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
(5)
(b)
The line emission spectrum observed in part (a) was produced by a hot gas.
(i)
The energy level diagram for the atoms that produced the line spectrum is shown in
the diagram below. Mark on the diagram a vertical arrow to show the electron
transition between the two levels that produced photons of energy 6.8 eV.
(ii)
The temperature of the gas was 5000K. Show that the mean kinetic energy of a gas
atom at this temperature is 0.65 eV.
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Page 14 of 50
(iii)
Describe how the atoms of a gas produce a line emission spectrum and explain why
the gas at a temperature of 5000K can produce a line of the wavelength calculated in
part (a)(i).
You may be awarded marks for the quality of written communication in your answer.
...............................................................................................................
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...............................................................................................................
(6)
(Total 11 marks)
Page 15 of 50
14
The diagram shows some of the energy levels of the mercury atom.
(a)
When electrons collide with mercury atoms, the atoms may be excited or may be ionised.
Explain what is meant by
(i)
excitation,
.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
ionisation.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(2)
Page 16 of 50
(b)
Determine the lowest frequency of emitted radiation with reference to the energy levels in
the diagram.
......................................................................................................................
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......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 4 marks)
15
The diagram shows some energy levels, in eV, of an atom.
Photons of specific wavelengths are emitted from these atoms when they are excited by
collisions with electrons.
You may be awarded marks for the quality of written communication in your answer.
(a)
Explain
(i)
what is meant by the process of excitation,
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
Page 17 of 50
(ii)
why the emitted photons have specific wavelengths.
.............................................................................................................
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.............................................................................................................
(5)
(b)
One of the emitted photons has an energy of 9.92 × 10–19 J.
(i)
Calculate the wavelength of this photon.
.............................................................................................................
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.............................................................................................................
.............................................................................................................
(ii)
Determine which transition is responsible for this emitted photon.
.............................................................................................................
.............................................................................................................
.............................................................................................................
(iii)
Draw an arrow on the energy level diagram above to show the transition responsible
for the emission of a photon with the shortest wavelength.
(7)
(Total 12 marks)
Page 18 of 50
16
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,
.............................................................................................................
.............................................................................................................
(ii)
the de Broglie wavelength.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(4)
Page 19 of 50
(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)
17
(a)
The diagram below shows some of the energy levels for an iron atom.
(i)
Draw another arrow on the diagram above to represent the smallest energy change
possible for an electron moving between two of the energy levels shown.
The electron energy change selected must result in energy being emitted from the
atom.
Label this arrow B.
(1)
Page 20 of 50
(ii)
In the diagram above, when the energy change labelled A occurs an X-ray photon is
emitted.
Show that the frequency of the photon is approximately 2 × 1018 Hz.
(3)
(b)
(i)
Radiation of frequency 2 × 1018 Hz has a wavelength of 1.5 × 10–10 m.
Calculate the speed of an electron that has a de Broglie wavelength of
1.5 × 10–10 m.
speed .........................................m s–1
(2)
(ii)
Explain why electrons of this wavelength would be suitable to investigate the
structure of a metallic crystal.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 8 marks)
Page 21 of 50
18
Some energy levels of an atom of a gas are shown in Figure 1.
Figure 1
When a current is passed through the gas at low pressure, a line spectrum is produced. Two of
these lines, which correspond to transitions from levels B and C respectively to the ground state,
are shown in Figure 2.
Figure 2
(a)
Describe what happens to an electron in an atom in the ground state in order for the atom
to emit light of wavelength 4.0 × 10–7 m.
You may be awarded marks for the quality of written communication in your answer.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(3)
Page 22 of 50
(b)
Determine the energy, in J, of
(i)
the photons responsible for each of the two lines shown in Figure 2,
.............................................................................................................
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.............................................................................................................
.............................................................................................................
.............................................................................................................
(ii)
levels B and C in Figure 1.
.............................................................................................................
.............................................................................................................
.............................................................................................................
energy of level B = ...............................................................................
energy of level C = ...............................................................................
