1
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)
2
The diagram below is an arrangement for analysing the light emitted by a source.
(a)
Suggest a light source that would emit a continuous spectrum.
........................................................................................................................
(1)
(b)
The light source emits a range of wavelengths from 500 nm to 700 nm. The light is incident
on a diffraction grating that has 10 000 lines per metre.
(i)
Calculate the angle from the straight through direction at which the first order
maximum for the 500 nm wavelength is formed.
Angle = ........................................
(3)
(ii)
Calculate the angular width of the first order spectrum.
Angular width ........................................
(1)
Page 1 of 63
(iii)
The detector is positioned 2.0 m from the grating. Calculate the distance between the
extreme ends of the first order spectrum in this position.
Distance = ........................................
(1)
(c)
The single slit is initially illuminated by light from a point source that is 0.02 m from the slit.
State and explain how the intensity of light incident on the single slit changes when the light
source is moved to a position 0.05 m from the slit.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 10 marks)
3
A white-light source illuminates a diffraction grating that has 6.30 × 105 lines per metre. The light
is incident normally on the grating.
(a)
Show that adjacent lines in the grating are separated by a distance of about 0.0016 mm.
(1)
Page 2 of 63
(b)
The table below shows the diffracting angles measured from the normal for the visible
spectral orders using this grating. The angles are given for the red and blue ends of each
spectrum.
(i)
First order
Second order
Third order
red
25.4°
59.0°
not possible
blue
15.0°
31.1°
50.0°
Use the value for the first order diffracting angle to calculate the wavelength of the red
light.
Wavelength of the red light ...............................
(3)
Page 3 of 63
(ii)
Describe carefully the appearance of the complete diffraction pattern on the screen.
You may draw a sketch of the pattern to help your explanation if you choose.
...............................................................................................................
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...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(4)
(Total 8 marks)
Page 4 of 63
4
(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 5 of 63
(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
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(4)
(Total 11 marks)
5
Light from a laser has a wavelength of 6.30 × 10–7 m. When the laser light is incident normally on
a diffraction grating the first order maximum is produced at an angle of 12°.
(a)
Calculate the spacing between the lines on the grating.
spacing of lines ..........................................
(2)
Page 6 of 63
(b)
Calculate the number of positions of maximum light intensity that are produced
when the laser light is incident on the grating.
Show your reasoning clearly.
number of positions ..........................................
(3)
(Total 5 marks)
6
A diffraction grating has 940 lines per mm.
(a)
Calculate the distance between adjacent lines on the grating.
distance between lines .............................................
(1)
(b)
Monochromatic light is incident on the grating and a second-order spectral line is formed at
an angle of 55° from the normal to the grating. Calculate the wavelength of the light.
wavelength .....................................
(3)
(Total 4 marks)
Page 7 of 63
7
A scanning photometer is a device in which the voltage across an LDR (light dependent resistor)
varies with the light intensity incident on the LDR.
Figure 1 shows a laser beam of wavelength 633 nm incidental normally on a diffraction grating.
The LDR of a scanning photometer is moved across the diffracted beam and produces the scan
shown in Figure 2. This shows the central bright fringe with one further maximum (the first order
image) on each side of it. The distance from the diffraction grating to the LDR is 265 mm.
Figure 1
Figure 2
Page 8 of 63
(i)
Show that the angle of the first order image measured from the straight through position is
approximately 3°.
(ii)
Calculate the number of lines per mm on the diffraction grating.
(Total 6 marks)
8
(a)
A student stands some distance away from a wall and claps her hands. She listens to the
echoes produced by the sound reflected from the wall and adjusts her rate of clapping so
that each clap coincides with the echo of the previous clap. A fellow student counts the
number of times she claps in 20 s.
(i)
There were 47 claps in 20 s.
Calculate the time between two successive claps.
time between two successive claps .......................................
(ii)
The speed of sound in the air is 340 m s–1. Calculate the distance of the student from
the wall.
distance from wall .................................................................
(3)
Page 9 of 63
(b)
The student goes on to carry out a diffraction experiment. She stands on one side of a
fence that consists of evenly-spaced wooden strips. The fence behaves as a diffraction
grating for sound waves. The figure below shows the arrangement.
(i)
She uses a loudspeaker to send a sound wave of frequency 2.4 kHz towards the
fence.
Calculate the wavelength of this sound wave.
speed of sound in air, v = 340 m s–1
wavelength ................................................
(ii)
A diffracted first-order maximum is observed at an angle of 28°. Calculate the spacing
between adjacent wooden strips.
strip spacing ..............................................
Page 10 of 63
(iii)
Determine the highest order of maximum that could be heard.
highest order .............................................
(6)
(Total 9 marks)
9
Red light of wavelength 7.00 × 10–7 m, incident normally on a diffraction grating, gave a first order
maximum at an angle of 75°.
(a)
Calculate the spacing of the diffraction grating.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(b)
Calculate the angle at which the first order maximum for violet light of wavelength
4.50 × 10–7 m would be observed.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(c)
At what angle or angles would a detector receive radiation which is of wavelength
7.50 × 10–7 m transmitted by the grating? Explain your answer.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 4 marks)
Page 11 of 63
10
(a)
Figure 1
In a laboratory experiment, monochromatic light of wavelength 633 nm from a laser is
incident normal to a diffraction grating. The diffracted waves are received on a white screen
which is parallel to the plane of the grating and 2.0 m from it. Figure 1 shows the positions
of the diffraction maxima with distances measured from the central maximum.
By means of a graphical method, use all these measurements to determine a mean value
for the number of rulings per unit length of the grating.
........................................................................................................................
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........................................................................................................................
........................................................................................................................
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Page 12 of 63
(6)
Page 13 of 63
(b)
Describe and explain the effect, if any, on the appearance of the diffraction pattern of
(i)
using a grating which has more rulings per unit length.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
using a laser source which has a shorter wavelength.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(iii)
increasing the distance between the grating and the screen.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
Page 14 of 63
(c)
Figure 2, below, shows the diffracted waves from four narrow slits of a diffraction grating
similar to the one described in part (a). The slit separation AB = BC = CD = DE = d and EQ
is a line drawn at a tangent to several wavefronts and which makes an angle θ with the
grating.
