Rotations for exploring paths
Question 10S: Short Answer
1.
Trip time 

distance
speed
16 .5 m
3  10 8 m s 1
 5.5  10  8 s.
2.
Number of rotations  rotations per second  number of seconds
 1 10 8 s 1  5.5  10 8 s
 5.5.
3. 
4.
Trip time 

distance
speed
24 m
3  10 8 m s 1
 8.0  10 8 s.
5.
Number of rotations  rotations per second  number of seconds
 1 10 8 s 1  8  10 8 s
 8.0.
6. 
7.
Trip time 

distance
speed
3m
3  10 8 m s 1
 1.0  10 8 s.
8.
Number of rotations  rotations per second  number of seconds
 5  1014 s 1  1 10 8 s
 5.0  10 6.
9. 
10.
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Trip time 

distance
speed
3 .3 m
3  10 8 m s 1
 1.1 10 8 s.
11.
Number of rotations  rotations per second  number of seconds
 5  1014 s 1  1.1 10 8 s
 5.5  10 6.
12. 
13. For the 2.2 m path
Number of rotations  trip time  frequency
2. 2 m
 2  1014 Hz 
3  10 8 m s 1
 1.47  10 6 rotations
For the 2.4 m path
Number of rotations  trip time  frequency
2. 4 m
 2  1014 Hz 
3  10 8 m s 1
 1.6  10 6.
Path lengths and arrow rotations
Question 30S: Short Answer
1.
distance  2 m 2  1m 2
 5 m2
 2.24 m.
2.
time 

distance
speed
2.24 m
3  10 8 m s 1
 7.45  10 9 s.
3.
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number  time  frequency
 (7.45  10 9 s)  (5  10 9 Hz )
 37 .3.
4.
distance  1m 2  3 m 2
 10 m 2
 3.16 m.
time 

distance
speed
3.16 m
3  10 8 m s 1
 1.05  10  8 s.
number  time  frequency
 (1.05  10 8 s)  (5  10 9 Hz )
 52 .7.
5.
total rotations  first segment rotations  second segment rotations
 37.3  52.7
 90.
6.
distance  5 m 2  20 m 2
 6.71m.
7.
time 

distance
speed
6.71m
3  10 8 m s 1
 2.24  10 8 s.
8.
number  time  frequency
 ( 2.24  10 8 s )  (5  10 9 Hz )
 111 .8.
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9.
Three paths on a mirror
Question 40S: Short Answer
1.
distance  0.7 m 2  0.8 m 2  0.3 m 2  0.8 m 2

1.13 m 2 
0.73 m 2
 1.92 m.
2.
time 

distance
speed
1.92 m
3  10 8 m s 1
 6.39  10 9 s.
3.
number  time  frequency
 ( 6.39  10 9 s )  (7  10 14 Hz )
 4 .47  10 6.
4.
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distance  0.4 m 2  0.8 m 2  0.8 m 2  0.6 m 2

0 .8 m 2 
1m 2
 1.89 m
distance
time 
speed
1.89 m

3  10 8 m s 1
 6.31  10 9 s
number  time  frequency
 ( 6.31  10  9 s )  (7  10 14 Hz )
 4.42  10 6.
5. A good answer would explain the links between the shorter distance, lower trip time and fewer
rotations.
6. A good answer would include references to the need to sum the contributions to all paths, and in
particular the need to sample neighbouring paths to those sampled, in order to see if the paths
come from regions of the mirror where arrows are tending to line up or curl up.
Paths through a block
Question 100S: Short Answer
1.
Path A
Path B
Path C
–10
s
1.26  10
–10
s
2.31  10
–10
s
6.41  10
s
4.21  10
Time exploring source to
block
1.00  10
s
1.03  10
Time exploring in block
2.92  10–10 s
2.56  10
Time exploring block to
detector
7.18  10
Total time for photon to
explore
4.64  10
–10
Path D
–10
s
6.67  10
–10
s
4.27  10
–10
Path E
–10
s
1.59  10
–10
s
2.08  10
–10
s
6.67  10
s
4.33  10
–10
–10
s
2.15  10
–10
s
1.92  10
s
7.18  10
s
4.76  10
–10
–10
–10
–10
2. Each top arrow must be rotated by the specified number of degrees anti-clockwise in the
appropriate box:
Path A
Path B
Path C
Path D
Path E
Arrow as leaves source