(5)
(Total 8 marks)
19
The diagram below shows part of an energy level diagram for a hydrogen atom.
n = 4 _________________ –0.85 eV
n = 3 _________________ –1.50 eV
n = 2 _________________ –3.40 eV
n = 1 _________________ –13.60 eV
(a)
The level, n = 1, is the ground state of the atom.
State the ionisation energy of the atom in eV.
answer = ................................... eV
(1)
(b)
When an electron of energy 12.1 eV collides with the atom, photons of three different
energies are emitted.
(i)
On the diagram above show with arrows the transitions responsible for these
photons.
(3)
Page 23 of 50
(ii)
Calculate the wavelength of the photon with the smallest energy. Give your answer to
an appropriate number of significant figures.
answer =.............................. m
(5)
(Total 9 marks)
20
(a)
A fluorescent tube is filled with mercury vapour at low pressure. In order to emit
electromagnetic radiation the mercury atoms must first be excited.
(i)
What is meant by an excited atom?
.............................................................................................................
.............................................................................................................
(1)
(ii)
Describe the process by which mercury atoms become excited in a fluorescent tube.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
Page 24 of 50
(iii)
What is the purpose of the coating on the inside surface of the glass in a fluorescent
tube?
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
The lowest energy levels of a mercury atom are shown in the diagram below. The diagram
is not to scale.
energy / J × 10–18
.................................................
n = 4 ________________________
0
–0.26
n = 3 ________________________ –0.59
n = 2 ________________________ –0.88
ground state n = 1 ________________________
(i)
–2.18
Calculate the frequency of an emitted photon due to the transition level n = 4 to level
n = 3.
answer = ........................................ Hz
(3)
(ii)
Draw an arrow on the diagram above to show a transition which emits a photon of a
longer wavelength than that emitted in the transition from level n = 4 to level n = 3.
(2)
(Total 12 marks)
Page 25 of 50
21
Figure 1 shows data for the variation of the power output of a photovoltaic cell with load
resistance. The data were obtained by placing the cell in sunlight. The intensity of the energy
from the Sun incident on the surface of the cell was constant.
Figure 1
Load resistance / Ω
(a)
Use data from Figure 1 to calculate the current in the load at the peak power.
(3)
Page 26 of 50
(b)
The intensity of the Sun’s radiation incident on the cell is 730 W m
cell has dimensions of 60 mm × 60 mm.
–2.
The active area of the
Calculate, at the peak power, the ratio
(3)
(c)
The average wavelength of the light incident on the cell is 500 nm. Estimate the number of
photons incident on the active area of the cell every second.
(2)
Page 27 of 50
(d)
The measurements of the data in Figure 1 were carried out when the rays from the sun
were incident at 90° to the surface of the panel. A householder wants to generate electrical
energy using a number of solar panels to produce a particular power output.
Identify two pieces of information scientists could provide to inform the production of a
suitable system.
........................................................................................................................
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........................................................................................................................
(2)
(Total 10 marks)
22
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(ii)
Explain the difference between excitation and ionisation.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
Page 28 of 50
(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 29 of 50
23
The diagram below shows the lowest three energy levels of a hydrogen atom.
(a)
An electron is incident on a hydrogen atom. As a result an electron in the ground state of
the hydrogen atom is excited to the n = 2 energy level. The atom then emits a photon of a
characteristic frequency.
(i)
Explain why the electron in the ground state becomes excited to the n = 2 energy
level.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(ii)
Calculate the frequency of the photon.
frequency = ......................................... Hz
(3)
Page 30 of 50
(iii)
The initial kinetic energy of the incident electron is 1.70 × 10–18 J.
Calculate its kinetic energy after the collision.
kinetic energy = ............................................ J
(2)
(iv)
Show that the incident electron cannot excite the electron in the ground state to the n
= 3 energy level.
(2)
(b)
When electrons in the ground state of hydrogen atoms are excited to the n = 3 energy
level, photons of more than one frequency are subsequently released.
(i)
Explain why different frequencies are possible.
...............................................................................................................
...............................................................................................................
(1)
(ii)
State and explain how many possible frequencies could be produced.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
...............................................................................................................