Figure 2
(i)
Explain why the waves advancing perpendicular to EQ will reinforce if superposed.
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 15 of 63
(ii)
Show that this will happen when sin θ =
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 15 marks)
11
(a)
A helium-neon laser produces monochromatic light of wavelength 632.8 nm which falls
normally on a diffraction grating. A first order maximum is produced at an angle of 18.5°
measured from the normal to the grating.
Calculate
(i)
the number of lines per metre on the grating,
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
the highest order which is observable.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(b)
When the grating is used with a different monochromatic source, the first order maximum is
observed at an angle of 17.2°.
Calculate the wavelength of this second source.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 8 marks)
Page 16 of 63
12
(a)
A double slit interference experiment is set up in a laboratory using a source of yellow
monochromatic light of wavelength 5.86 × 10–7 m. The separation of the two vertical
parallel slits is 0.36 mm and the distance from the slits to the plane where the fringes are
observed is 1.80 m.
(i)
Describe the appearance of the fringes.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the fringe separation, and also the angle between the middle of the central
fringe and the middle of the second bright fringe.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(iii)
Explain why more fringes will be seen if each of the slits is made narrower, assuming
that no other changes are made.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(8)
(b)
Light of wavelength 5.86 × 10–7 m falls at right angles on a diffraction grating which has 400
lines per mm.
(i)
Calculate the angle between the straight through image and the first order image.
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 17 of 63
(ii)
Determine the highest order image which can be seen with this arrangement.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
(c)
Give two reasons why the diffraction grating arrangement is more suitable for the accurate
measurement of the wavelength of light than the two-slit interference arrangement.
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 15 marks)
13
Figure 1
Red light from a laser is passed through a single narrow slit, as shown in Figure 1. A pattern of
bright and dark regions can be observed on the screen which is placed several metres beyond
the slit.
(a)
The pattern on the screen may be represented as a graph of intensity against distance
along the screen. The graph has been started in outline in Figure 2. The central bright
region is already shown. Complete this graph to represent the rest of the pattern by
drawing on Figure 2.
Page 18 of 63
Figure 2
(4)
(b)
State the effect on the pattern if each of the following changes is made separately.
(i)
The width of the narrow slit is reduced.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
With the original slit width, the intense red source is replaced with an intense source
of green light.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(3)
(Total 7 marks)
14
Using a diffraction grating with monochromatic light of wavelength 500 nm incident normally, a
student found the 2nd order diffracted maxima in a direction at 30° to the central bright fringe.
What is the number of lines per metre on the grating?
A
2 × 104
B
2 × 105
C
4 × 105
D
5 × 105
(Total 1 mark)
15
Monochromatic light of wavelength 590 nm is incident normally on a plane diffraction grating
having 4 × 10 5 lines m−1 . An interference pattern is produced. What is the highest order visible
in this interference pattern?
Page 19 of 63
A
2
B
3
C
4
D
5
(Total 1 mark)
16
A narrow beam of monochromatic light falls on a diffraction grating at normal incidence. The
second order diffracted beam makes an angle of 45° with the grating. What is the highest order
visible with this grating at this wavelength?
A
2
B
3
C
4
D
5
(Total 1 mark)
17
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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(ii)
Calculate the energy, in eV, of a photon of this wavelength.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(5)
Page 20 of 63
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
Page 21 of 63
(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.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
(6)
(Total 11 marks)
18
The diagram below shows a section of a diffraction grating. Monochromatic light of wavelength λ
is incident normally on its surface. Light waves diffracted through angle θ form the second order
image after passing through a converging lens (not shown). A, B and C are adjacent slits on
the grating.
(a)
(i)
State the phase difference between the waves at A and D.
.............................................................................................................
(ii)
State the path length between C and E in terms of λ.
.............................................................................................................
Page 22 of 63
(iii)
Use your results to show that, for the second order image,
2λ = d sin θ, where d is the distance between adjacent slits.
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
.............................................................................................................
(3)
(b)
A diffraction grating has 4.5 × 105 lines m–1. It is being used to investigate the line spectrum
of hydrogen, which contains a visible blue-green line of wavelength 486 nm. Determine the
highest order diffracted image that could be produced for this spectral line by this grating.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
(Total 5 marks)
19
Light of wavelength λ is incident normally on a diffraction grating of slit separation 4λ. What is the
angle between the second order maximum and third order maximum?
A
14.5°
B
18.6°
C
48.6°
D
71.4°
(Total 1 mark)
Page 23 of 63
20
Light of wavelength λ is incident normally on a diffraction grating for which adjacent lines are a
distance 3λ apart. What is the angle between the second order maximum and the straightthrough position?
A
9.6°
B
20°
C
42°
D
There is no second order maximum.
(Total 1 mark)
Page 24 of 63
21
The diagram above shows the first four diffraction orders each side of the zero order when a
beam of monochromatic light is incident normally on a diffraction grating of slit separation d. All
the angles of diffraction are small. Which one of the patterns, A to D, drawn on the same scale,
is obtained when the grating is exchanged for one with a slit separation
?
A
B
C
D
(Total 1 mark)
Page 25 of 63
22
A diffraction grating has 300 lines per mm. It is illuminated with monochromatic light of
wavelength 540 nm.
Calculate the angle of the 2nd order maximum, giving your answer to the appropriate number of
significant figures.
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
...............................................................................................................................
angle .......................................... degrees
(Total 4 marks)
23
A narrow beam of monochromatic light of wavelength 590 nm is directed normally at a diffraction
grating, as shown in the diagram below.
(a)
The grating spacing of the diffraction grating is 1.67 × 10–6 m.
(i)
Calculate the angle of diffraction of the second order diffracted beam.
answer .................................... degrees
(4)
(ii)
Show that no beams higher than the second order can be observed at this
wavelength.
(3)
Page 26 of 63
(b)
The light source is replaced by a monochromatic light source of unknown wavelength.
A narrow beam of light from this light source is directed normally at the grating.
Measurement of the angle of diffraction of the second order beam gives a value of 42.1°.
Calculate the wavelength of this light source.
answer ....................................... m
(2)
(Total 9 marks)
24
In a double slit system used to produce interference fringes, the separation of the slits is s and
the width of each slit is x. L is a source of monochromatic light. Which one of the following
changes would decrease the separation of the fringes seen on the screen?