Arrow as enters block
0
230
148
268
305
Arrow as leaves block
28
203
37
240
332
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s
s
–10
s
–10
s
Arrow as arrives at detector
Path A
Path B
Path C
Path D
Path E
9
83
46
120
313
3. No. A good answer would explain the links between groups of paths delivering arrows that line up
or curl up and the consequent need for at least representative sample of paths.
4. Path C. This has the lowest trip time. A good answer would explain the links between least time
and changes in trip time.
A quantum lens
Question 120S: Short Answer
1. Top path measures 12 cm:
distance
trip time 
speed
0.12 m

3.0  10 8 m s 1
 4.0  10 10 s.
2. The same so:
trip time  4.0  10 10 s.
3. Middle path except for glass measures 11 cm:
distance
trip time 
speed
0.11 m

3.0  10 8 m s 1
 3.7  10 10 s.
4. The difference in trip time between (2) and (3):
difference in trip time  3.1 10 11 s.
5. Find the speed for the 0.5 cm thickness of lens:
distance
speed 
time
0.005 m

3.69  10 10 s
 1.63  10 8 m s 1.
6. Refractive index is the ratio of speeds:
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n

c vacuum
cn
3 .0  10 8 m s 1
1 .63  10 8 m s 1
 1 .86 .
This is high, compared with everything except diamond, so this design could be expensive. Try a
thicker lens, so using material of lower refractive index.
7. A good answer would explain that the same trip times lead to the same number of rotations for
arrows from all paths, so forcing the arrows to line up, summing to a large amplitude, and thus
yielding a large probability of finding a photon at that point.
8. Not at all. Arrows of identical frequency would complete identical numbers of rotations. Changing
the frequency of all the arrows in the calculation still leaves them lined up – an example of global
symmetry!
9. A good answer would show that this can only be the case if refractive index depends on
frequency.
A quantum view of diffraction
Question 140S: Short Answer
1. C
2. B and D
3. Arrows:
amplitude: zero
4. Arrows:
amplitude:
5. The intensity is proportional to amplitude.
At D the intensity is zero and so the amplitude must also be zero. Therefore the arrows must add
up to zero.
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At C the intensity is greater than at E. So the amplitude at C must be greater than the amplitude
at E. Arrows must curl up less and line up more.
6. A and E both move further away from C. The bright patch at C becomes broader.
Electron volts – a new unit
Question 145S: Short Answer
1. See table below
Metal
Work function / eV
Work function / J
Zinc
3.63
5.81 × 10-19
Platinum
5.36
8.58 × 10-19
Caesium
1.95
3.12 × 10-199
Copper
4.65
7.44 × 10-19
Sodium
2.36
3.78 × 10-19
2.
E  hf  (6.6  10 34 J s)  (1.1 1015 Hz)  7.3  10 19 J
E
7.26  10 19 J
1.6  10 19 J/eV
 4.5 eV
3.
E  VQ  5000 V  e  5000 eV
5000 × 1.6 × 10-19 J = 8.0 × 10–16 J
Arrows for electrons
Question 150S: Short Answer
1. Kinetic energy EK = 1000 eV × 1.602 × 10-19 C = 1.602 × 10-16 J
2. Speed v = (2 EK / m) = 1.88 × 107 m s-1
3. de Broglie wavelength = h / mv = 6.626 × 10-34 J s / (9.11 × 10-31 kg × 1.88 × 107 m s-1) = 3.88 ×
10-11 m = 0.0388 nm
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4. The wavelength is smaller than, but comparable to the size of an atom. Thus diffraction effects
can be large.
5. The wavelength is smaller, so that diffraction angles are smaller for the same sized obstacle or
slit.
Electrons through gratings
Question 160S: Short Answer
1. The wavelength decreases, because  = h / mv .
2. The distance to complete one turn is equal to the wavelength, so it decreases.
3. There are a greater number of turns in both paths, because the distance to make one turn is
smaller.
4. The path difference must be reduced, so that it is the same as before if measured in numbers of
smaller wavelengths or distance for one arrow turn.
5. Point P must move downwards, reducing the length of both paths and the difference in length
between them.
Understanding the photoelectric effect
Question 50C: Comprehension
1. E = hf therefore f = E / h and c = f  therefore  = c / f = ch / E.
For example A:
fA 
7.2  10 19 J
6.6  10 34 J s
 A  c / fA 
 1.1 1015 Hz
ch (3 x 10 8 m s 1 ) x (6.6 x 10 34 J s)

 2.8  10 7 m  280 nm (to 2 s.f.)
EA
7.2 x 10 19 J
For example B:
fB 
B 
5.1 10 19 J
6.6  10 34 J s
 7.7  1014 Hz
c
ch (3  10 8 m s 1 )  (6.6  10 34 J s)