(2)
(Total 12 marks)
Page 31 of 50
Mark schemes
1
Correct substitution ignoring powers of 10 in hc / λ
C1
Photon energy = 3.0(3) × 10−19J
Photon energy in eV = 1.9 eV gets 3 marks
A1
Conversion of −3.4 eV to J (5.44 × 10−19 seen)
C1
Answer −2.4 × 10−19 J (must have negative sign)
−8.4(8.5) × 10−19 J gets 3 marks
A1
[4]
2
3
D
1
B
[1]
4
(a)
(1)
= 2.1× 106 m s–1 (1)
2
(b)
(i)
difference between E2 and E0 = 1.94 × 10–18 J (1)
which is less than the electron kinetic energy (1)
(ii)
(E2 – E1) = 3.06 × 10–19 (1) (=
λ=
)
(1) = 6.5 × 10–7 m (1)
in visible [or red] region (1)
6
(c)
for ionisation, p.d. =
(1) =13.6 V (1)
2
[10]
Page 32 of 50
5
6
B
[1]
(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 33 of 50
(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]
7
(a)
only certain energies [or energy changes] allowed (1)
a line [or photon] corresponds to transition between levels (1)
each transition [or energy change]
corresponds to a definite wavelength [or frequency] (1)
3
(b)
(i)
Eion = 10.4× 1.6 × 10–19 (or10.4 eV) (1)
= 1.66 × 10–18 (J) (1)
(ii)
E
(1)
= 1.40 × 10–18 J (1)
= 8.8 eV (1)
which is from 1.6 to 10.4 (1)
(iii)
ultra-violet (1)
7
[10]
8
(a)
need for excitation (1)
electrons in an atom can only exist at definite/discrete energy levels / orbits (1)
an electron falls from one level to another (1)
photon emitted (1)
photon has definite wavelength (1)
The Quality of Written Communication marks were awarded primarily for the quality
of answers to this part
(5)
(b)
(i)
an electron is removed from a (neutral) atom (1)
(ii)
2.2 × 10–18 (J) (1)
Page 34 of 50
(iii)
(fmin = E/h gives) fmin = 2.2 × 10–18/6.6(3) × 10–34 (1)
(allow e.c.f from result of (b)(ii)) (1)
= 3.3(2) × 1015 Hz (1)
(iv)
ionised electron gains kinetic energy (or electron breaks free of atom) (1)
(5)
(c)
(1) (= 4.52 × 1014 Hz)
(λ = c/f gives) λ = 3.0 × 108/4.52 × 1014 (1) (allow e.c.f. for f)
= 6.6(3) × 10–7 m (1)
(3)
[13]
9
(a)
an electron is removed from the atom (1)
(b)
2.18 × 10–18(J) (1)
(c)
(single photon):electron loses energy [or falls] from level n = 3 to n = 1
and emits a single photon (1)
(two photons): electron falls from level n = 3 to n = 2, emitting a photon (1)
followed by a fall from level n = 2 to n = 1, emitting another photon (1)
The Quality of Written Communication marks are awarded for the quality of answers
tothis question.
(f)
level n = 5 to the ground state [or E5 → E1] (1)
(e)
(use of hf = E1 – E5 gives) f =
(1)
= 2.47 × 1015 Hz (1)
[8]
10
(a)
electrons move(or excited) from one energy level(or orbit) to another (1)
emitting or absorbing a definite frequency / wavelength / colour (1)
or photon energy(of electromagnetic radiation) (1)
The Quality of Written Communication marks were awarded primarily for the quality
of answers to this part
(2)
(b)
(i)
Ei = 5.2 (eV) (1) × 1.6 × 10–19
= 8.3 × 10–19(J) (1)
(allow e.c.f. if incorrect value of energy in eV)
Page 35 of 50
(ii)
(f =
gives) f =
(1)
= 4.9 × 1014 Hz (1)
(iii)
(ΔE = hf gives) E = 6.63 × 10–34 × 4.9 × 1014 (1)
= 3.2 × 10–19 (J) (1)
(allow e.c.f. from (ii))
(iv)
line drawn from B to D (1)
(v)
D to E (1)
(vi)
B to C (1)
(9)
[11]
11
(a)
(i)
pd across resistor (= 3.0 – 2.2) = 0.8 (V) (1)
(use of V = IR gives) R =
(ii)
= 23 Ω (1) (22.9 Ω)
charge flow in 1 s = 0.035 (C) (1)
no. of electrons (in 1 s)
= 2.2 × 1017 (1)
(2.19 × 1017)
4
(b)
(i)
(use of E = hf =
gives) E =
(1)
= 3.1(3) × 10–19 J (1)
(ii)