A
moving the screen closer to the double slits
B
decreasing the width, x, of each slit, but keeping s constant
C
decreasing the separation, s, of the slits
D
exchanging L for a monochromatic source of longer wavelength
(Total 1 mark)
Page 27 of 63
25
For a plane transmission diffraction grating, the diffraction grating equation for the first order
beam is:
λ = d sin θ
(a)
The figure below shows two of the slits in the grating. Label the figure below with the
distances d and λ.
(2)
(b)
State and explain what happens to the value of angle θ for the first order beam if the
wavelength of the monochromatic light decreases.
......................................................................................................................
......................................................................................................................
......................................................................................................................
......................................................................................................................
(2)
Page 28 of 63
(c)
A diffraction grating was used with a spectrometer to obtain the line spectrum of star X
shown in the figure below. Shown are some line spectra for six elements that have been
obtained in the laboratory.
Place ticks in the boxes next to the three elements that are present in the atmosphere of
star X.
(2)
(d)
The diffraction grating used to obtain the spectrum of star X had 300 slits per mm.
(i)
Calculate the distance between the centres of two adjacent slits on this grating.
answer = ................................. m
(1)
Page 29 of 63
(ii)
Calculate the first order angle of diffraction of line P in the figure above.
answer = ........................ degrees
(2)
(Total 9 marks)
26
A single slit diffraction pattern is produced on a screen using a laser. The intensity of the central
maximum is plotted on the axes in the figure below.
(a)
On the figure above, sketch how the intensity varies across the screen to the right of the
central maximum.
(2)
(b)
A laser is a source of monochromatic, coherent light. State what is meant by
monochromatic light ....................................................................................
......................................................................................................................
coherent light ...............................................................................................
......................................................................................................................
(2)
Page 30 of 63
(c)
Describe how the pattern would change if light of a longer wavelength was used.
......................................................................................................................
......................................................................................................................
(1)
(d)
State two ways in which the appearance of the fringes would change if the slit was made
narrower.
......................................................................................................................
......................................................................................................................
(2)
(e)
The laser is replaced with a lamp that produces a narrow beam of white light. Sketch and
label the appearance of the fringes as you would see them on a screen.
(3)
(Total 10 marks)
27
(a)
A laser emits monochromatic light.
Explain the meaning of the term monochromatic light.
........................................................................................................................
........................................................................................................................
(1)
Page 31 of 63
(b)
The diagram below shows a laser emitting blue light directed at a single slit, where the slit
width is greater than the wavelength of the light. The intensity graph for the diffracted blue
light is shown.
The laser is replaced by a laser emitting red light.
On the axes shown in the diagram above sketch the intensity graph for a laser emitting red
light.
(2)
(c)
State and explain one precaution that should be taken when using laser light
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
Page 32 of 63
(d)
The red laser light is replaced by a non-laser source emitting white light.
Describe how the appearance of the pattern would change.
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(3)
(Total 8 marks)
28
The figure below shows a spectrometer that uses a diffraction grating to split a beam of light into
its constituent wavelengths and enables the angles of the diffracted beams to be measured.
(a)
Give one possible application of the spectrometer and diffraction grating used in this way.
........................................................................................................................
........................................................................................................................
(1)
(b)
(i)
When the spectrometer telescope is rotated from an initial angle of zero degrees, a
spectrum is not observed until the angle of diffraction θ is about 50°. State the order
of this spectrum.
...............................................................................................................
(1)
Page 33 of 63
(ii)
White light is directed into the spectrometer. Light emerges at A and B. State one
difference between the light emerging at B compared to that emerging at A.
...............................................................................................................
...............................................................................................................
...............................................................................................................
(1)
(c)
The angle of diffraction θ at the centre of the observed beam B in the image above is 51.0°
and the grating has 1480 lines per mm.
Calculate the wavelength of the light observed at the centre of beam B.
wavelength .......................................... m
(3)
(d)
Determine by calculation whether any more orders could be observed at the wavelength
calculated in part (c).
(2)
(Total 8 marks)
29
A diffraction pattern is formed by passing monochromatic light through a single slit. If the width of
the single slit is reduced, which of the following is true?
Width of central
maximum
Intensity of central
maximum
A
unchanged
decreases
B
increases
increases
C
increases
decreases
D
decreases
decreases
(Total 1 mark)
Page 34 of 63
30
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)
31
Read through the following passage and answer the questions that follow it.
Measuring the speed of sound in air
5
10
(a)
After the wave nature of sound had been identified, many attempts were made to
measure its speed in air. The earliest known attempt was made by the French
scientist Gassendi in the 17th century. The procedure involved timing the interval
between seeing the flash of a gun and hearing the bang from some distance away.
Gassendi assumed that, compared with the speed of sound, the speed of light is
infinite. The value he obtained for the speed of sound was 480 m s–1. He also
realised that the speed of sound does not depend on frequency.
A much better value of 350 m s–1 was obtained by the Italian physicists Borelli and
Viviani using the same procedure. In 1740 another Italian, Bianconi, showed that
sound travels faster when the temperature of the air is greater.
In 1738 a value of 332 m s–1 was obtained by scientists in Paris. This is remarkably
close to the currently accepted value considering the measuring equipment
available to the scientists at that time. Since 1986 the accepted value has been
331.29 m s–1 at 0 °C.
Suggest an experiment that will demonstrate the wave nature of sound (line 1).
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(b)
Using Gassendi’s value for the speed of sound (line 6), calculate the time between seeing
the flash of a gun and hearing its bang over a distance of 2.5 km.
time = ........................ s
(1)
Page 35 of 63
(c)
Explain why it was necessary to assume that ‘compared with the speed of sound, the
speed of light is infinite’ (line 5).
........................................................................................................................
........................................................................................................................
........................................................................................................................
(1)
(d)
Explain one observation that could have led Gassendi to conclude that ‘the speed of sound
does not depend on frequency’ (line 7).
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(e)
Explain how the value obtained by Borelli and Viviani was ‘much better’ than that obtained
by Gassendi (line 8).
........................................................................................................................
........................................................................................................................
(1)
(f)
The speed of sound c in dry air is given by
where θ is the temperature in °C, and k is a constant.