(use of P = VI gives) P (= 2.2 × 0.035) = 0.077 (W)
[or use of P = I2R with R
= 63 (Ω)]
maximum no. of photons emitted per sec. =
= 2.5 × 1017 (1) (2.48 × 1017)
(allow C.E. for value of E from (i) and value of P from (ii))
4
[8]
12
(a)
(i)
a (mercury) atom in which an orbiting electron is raised
to a higher (energy) level or orbit (1)
(ii)
by electron collision (1)
with an electron accelerated by the high voltage (of the tube) (1)
3
Page 36 of 50
(b)
the powder absorbs light / photons (emitted from the mercury) (1)
powder atoms are excited and emit light / photons (1)
of different wavelengths (to those received) (1)
any other relevant statement such as,
electrons cascade down energy levels, emitting many wavelengths,
or the spectral lines are broadened (1)
max 2
(c)
(i)
(use of hf = E1 - E2 gives) f =
(1)
= 5.0 × 1014Hz (1)
(ii)
line joining level n = 3 to level n = 2 with arrow pointing down (1)
3
[8]
13
(a)
(i)
(use of d sin θ = nλ gives)
2λ = d sin 35.8° (1)
(= 1.67 × 10–6)
= 4.9 × 10–7m (1)
(4.87 × 10–7m)
(ii)
E (= hf = 6.63 × 10–34 × 6.16 × 1014) = 4.1 × 10–19(J) (1) (4.0(8) × 10–19(J))
= 2.6 (eV) (1)
(2.55 (eV)
(for E = 4.1 × 10–19(J) = 2.56 (eV)
5
Page 37 of 50
(b)
(i)
from C to A (1)
(ii)
(use of Ek = 3 / 2kT gives)
Ek = 1.5 × 1.38 × 10–23 × 5000 = 1.0(4) × 10–19J
[or = 0.64(7) eV] (1)
(iii)
some gas atoms have enough kinetic energy to cause excitation by
collision (1)
photons (of certain energies) only released when de-excitation
or electron transfer to a lower level, occurs (1)
gas atoms have a spread of speeds / kinetic energies (1)
mean Ek (of gas atoms) proportional to T (1)
excitation can occur to level C (1)
de-excitation from C to B produces 2.6 eV photon / light
of this wavelength (1)
(max 6)
QWC 1
[11]
14
(a)
(i)
an electron moves up from one energy level to another (1)
(ii)
an electron is removed from an atom (1)
2
(b)
(use of hf = E2 – E1 gives)
f = (2.56 – 1.92) × 10–19 (1)/6.63 × 10–34
= 9.65 × 1013 Hz
(allow C.E. for incorrect ΔE)
2
[4]
15
(a)
(i)
an electron/atom in an energy level/state or an orbiting electron (1)
is given energy (1)
to move to a higher level or orbit (1)
(ii)
electromagnetic radiation is emitted when an electron falls (1)
from one fixed level to another fixed level (1)
giving the photon a discrete amount of energy (1)
max 5
QWC 2
(b)
(i)
(use of E = hf gives) f =
(use of c = f
gives)
( =1.5 × 1015 (Hz))
(1)
= 1.5
(1)
= 2.0 × 10–7 m (1)
Page 38 of 50
(ii)
energy (in eV)
= 6.2 (eV) (1)
transition from n = 2 to n = 1 (1)
(iii)
line between n = 4 and n = 1 (1)
direction from 4 to 1 (1)
7
[12]
16
(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
(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]
17
(a)
(i)
–0.66 to –0.72keV line marked as B downward arrow
B1
Page 39 of 50
(ii)
uses 7.06 (eV) (condone negative sign)
B1
attempts to multiply by 1.6 × 10–16 (condone incorrect
power of 10) and to divide by 6.63 × 10–34
B1
1.7(0) × 1018 (Hz) cao
B1
4
(b)
(i)
λ = h/mv or λ = h/p or correct substitution
C1
4.4(2) × 106 (m s–1) [4.8(5) with h to 2 sf]
A1
(ii)
same order of magnitude as atomic spacing
B1
produces wide diffraction angle/good diffraction
B1
4
[8]
18
(a)
an electron is excited/promoted to a higher level/orbit (1)
reason for excitation: e.g. electron impact/light/energy externally
applied (1)
electron relaxes/de-excited/falls back emitting a photon/
em radiation (1)
wavelength depends on the energy change (1)
Max 3
QWC 1
Page 40 of 50
(b)
(i)
use of E = hf gives) E =
=
(1)
= 5.0 × 10–19 (J) (1)
(4.95 × 10–19 (J))
and
(ii)
= 9.9 × 10–19 (J) (1)
(energy of) level B = – 1.5 × 10–18 (J) (1)
level C = (–) 1.0 × 10–18 (J) (1)
5
[8]
19
(a)
ionisation energy = 13.6eV (1)
1
(b)
(i)
(ii)
energy in Joules = 1.90 (1) × 1.6 × 10–19 = 3.04 × 10–19 (J) (1)
(use of E = hc/λ)
3.04 × 10–19 = 6.63 × 10–34 × 3 × 108/λ (1)
(working/equation must be shown)
λ = 6.54 × 10–7 m (1)(1) (2 or 3 sf for second mark)
(accept 0.65 which gives an answer of λ = 1.91 × 10–6 m)
8
[9]
Page 41 of 50
20
(a)
(i)
an electron/atom is at a higher level than the ground state (1)
or electron jumped/moved up to another/higher level
1
(ii)
electrons (or electric current) flow through the tube (1)
and collide with orbiting/atomic electrons or mercury atoms (1)
raising the electrons to a higher level (in the mercury atoms) (1)
3
(iii)
photons emitted from mercury atoms are in the ultra
violet (spectrum) or high energy photons (1)
these photons are absorbed by the powder or powder changes
frequency/wavelength (1)
and the powder emits photons in the visible spectrum (1)
incident photons have a variety of different wavelengths (1)
max 3
(b)
(i)
(use of E = hf)
–0.26 × 10–18 – 0.59 × 10–18 (1) = 6.63 × 10–34 × f (1)
f = 0.33 × 10–18/(6.63 × 10–34) = 5.0 × 1014 (Hz) (1)
3
(ii)
one arrow between n = 3 and n = 2 (1) in correct direction (1)
2
[12]
21
(a)
Peak power = 107 / 108 mW and load resistance = 290 / 310 Ω ✓
1
Use of power = I2R with candidate values✓
1
0.0186 – 0.0193 A ✓
1
(b)
Area of cell = 36 x 10-4 m2 and solar power arriving = 730 × (an area)✓
1
seen✓
1
0.041 (correct answer only; lose if ratio given unit) ✓
1
Page 42 of 50
(c)
energy of one photon =
= 4.0 ×10-19J✓
1
Number of photons =
= 6.6 × 1018 s-1✓
1
(d)
Two from
Intensity of the sun at the Earth’s surface
Average position of the sun
Efficiency of the panel
Power output of 1 panel
Weather conditions at the installation=
✓✓
Allow other valid physics answers=
2
[10]
22
(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 43 of 50
23
(a)
(i)
absorbs enough energy (from the incident) electron( by collision) OR incident electron
loses energy (to orbital electron)
exact energy / 10.1((eV) needed to make the transition / move up to level 2
For second mark must imply exact energy
2
(ii)
(use of E2 –E1) = hf
−3.41 − − 13.6 = 10.19
energy of photon = 10.19 × 1.6 × 10−19 = 1.63 × 10−18 (J)
6.63 × 10−34 × f = 1.63 × 10−18
f = 2.46 × 1015(Hz)
(accept 2.5 but not 2.4)
CE from energy difference but not from energy conversion
3
(iii)
Ek = 1.7 × 10−18 − 1.63 × 10−18
= 7.0 × 10−20 J
2
(iv)
energy required is 12.09 eV / 1.9 × 10−18
energy of incident electron is only 10.63 eV / energy of electron less than this (1.7 ×
10−18 J)
State and explain must have consistent units i.e. eV or J
2
(b)
(i)
Electrons return to lower levels by different routes / cascade / not straight to ground
state
1
(ii)
3
n=3 to n=1 or n=3 to n=2 and n=2 to n=1
no CE from first mark
2
[12]
Page 44 of 50
Examiner reports
1
4
Many were successful in obtaining the energy of the emitted photon and / or were able to convert
3.4 eV to J. Calculating the energy of the initial energy level proved more of a problem and there
were many answers with the correct numerical value but no negative sign or answers where
candidates calculated the energy level the electron would have to move into rather than from.