Calculate a value for k using data from the passage.
k = ........................ m s–1 K–½
(2)
Page 36 of 63
(g)
State the steps taken by the scientific community for the value of a quantity to be ‘accepted’
(line 13).
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
(2)
(Total 10 marks)
Page 37 of 63
Mark schemes
1
2
A
[1]
(a)
filament lamp / sun etc.
B1
(1)
(b)
(i)
d = 1.0 × 10–4 m
C1
use of λ = dsin θ or substituted values
C1
θ1 = 0.286° / 0.29°
A1
(3)
(ii)
Δθ = 0.115° (c.a.o.)
B1
(1)
(iii)
width = 4.0 × 10–3 m or 3.9 × 10–3 m (e.c.f. for 2 × sin (b(ii))
or 2 × tan (b(ii)); allow 1 s.f.)
B1
(1)
(c)
lower intensity
C1
because energy spreads
C1
use or statement of inverse square law
C1
ratio 0.16 or falls by factor of 6.25 c.a.o.
A1
(4)
[10]
Page 38 of 63
3
(a)
Separation = 1/630000
B1
1
(b)
(i)
quote nλ = d sinθ
C1
λ = 1.59 × 10–6 × sin(25.4)
C1
= 6.8 × 10–7 m or 6.8 × 10–4 mm
A1
3
(ii)
Central maximum/zeroth order mentioned
B1
Central maximum is white
B1
Describe/draw 1st/2nd orders colours in correct order
B1
Third order overlap
symmetry of pattern
dispersion change
fainter away from centre
B1
max 4
[8]
Page 39 of 63
4
(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 40 of 63
(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]
5
(a)
d sin θ = n λ or d sin 12 = 6.3 x 10–7
C1
3.0 x 10–6 m
A1
2
(b)
n sin 90 = (≤) 3.0 × 10–6 / 6.3 × 10–7 or n = 4.8
Allow for approach using different n values
even if unsuccessful
C1
number of orders visible = 4
A1
Total maxima = twice their maximum order + 1
B1
3
[5]
Page 41 of 63
6
(a)
1.06 × 10-6 m
B1
1
(b)
use of nλ = dsinθ with n = 2 and d = ans to (a)
C1
λ = 1.06 × 10-6 × sin 55º/2
C1
4.34 × 10-7 m
A1
3
[4]
7
(i)
use of (26 ± 2) mm for 2θ or (13 ± 1) mm for θ
M1
range 2.60 → 3.02
A1
(ii)
nλ= d sin θ seen
M1
substitution with correct powers
1.20 × 10−5 → 1.40 × 10−5
A1
1/d or number of lines m−1 calculated
71400 → 833300 needs unit
B1
conversion to number of lines mm−1
71(.4) → 83(.3) needs unit
B1
[6]
Page 42 of 63
8
(a)
(i)
20/47 = 0.43 s [0.426]
B1
(ii)
distance = 340 × 0.43 condone doubling for this mark
[v = fλ solutions score zero]
e.c.f. possible from 11 (a) (i)
M1
72 m [72.3] [73.1]
A1
3
(b)
(i)
λ = 340/2400 condone power of ten error in this mark
C1
= 0.14(2) m
A1
(ii)
uses d = nλ/sin θ
do not allow working from sinθ = λ/b; this scores 0
overall
B1
= 0.14[2]/ sin 28° =0.30(2) m e.c.f. possible
B1
(iii)
calculates n = 2.14 when θ = 90° if states
n ≈ 2 then allow this mark
M1
highest n is 2
A1
6
[9]
9
(a)
(1)
(1)
Page 43 of 63
(1)
(b)
(1)
> 1 (or sin θ > 1) (1)
(c)
0° (or straight through position) because no first order line (1)
(2)
[4]
10
(a)
x/m
sin θ
1
0.173
0.086
2
0.316
0.156
3
0.499
0.242
4
0.687
0.325
5
0.860
0.395
1
If angles only calculated 1 / 2
at least 4 points plotted correctly (1)
best straight line (1)
gradient calculated from suitable triangle, 50% of each axis (1)
correct value from readings (1)
appropriate use of d sin θ = nλ (1)
hence N (rulings per metre) = 1.25 × 105 m–1 (1.1 to 1.4 ok) (1)
max 2 / 6 if no graph and more than one data set used correctly, 1 / 6 only one set
if tan calc but plotted as sin, mark as scheme
tan or distance plotted, 0 / 6
max 6
(b)
(i)
maxima wider spaced [or pattern brighter] (1)
sin θ or θ increases with N [or light more concentrated] (1)
Page 44 of 63
(ii)
maxima spacing less (1)
sin θ or θ decreases with λ [or statement] (1)
(iii)
maxima wider spaced [or pattern less bright] (1)
same θ but larger D [or light more spread out] (1)
6
(c)
(i)
waves in phase from (1)
any sensible ref to coherence (1)
whole number of wavelengths path difference (1)
(ii)
use of geometry to show that sin θ =
max 3
[15]
11
(a)
(i)
(since dsinθ = nλ)dsin18.5° = 632.8 × 10–9 (1)
d = 1.99 × 10–6 (1)
number of lines per metre =
(ii)
= 5.01 × 105 (1)
nλ = 1.99 × 10–6 sin90° (1)
n=–
= 3.1(5) (1)
hence highest order is third (1)
(6)
(b)
= 590 nm(1)
(2)
[8]
12
(a)
(i)
vertical or parallel (1)
equally spaced (1)
black and yellow [or dark and light] bands (1)
(ii)
= 2.9 × 10–3 m (1)
tan θ =
(1) gives θ = 0.18° (1)
Page 45 of 63
(iii)
narrower slits give more diffraction (1)
more overlap (so more fringes) (1)
fringes same width (1)
(max 8)
(b)
(i)
× sin θ = 5.86 × 10–7 (1)
θ = 13.6° (1)
(ii)
θ = 90° and correctly used (1)
= 4.3 ∴ 4th order (1)
(5)
(c)
brighter images (1)
large angles (1)
sharper (or narrower) lines (1)
(max 2)
[15]
13
(a)
graph to show:
maxima of successively smaller intensity (1)
subsidiary maxima / minima equally spaced (1)
(at least two each side of central axis)