All parts of this question were generally answered well. In part (a) some candidates used an
electron mass of 0.00055 u, and some candidates forgot to take the square root.
In part (b) a few candidates calculated the frequency and then did not proceed to evaluate λ. A
more common error was to forget to state the region of the electromagnetic spectrum after
calculating the wavelength value correctly.
Part (c) was done well.
6
(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.
(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
(c)
used λ =
7
In part (a) marks were often lost because of carelessly worded answers. It is difficult to believe
that some candidates have seen line spectra as they seem convinced that spectra is singular.
The most common score was probably one mark. Only a small proportion of candidates referred
to the electron having certain allowed energy levels. Most candidates knew that a photon, or less
often a line, was emitted when an electron dropped from one energy level to another, but few
unambiguously related the difference in energy of the levels to the frequency or wavelength of
the emitted line. There was some confusion with absorption spectra.
Part (b) proved to be much easier and many candidates scored 6 or 7 marks. A few candidates
thought that the ionisation energy was 1.6eV, or forgot to change to J. Most candidates obtained
8.8 eV in part (b)(ii), but some candidates dropped a mark by not recognising the wavelength as
being in the ultraviolet. The commonest wrong answer was ϒ rays.
Page 45 of 50
8
9
This question proved to be a good discriminator. In part (a) the absorption spectrum was often
confused with the emission spectrum. Part (b) was answered reasonably well but in part (iv) the
weaker candidates assumed that the electron would either simply move up an energy level or not
move from its current orbit. Part (c) involved a two-stage calculation but at least a quarter of the
candidates stopped after calculating the frequency.
Most candidates were aware of the answer to part (a) but failed to gain the mark by not stating
explicitly that the electron was removed from the atom. Answers such as, ‘the atom goes to level
n = ∞ ‘ were common. In part (b) the energy required to ionise the atom was often quoted as a
negative quantity.
Part (c) tested candidates’ communication skills. Most candidates used much of the available
space to describe the process of excitation but many were not clear as to how the two photons
were produced. It was common to read statements such as ‘If the energy drop is very large two
photons, not one, are produced’ or ‘Sometimes the single photon is replaced by two, each
sharing the energy equally’
The transition between energy states in part (d) was often given the wrong way round with an
electron in the ground state moving to level n = 5. In part (e) the frequency of the emitted photon
was calculated successfully by most candidates with the remaining candidates falling into two
groups; those who failed to use the factor of 10–18 in the energy value and those who failed to
use the equation ΔE = hf.
10
Part (a) tested the ability of candidates to explain a sequence of events. This was quite difficult in
itself and the answers indicated clearly which were the good candidates. The idea of electrons
moving between allowed orbits was understood by the majority of candidates but very few linked
this to the production or absorption of specific wavelengths.
Most candidates experienced difficulty with one or more sections of part (b) and a good range of
marks was obtained. In part (i) the conversion from electron volts to joules was usually incorrect
or else the ionisation energy was not recognised as being 5.2 eV. In part (ii) the expression f = c /
λ was usually given but often not evaluated. It was also common to see the approximate answer
of 5.0 × 1014 Hz being given as the final answer to the calculation. Again, in part (iii) the
approximate value of the frequency was used by many candidates even though they had
calculated an accurate value in the previous section. Of the candidates who were penalised for
incorrect use of significant figures, about 90% of them were penalised in this section. Part (iv)
required a line to be drawn on the diagram representing the emission of a photon. More often
than not the line did not show the correct direction. The final two parts performed better but a
common answer to part (vi) was a transition from A to E rather than the correct transition from B
to C.