width of subsidiary sections half width of central section (1)
symmetrical pattern each side of central axis (1)
4
(b)
(i)
broader maxima or pattern (1) [or fringes wider apart]
dimmer pattern (1)
(ii)
maxima are closer (1) [or narrower fringes]
green and dark regions (1)
max 3
[7]
14
15
16
D
[1]
C
[1]
A
[1]
Page 46 of 63
17
(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
(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]
18
(a)
(i)
0, 2π or 4π [or 0, 360° or 720°] (1)
(ii)
4λ (1)
(iii)
sin θ =
[or sin θ =
(1)
]
CE = 4λ and AC = 2d (1) (hence result)
[or BD = 2λ and AB = d]
max 3
Page 47 of 63
(b)
(limiting case is when θ = 90° or sin θ = 1)
(1)
(= 4.6)
highest order is 4th (1)
2
[5]
19
20
21
22
B
[1]
C
[1]
D
[1]
sinθ = nλ/d in this form/correct calculations of d/d = 1/300
C1
substitutes correctly – condone powers of 10
C1
18.9
C1
2 or 3 sf only
A1
[4]
23
(a)
(i)
= 590 × 10–9m (1)
(using d sin θ = nλ gives)
(1) = 0.707 or
7.07 × 108 if nm used (1)
θ = 45.0° (1) (accept 45°)
Page 48 of 63
(ii)
(sin θ ≤ 1) gives
≤ 1 or n ≤
or =
(1) = 2.83 (1)
so 3rd order or higher order is not possible (1)
alternative solution:
(substituting) n = 3 (into d sin θ = nλ gives) (1)
sin θ (
) = 1.06 (1)
gives ‘error’/which is not possible (1)
7
(b)
(using d sin θ = nλ gives)
2 λ = 1.67 × 10–6 × sin 42.1 (1)
λ(= 0.5 × 1.67 × 10–6 × sin 42.1) = 5.6(0) × 10–7 m (or 560 nm) (1)
2
[9]
24
25
A
[1]
(a)
λ correct (1)
d correct (1) arrow or line needed, both ends extending beyond
central black line
2
(b)
angle θ gets smaller (1)
because path difference gets smaller/d constant, (λ smaller) so
sin θ smaller (1)
max 1 for correct explanation for λ increasing
2
(c)
boxes 1,5,6 (1)(1)
two correct 1 mark
4 ticks max 1
5 or 6 ticks gets 0
2
Page 49 of 63
(d)
(i)
3.3 × 10–6 m (1) (1/300 = 3.33 × 10–3 mm, 3300 nm) DNA 1 sf here
DNA 1/300 000 as answer
accept 3 1/3 × 10–6, 3.33 × 10–6 recurring, etc
1
(ii)
(sin θ =)
(1)
correct wavelength used and seen (545 to 548 × 10–9)
and 9.4 to 9.6 (°) (1) ecf (d) (i), for correct wavelength only
(545 to 548 × 10–9)
2
[9]
26
(a)
3 subsidiary maxima in correct positions (1)
intensity decreasing (1)
2
(b)
a single wavelength (1)
constant phase relationship/difference (1)
2
(c)
maxima further apart/central maximum wider/subsidiary maximum
wider/maxima are wider (1)
1
(d)
wider/increased separation (1)
lower intensity (1)
2
Page 50 of 63
(e)
distinct fringes shown with subsidiary maxima (1)
indication that colours are present within each subsidiary maxima (1)
blue/violet on the inner edge or red outer for at least one subsidiary
maximum (1)
(middle of) central maximum white (1)
3
[10]
27
(a)
single frequency (or wavelength or photon energy)
not single colour
accept ‘ very narrow band of frequencies’
1
(b)
subsidiary maxima (centre of) peaks further away from centre
For second mark: One square tolerance horizontally. One whole
subsid max seen on either side.
subsidiary maxima peaks further away from centre AND central maximum twice width of
subsidiaries AND symmetrical
Central higher than subsid and subsid same height + / − 2 squares.
Minima on the x axis + / − 1 square.
Must see 1 whole subsidiary for second mark
2
(c)
ONE FROM:
•
•
•
•
•
•
don't shine towards a person
avoid (accidental) reflections
wear laser safety goggles
'laser on' warning light outside room
Stand behind laser
other sensible suggestion
allow green goggles for red laser, ‘high intensity goggles’, etc.
not ‘goggles’, ‘sunglasses’
eye / skin damage could occur
2
Page 51 of 63
(d)
3 from 4
•
•
•
•
central white (fringe)
each / every / all subsidiary maxima are composed of a spectrum (clearly stated or
implied)
each / every / all subsidiary maxima are composed of a spectrum (clearly stated or
implied) AND (subsidiary maxima) have violet (allow blue) nearest central maximum
OR red furthest from centre
Fringe spacing less / maxima are wider / dark fringes are smaller (or not present)
allow ‘white in middle’
For second mark do not allow ‘there are colours’ or ‘there is a
spectrum’ on their own
Allow ‘rainbow pattern’ instead of spectrum but not ‘a rainbow’
Allow ‘rainbow pattern’ instead of spectrum but not ‘a rainbow’
If they get the first, the second and third are easier to award
Allow full credit for annotated sketch
3
[8]
28
(a)
one of:
(spectral) analysis of light from stars
(analyse) composition of stars
chemical analysis
measuring red shift \ rotation of stars ✓
insufficient answers:
‘observe spectra’, ‘spectroscopy’, ‘view absorption \ emission spectrum’, ‘compare
spectra’, ‘look at light from stars’.