Page 46 of 50
11
Part (a) proved to be very accessible and many candidates scored full marks. Most candidates
calculated the resistor pd as 0.8 V and then calculated the resistance, as expected. Other
candidates however, calculated the total circuit resistance, then the diode resistance and
obtained the required resistance by subtraction. In this particular problem some candidates used
an incorrect pd and were not awarded any credit. Many clear and correct answers were seen in
part (ii).
The energy of the photon was calculated correctly in part (b) by many candidates, but some
failed to score because the wavelength was taken as 1 / f or because the energy was taken to be
½QV. The general principle behind the question in part (ii) was understood by most candidates
and many correct answers were seen. A small minority of candidates however, calculated and
used the power supplied to the resistor and not the diode.
12
Candidates, generally, did not perform well in parts (a) and (b), both of which required
descriptions and explanations. In part (a) the description of an excited mercury atom very often
contained far too much detail e.g. the description often included a discussion of the subsequent
relaxation of the atom. Alternatively, too little information was often the problem, e.g. statements
such as ‘the atom is given extra energy’. Other candidates referred to a change in orbit without
giving the direction of the change. However, on balance, candidates did seem to know what an
excited atom was. Part (a)(ii) discriminated well. Many candidates simply repeated the answer to
part (a)(i) using different words rather than describe how excitation takes place in a particular
environment.
Answers to part (b) included all sorts of reasons, ranging from ‘it keeps the heat in’, ‘it keeps
mercury safe’ and ‘it reflects electrons’, to the correct answer. On the positive side almost all
candidates attempted parts (a) and (b) but it was obvious to the examiners that this section of the
Specification had, in general, not been taught to any depth.
The majority of candidates performed well in part (c) and calculated the correct frequency. The
usual error was omitting the powers of 10 when transferring the values of the energies. Also,
many candidates failed to draw the arrow in the correct position in part (c)(ii).
13
Many candidates scored full marks in part (a), although some lost a mark as a result of not
converting d from mm into m and also giving 4.87 × 10−4 m as the wavelength. In part (ii), a
significant number of candidates failed to convert correctly from joules into eV, often multiplying
rather than dividing by 1.6 × 10–19.
In part (b) (i), most candidates knew that the required transition occurred between energy levels
A and C, but many indicated that the transition was either upwards or in both directions. Again,
most candidates were able to show that the mean kinetic energy in part (ii) was 0.65 eV. In part
(iii), only a small number of candidates realised that excitation of the gas atoms is due to
collisions between the atoms, and that the spread of speeds of the gas atoms at 5000 K resulted
in collisions which caused electrons in the atoms to move to the higher energy levels, including
C. Some did realise that a 2.6 eV photon was emitted when an electron transition occurred from
C to B. Sadly, the only credit gained in part (iii) by most candidates was for stating that a photon
was emitted when de-excitation of a gas atom occurred. Some candidates thought that excitation
in a heated gas is due to atoms absorbing photons rather than collisions between gas atoms.
Page 47 of 50
14
The topics in part (a), namely excitation and ionisation, have been examined before and, as on
previous occasions, candidates tried to explain the full excitation and relaxation processes
instead of simply stating that an electron is promoted up to a different energy level in the
excitation process. Ionisation was understood much better, but too many students thought the
incoming electron was captured by the atom. Although it is appreciated that temporary negative
ions or resonances occur for extremely short time periods no marks were awarded for electron
capture because it would be outside the experience of candidates.
The most common error in part (b) was not identifying the correct change in energy. Also, the
10–19 factor in the given energies was omitted in many of the calculations.
15
The explanation of the excitation process in part (a)(i) was done well. There were very few
references to ionisation, which has occurred in previous papers, and also practically no
candidates referred to an electron leaving a metal. There was a tendency to continue the
explanation into the relaxation process, which was unnecessary. In part (ii) several candidates
worked backwards to the expected answer. These candidates typically wrote that the wavelength
was fixed because the frequency was fixed because the energy was fixed. It was easy for these
candidates to miss the relevant marking points. Very few candidates gave clear statements about
the energy levels occurring at discrete energies or that an electron drops down an energy level
when a photon was emitted.