Allow : measuring wavelength or frequency from a named source of
light
Allow any other legitimate application that specifies the source of
light. E.g.
absorbtion \ emission spectra in stars,
‘observe spectra of materials’
1
(b)
(i)
first order beam
first order spectrum
first order image
✓
Allow ‘n = 1’ , ‘1’ , ‘one’, 1
st
1
(ii)
the light at A will appear white (and at B there will be a spectrum)
OR greater intensity at A ✓
1
Page 52 of 63
(c)
( d = 1 / (lines per mm × 103)
= 6.757 × 10−7 (m) OR 6.757 × 10−4 (mm) ✓
( nλ = d sin θ )
= 6.757 × 10−7 × sin 51.0 ✓ ecf only for :
• incorrect power of ten in otherwise correct calculation of d
• use of d = 1480, 1.48, 14.8 (etc)
• from incorrect order in bii
= 5.25 × 10−7 (m) ✓ ecf only for :
• incorrect power of ten in otherwise correct d
• from incorrect order in bii
Some working required for full marks. Correct answer only gets 2
Power of 10 error in d gets max 2
For use of d in mm, answer =
5.25 × 10−4 gets max 2
n = 2 gets max 2 unless ecf from bii
use of d = 1480 yields wavelength of 1150m
3
(d)
n = d (sin90) / λ
OR n = 6.757 × 10−7 / 5.25 × 10−7 ✓ ecf both numbers from c
= 1.29 so no more beams observed ✓ or answer consistent with their working
OR
2 = d (sinθ) / λ OR sinθ = 2 × 5.25 × 10−7 / 6.757 × 10−7 ✓ ecf both numbers from c
sinθ = 1.55 (so not possible to calculate angle) so no more beams ✓
OR sin−1(2 × (their λ / their d) ) ✓
(not possible to calculate) so no more beams ✓ ecf
Accept 1.28, 1.3
Second line gets both marks
Conclusion consistent with working
2
[8]
29
30
31
C
[1]
D
[1]
(a)
Suitable experiment eg diffraction through a door / out of a pipe ✓
1
(b)
Using c = d / t
t = 2 500 / 480 = 5.2 s ✓
1
Page 53 of 63
(c)
(Measured time is difference between time taken by light and time taken by sound)
Calculation assumes that light takes no time to reach observer, ie speed is infinite ✓
Do not allow “could not know speed of light”
1
(d)
Sound from gun is a mixture of frequencies. ✓
Alternative for 1st mark ‘(so speed is independent of frequency) the
sound of the gun is similar when close and far away’
1
All the sound reaches observer at the same time, ✓
1
(e)
More accurate, as it is closer to the accepted value. ✓
1
(f)
When θ = 0 °C
c = 331.29 m s–1
1
Therefore
331.29 = k √273.15 ✓
k = 20.045 ✓
1
(g)
The method and value are published ✓
1
other scientists repeat the experiment using the same method ✓
1
[10]
Page 54 of 63
Examiner reports
2
(a)
Many candidates incorrectly gave laser or sodium vapour lamp as a source producing a
continuous spectrum.
(b)
(i)
Many candidates correctly calculated the first angle.
(ii)
Many simply doubled the answer to part (i) to calculate the angular dispersion. Use of
the sine or tangent was relatively uncommon and even with error carried forward few
candidates gained the mark for the width of the first order spectrum.
(c)
3
Most candidates recognised that the intensity would fall because the energy was being
spread out over a larger area, many candidates recognised that this was an inverse square
relationship but only the strongest calculated the correct factor.
This question examined topics in diffraction-grating theory.
(a)
Many could show successfully how the grating spacing was calculated.
(b)
(i)
Again many could carry out the simple calculation accurately. Common errors were to
treat n in the equation nλ = sin θ not as ‘1’ but as 6.30 × 105 lines per metre, and to
carry through the calculation in millimetres, but express the answer in metres.
4
5
(a)
(ii)
Judging by the poor quality of many answer, few candidates have ever seen a
diffraction grating pattern during their course of study. There were a number of
possible scoring points available featuring the appearance of the central maximum,
the order of colours (which could be deduced from the question), the overlap of
orders, and so on. Only a few candidates scored more than one or two marks.
(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)
Some incorrectly quoted and used sin θ = λ/b but most used the appropriate formula.
Forgetting a unit cost some the second mark.
Page 55 of 63
(b)
6
7
8
Relatively few gained all three marks here. Those who were on the right track obtained 4.8
and rounded down to 4 orders but then forgot about the symmetry of the pattern and the
central maximum. There was a significant proportion who simply divided 180° by 12°.
Large numbers of candidates scored well on this part, apart from the inevitable difficulties with
powers of ten, but let themselves down by omitting the unit or by writing mm for m (or
vice-versa). Even more able candidates need to pay attention to checking these matters.
Examiners were looking not just for a final answer but for a clear exposition of the calculation in
this part. Again, as in an earlier question, it is apparent that some can recognise the equation to
use but not which quantity to substitute for the various symbols.
(i)
Few candidates used both first order images to calculate 26 and hence 6. Many went
straight into either the diffraction grating equation or single slit equation and used incorrect
values for the variables.
(ii)
Having used the diffraction grating equation in (i) many candidates made incorrect use of
the double slit equation here or else used the diffraction grating equation again but
calculated n as number of lines per mm. Only a limited number were able to progress all
the way through this calculation.
(a)
(i)
Only about half could calculate the time between two successive events given
the total number of time intervals and the total time. Many calculated the reciprocal of
the answer they required.
(ii)
The calculation of the speed of the wave was equally tricky for the candidates with
about half remembering that the echo consists of a sound wave travelling to and from
the reflecting surface.
(i)
The second part of the question featured calculations relating to the diffraction
grating, in this case in the unusual context of a sound wave incident on a slatted
wooden fence. The first of these was a straightforward calculation of wavelength
using v = fk. Again, many (about a quarter of the total this time) carried this through
with a reciprocal error.
(ii)
Candidates were expected to use an appropriate equation (on the data sheet) to
calculate the spacing between the vertical fence slats. A large number used the
equation for the angle of the first minimum diffraction at a single aperture (to the
extent of using the wrong symbol for the slit spacing and omitting any mention of n,
the diffraction order). This was an error in physics and was therefore not credited.
(b)
Page 56 of 63
(iii)
10
Failures in part (ii) were all the more surprising given the strong lead that candidates
could have used here. Many realised what they had to do, calculated the value of the
order for a diffracting angle of 90° (to yield an answer of 2.14) and then recognised
that the diffraction order of 2 was the largest that could be observed. Candidates who
failed to gain full credit included those who quoted 2.14 for the final answer and those
who rounded up to 3.
This question attracted the smallest proportion of answers. For those candidates who attempted
the question, the answers to part (a) fell mainly into one of two categories. In the first group were
candidates who had obviously either seen or done the experiment or who at least were able to
interpret the diagram correctly. This group of candidates could calculate the diffraction angles,
plot an appropriate graph and use it to determine the final answer, scoring full marks. The second
group consisted of those candidates who confused the situation with a Young double slit
experiment. Although these candidates could plot a graph, there was no way that it could be
related to any sensible means of determining the number of rulings per metre on a grating. It was
impossible to award any marks to answers of this kind. There were some candidates who made
some progress with the question but whose answers fell a long way short of the ideal. For
example, some calculated the tangent of the diffraction angle but interpreted this as sine. Others
did not plot a graph at all but calculated at least one value from the data. There was some merit
in answers of this kind but, at best, they were worth one or two marks out of six.
Success with part (b) hinged upon the recognition of the relevance of the diffraction grating
equation. Many, but not all, of those candidates who had interpreted the question as relating to
the double slit experiment, continued with the same theme and few qualified for any marks.
There was a minority of very good answers in which the grating equation was clearly stated and
the effect of changing the appropriate variables was analysed. Some otherwise quite good
answers were spoiled by imprecise use of language such as “more (or less) diffraction” or “the
pattern becomes less”. Some candidates thought that, in answer to part (b)(iii), the pattern
remained unchanged because the diffraction angle did not alter.
Answers to part (c) were generally weak and many candidates left the answer space blank. It
had been hoped that the diagram would provide a useful prompt, but this was not the case. Of
those candidates who answered the question, many scored a mark for some reference to either
“waves in phase” or “coherence”. Few candidates were able to put together the logical steps
stating the waves from each slit to have a whole number of wavelengths path difference, so
being in phase and reinforcing when superposed. In answer to part (c)(iii), most candidates put n
= 1 in the grating equation and did not use the diagram to show, for example, that a line drawn
from D perpendicular to EQ has a length equal to the wavelength. The equation then follows
naturally from this diagram.
11
This question produced a substantial number of fully correct answers from candidates who had
clearly been well prepared on the topic of diffraction.
Page 57 of 63
In part (a)(i) most candidates calculated d correctly and went on to give a correct value of N. The
most common errors were to convert nanometres into metres incorrectly incurring an arithmetical
error penalty, or to give the wrong number of significant figures and incur a significant figure
penalty. In part (a)(ii) some candidates correctly stated that the highest order was the third but did
not support their answer with proper argument and calculation.
Part (b) was answered well by nearly all candidates.
12
Descriptions of the fringes in part (a)(i) were generally poor. In spite of the emphasis on the word
vertical, hardly any candidates referred to the vertical lines which result. Furthermore, simple
references to dark and bright equally-spaced bands were very rare. Confusion with the single
diffraction pattern was common and some candidates contradicted themselves by referring to
equally- spaced fringes at the same time as showing a diagram of the single slit pattern. Some of
the descriptions made it difficult to believe that the writer had ever seen double slit interference
fringes.
The large majority of candidates correctly calculated the fringe separation but failed to determine
the angle. The word ‘angle’ frequently triggered the diffraction grating equation and, for those
candidates who did attempt a simple geometrical calculation, many calculated the angle only to
the first bright fringe and not the second. Such candidates penalised themselves by failing to
read the question carefully. Several candidates who had calculated d correctly were quite
prepared to offer a maximum order of thousands as their answer to part (b)(ii).
Answers to part (a)(iii) showed a general weak understanding of the relative parts played by
interference and diffraction in this part of the question. One group of candidates failed to score
any marks because they confused variation of slit width with slit separation and gave their
answer in terms of the latter. For those candidates who did consider the narrowing of each slit,
many scored only a single mark for recognising that the light at each slit will be diffracted through
a larger angle. Although answers were often given in rather vague terms such as “more
diffraction”. To gain full credit, it was necessary to refer to the greater area of overlap of the
diffracted beams and to recognise that more fringes are seen in this increased area because the
fringe width is unchanged. Hardly any candidates made the last of these two points.
Calculations in parts (b)(i) and (b)(ii) were done well and many candidates scored full marks.
Common errors in part (b)(i) included incorrect conversion of units from 400 lines per millimetre,
poor arithmetic and the often seen confusion between the grating constant, d, and the number of
rulings per unit length, N.
Good answers to part (c) were rare and the two systems were not often compared from a
position of secure knowledge. Very few candidates were able to state that the grating is designed
to produce fewer lines which are brighter and more widely spaced. lndeed, many candidates
stated that the grating produced more fringes. Most answers concentrated on the accuracy of the
readings rather than on how the readings were to be obtained. Answers were, in general, very
vague.
Page 58 of 63
13
The main problem in candidates’ answers to this question was confusion between single slit
diffraction and double slit interference. The completion of the sketch graph in part (a) regularly
gained full marks for the many candidates who appreciated its salient features. Equally spaced
markers on the horizontal axis were intended to guide candidates into showing the subsidiary
minima at equal spacing with the separation between consecutive minima being half the width of
the central maximum. Candidates who thought that the separation between subsidiary minima
was the same as the width of the central peak usually achieved only half marks on this part. The
relatively low intensity of all the subsidiary maxima is not well known and no marking point was
available for this aspect.
When marking part (b) the examiners viewed with great suspicion any answer which relied on the
Young’s slits equation, λ = (ws / D). The fact that narrowing the slit causes the pattern to broaden
was usually appreciated in part (i), but few candidates stated that the pattern is also dimmer. In
part (ii) many candidates had difficulty in deciding whether red light or green light has the longer
wavelength and the wrong choice, of course, caused many incorrect answers to the way the
separation of the maxima changes. The other point the candidates could have made is that the
colour of the pattern changes from red to green. It is worth pointing out that the marks available
for parts of questions are given in the question paper and that these are intended to help
candidates. In part (b), three marks were available and this would indicate that a minimum of
three points are required in a complete answer.
17
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.
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Knowledge of the derivation of d sin θ = nλ for the diffraction grating is required by section 13.1.7
of the Specification. Fundamental to this derivation, is familiarity with the concepts of phase and
path difference. Part (a) proved to be an effective test of candidates’ understanding in these
areas, and the question seemed to strike many candidates with apprehension: blank spaces
were fairly common and ridiculous answers very frequent. Phase difference was particularly
badly known, with many answers to part (i) expressed in terms of λ A correct answer of 4λ in part
(ii) became almost a prerequisite for a successful approach to part (iii). Clearly 2λ = d sinθ can be
shown by inserting n = 2 into the standard formula, but this was not the target of part (iii) and no
marks could be awarded for such a trivial response.