Part (b) again proved to be a good discriminator. In part (i) several candidates attempted to use
the de Broglie relationship instead of the usual λ = ch/E. In parts (ii) and (iii) only about half the
candidates gave the correct transition and the correct direction. Most of them successfully
converted the energy of the transition from joules into eV.
16
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.
Page 48 of 50
17
In part (a) (i) a good number of candidates correctly identified the energy change B. A significant
minority penalised themselves by marking the arrow upwards. Others were penalised for making
the change on top of the change A.
Most candidates attempting part (a) (ii) identified the energy changes as being 7.06 keV correctly
and many then went on to convert this energy into joules (some forgetting the factor of 1000). Of
those that did not convert the energy into joules, many divided their answer by the Planck
constant. A significant number of candidates interpreted this question as being an example of the
photoelectric effect.
Although many candidates gained the correct answer to part (b) (i) using the de Broglie equation,
a large proportion managed to juggle c = f λ to obtain a speed of 3.00 × 108 m s–1. It was also
common for the less able candidates to simply write down both these equations to gain no credit.
Part (b) (ii) was not done well and many candidates either completely missed it out or else said
no more than the wavelength was small. Few candidates recognised that for effective diffraction
the atomic spacing in the crystal needs to be of the same order of magnitude as the de Broglie
wavelength of the incident particles.
18
The description of how an excitation spectrum was produced was generally done well in part (a),
but many candidates omitted the fact that energy had to be provided by some means to start the
process.
In part (b) (i), candidates who were aware of the equation E = hc/λ completed the calculations
correctly, but there were a few significant figure errors and several candidates had trouble with
powers of 10. The determination of the energy levels B and C in part (b) (ii) caused considerable
difficulty, with only a minority of candidates obtaining the correct answers. The connection
between the photon energies and the energy levels was not obvious to most candidates.
19
Correct responses for part (a) were common, although a significant minority of candidates did
give the answer 12.75 eV which is the energy change from level one to level four.
The other parts of the question were answered well by the majority of candidates, with the only
common errors occurring in the transition diagram, where often too many arrows were drawn or
the arrows were shown in the wrong direction or there was no arrow at all. The calculation in part
(b) (ii) was approached with confidence by many candidates and most appreciated that they
were required to limit the number of significant figures in their answers.
Page 49 of 50
20
Part (a) proved to be quite discriminating and less able candidates found it hard to explain the
process by which mercury atoms become excited in a fluorescent tube. There was also evidence
to suggest that some candidates think that excitation only occurs due to the absorption of
photons and seemed unaware that it can also happen by electron collision. Most candidates
seemed to appreciate that the mercury atoms emitted photons that were in the ultraviolet part of
the spectrum and that the coating changed the frequency of these although there was a
tendency to describe these photons as photons of light or coloured light rather than visible light.
Part (b) was answered well and the only common error was a failure to appreciate that the
energy levels were in Joules and that the value adjacent to each level needed to be multiplied by
10–18. A minority of candidates either emitted this factor or assumed that the energies were in
electron volts and multiplied them by 1.6 × 10–19.
22
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.
23
This question required candidates to be familiar with discrete energy levels and excitation by
electron collision. This is a topic which has caused problems in the past and it is clear that the
ideas involved continue to trouble candidates.
In part (a) they were required to explain the process of excitation and less than 20% of
candidates were awarded full marks for their answers. Many were able to explain the energy
transfer that took place between the electrons but very few were able to explain convincingly that
an exact amount of energy had to be transferred. It was also quite common to see answers
referring to excitation due to photon absorption rather than electron collision. In part (a) (ii)
candidates were required to calculate the frequency of the photon emitted when an electron
drops to the ground state. This was generally done well although nearly a third of candidates
failed to convert the energy in electron volts to joules and were therefore limited to one mark. The
remaining parts of (a) were concerned with the energy of the incident electron. This question
proved to be quite discriminating and only the stronger candidates managed to score full marks.
Part (b) also turned out to be very discriminating and only about half of candidates were able to
explain why hydrogen atoms, whose electrons had been excited to level 3, were able to emit
photons of three different frequencies.
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