Several recent questions about the diffraction grating in the Unit 4 Section A papers have
covered areas similar in content to part (b), and candidates answers to this part were usually
much more satisfactory than those in part (a). There was some confusion between the number of
lines per metre (4.5 × 105) and the grating spacing (2.2 × 10–6 m). A small number of candidates
took the numbers from their calculations too literally, quoting their final answers for the order as
4.57, whilst others failed to comprehend that this meant that the highest order would be the
fourth rather than the fifth.
22
23
Many candidates answered this question well. The most common error was to use the value for
the number of lines per mm as the grating spacing. Otherwise, a few candidates had their
calculators set in radians rather than degrees. Significant figures were coped with well.
There were common mistakes to part (a) (i), such as failing to put n = 2. Some candidates
thought n was the refractive index and for this reason put n = 1. A significant number did not
convert from nm to m. Part (a) (ii) was done very well by the majority of candidates, either by
substituting in 90° or n = 3. Most were successful in finding the wavelength to part (b).
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Many candidates did not seem at all familiar with the use of this diagram in the derivation of the
grating equation in part (a) and the placing of the labels was often completely random. A large
number did not attempt to label the diagram and half of all candidates did not score any marks.
Many who scored one mark had labelled the wavelength correctly but did not accurately indicate
the ‘line spacing’ with a suitable arrow or line.
Most candidates gained the first mark in part (b) for realising that sin θ decreased so θ would
decrease. Many candidates failed to gain the second mark by not stating that d remained
constant. Very few candidates attempted to explain in terms of path difference.
The majority of candidates had no problem matching up the spectral lines in part (c).
In part (d) (i), about half of all candidates were unable to convert lines per mm to line spacing and
there was considerable confusion with powers of ten. Many candidates did not convert to metres
and many also rounded to one significant figure.
In part (d) (ii) it was expected that the candidate would read an accurate value off the scale.
However, many chose a value to the nearest 10 nm, typically 550 nm. In this situation, it is
always best to interpolate when reading off the scale. The uncertainty in this reading can then be
expressed by giving the final answer to two significant figures. Line P is somewhere between 545
and 548 nm.
26
Most candidates gained at least one mark in part (a) for showing that the intensity of peaks
reduced with distance from the centre. However, many did not recall the key difference between
the pattern for single and double slits – the single slit pattern has a central maximum which is
double the width of the subsidiary maxima.
There were many correct definitions of monochromatic and coherent in part (b). A few stated
‘same colour’ for monochromatic and ‘in phase’ for coherent. Neither of these were accepted.
In part (c), many candidates incorrectly used the equation for two slits to show that the maxima
were further apart. This was not penalised since an explanation was not asked for.
Many candidates got part (d) the wrong way around, saying that the fringes would be more
closely spaced and more intense. There seemed to be some guess work evident here.
Candidates need to be able to describe the appearance of the single slit pattern and be aware of
how it will change for different wavelengths, slit widths and for monochromatic and white light.
Some teachers introduce the equation for the single slit although it is not in the specification. This
is not necessary but can certainly help the more mathematically minded students. To illustrate
the change in the pattern, a simple demonstration can be carried out with a red and a green laser
shone through the same slit onto a screen.
A pleasing number of candidates produced very detailed and high quality answers to part (e),
with many gaining all three marks. Some drew a graph of intensity, which did not gain a mark on
its own.
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(a)
In general, this was a well answered question apart from a tendency for candidates to add
extra detail, e.g. ‘single wavelength and coherent’ ; this loses the mark. As does: ‘single
wavelength / colour’ ; because this implies that monochromatic could be just a single colour.
However, ‘light of a single wavelength and therefore a single colour’ would be acceptable. It
is therefore best to learn the appropriate definition and not add any further detail.
(b)
Many candidates did not know what to do on this question.
The red light subsidiary maxima were often shown closer to the central maximum than the
blue.
Perhaps single slit diffraction tends to be a little overlooked because the specification does
not require any mathematical description. Nevertheless, students should be shown images
of the single slit pattern and how it changes for different wavelengths. Images are readily
available on the internet via any search engine.
(c)
When talking about laser safety, it is not acceptable to say simply ‘wear goggles’ . One must
say ‘laser safety goggles’ , ‘laser safety glasses’ , or ‘laser safety eyewear’ . Standard
laboratory goggles would not afford any significant protection against laser light.
(d)
Only a few candidates were able to describe the pattern accurately. Answers tended to be
vague and ambiguous. Only a small number decided to add a sketch to clarify their answer
and this approach should be encouraged. Again, perhaps the single slit has been
overlooked by some in favour of the ‘more difficult’ double slit and grating.
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(a)
There were some rather vague answers here such as ‘ To calculate the wavelength of a
light’ or ‘ to look at the light from stars’. There needed to be a little more than this to get the
mark, i.e. a specific example such as ‘analyse the elements present in the atmosphere of a
star’ or explain that the composition of a material or gas can be determined.
(b)
The candidates who knew this often lacked detail in their answer, e.g. ‘it would be dimmer’.
Some thought there would only be one colour at B rather than a spectrum.
Quite a few thought that the wavelength at B would be different from A due to the
increased angle.
Some candidates thought that the light at B would be composed of different wavelengths
and the white light at A would be a single wavelength.
(c)
This was a fairly standard exam question but surprisingly there were few correct answers.
Students seemed to be poorly prepared for this question and confusion reigned regarding
the meaning of the terms in the grating equation. Use of the lines per mm as the line
spacing (d = 1480) was very common.
There was also confusion between line spacing, d, and order, n. Some used 1480 for d and
for n.
Candidates often used 1 / 1480 and then failed to convert this into metres.
(d)
There were a surprising number of candidates who did not attempt this question.
Even if they felt they had the wrong numbers for wavelength and line spacing in part (c),
candidates simply needed to divide their d by their λ, and if greater than 1, conclude that no
further orders are possible.
There was also some confusion over the method required, e.g. some used the angle given
in part (c), (51°), and calculated a new wavelength that would give a second order at that
angle.
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