 3.9  10 7 m  390 nm (to 2 s.f.)
fB EB
5.1 10 19 J
Note that the answer for example B is just outside the visible range.
2. Due to the quantum nature of the incoming radiation, an electron must absorb a whole photon of
light energy. It cannot take from the photon just what it needs, but absorbs the whole quantum of
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energy.
3. Potential energy = charge × potential difference. Thus E p = eV s
4. = hf – eV s
= ((6.6 × 10–34 J s) × (1.1 × 1015 Hz)) – ((1.6 × 10–19 C) × 0.88 V)
= 5.8 × 10–19 J.
5. The minimum frequency corresponds to minimum energy to free an electron, i.e. the electron will
have zero kinetic energy when it is free.
Hence:
hf  
f   / h  ( 4.64  10 19 J) /(6.6  10 34 J s)  7.0  1014 Hz
6. V s = ( h / e ) f – / e  gradient = h / e hence h = e × gradient.
Intercept on vertical (V ) axis = – / e hence = – e × intercept.
Note that the intercept will be negative so the value of the work function will be positive.
Mirrors for precision engineering
Question 90C: Comprehension
1. Photolithography is the process of making semiconductor chips using photographs as templates.
2. At 13 nm:
c
f 


3  10 8 m s 1
13  10  9 m
 2.3  10 16 Hz.
At 157 nm:
c
f 


3  10 8 m s 1
157  10  9 m
 1 .9  10 15 Hz.
3. At 13 nm:
E  hf
 6.634  10  34 J s 1  9.3  10 15 Hz
 1.53  10 18 J.
At 157 nm:
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E  hf
 6.634  10 34 J s 1  1.9  10 15 Hz
 1.27  10 18 J.
4. A good answer would explain these links: start with higher frequency, that this leads to more rapid
curling up for small differences in trip time, therefore leading to better image definition (higher
resolution) and point out that higher-energy photons require fewer of them, possibly reducing the
exposure time.
5.
mask as source
Silicon as detector
6. A good answer would suggest that the curvature ensures that a large number of paths from the
source (one point on the mask) to the detector (one point on the silicon) produce arrows that line
up, thus ensuring a high chance of photons arriving at that point.
7. From the passage: 1nm measures out five atoms, so each atom is 0.2 nm across:
distance
trip time 
speed

0.2  10 9 m
3.0  10 8 m s 1
 6.7  10  21 s.
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An ancient lens
Question 130C: Comprehension
1.
3.1 cm
or
4.1 cm
20 cm
0.6 cm
26.5 cm
2. Trip time through the top of the lens (a glancing path) must equal the trip time through the middle.
Calculate source and detector positions from focal length:
Top path: distance / speed = trip time.
Middle path: same calculation for air and glass separately, then add up.
3. A good answer would explain that the same trip times lead to the same number of rotations for
arrows from all paths, so forcing the arrows to line up, summing to a large amplitude, and thus
yielding a large probability of finding a photon at that point.
4. A good answer would start by stating that arrows of identical frequency would complete identical
numbers of rotations. Changing the frequency of all the arrows in the calculation still leaves them
lined up – an example of global symmetry! So the lens would seem to work for all colours –
provided that the slowing down of the rate of exploration in the glass is the same for all
frequencies.
5. A good answer would show that this could only be the case if refractive index depends on
frequency.
Uncertainties in measuring the Planck constant h
Question 100D: Data Handling
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measuring Planck’s constant
5×10–19
4×10–19
3×10–19
2×10–19
1×10–19
0
0
2×1014 4×1014 6×1014 8×1014
photon frequency / Hz
2. A lot of human judgement comes into interpreting these best fit lines; expect your estimate for h to
be close to (8 ± 2) × 10–34 J s.
Yes; the estimate includes the accepted value of the Planck constant.
3. The best fit linear trendline has the equation y = 7.3 × 10–34x – 9.9 × 10–20.
You might judge that the overall uncertainty in the experiment is about ± 15–20%, not being too
optimistic about combining uncertainties in energy and frequency values.
By this method h = (7.3 ± 1.5) × 10–34 J s.
This estimate does include the accepted value of the Planck constant.
4. Some systematic error must be present. Although the graph has an acceptable gradient, agreeing
with h within the estimated uncertainty, the graph line is ‘too low’. It has a negative y intercept,
when it should pass through the origin.
The manufacturer’s wavelength values could be systematically too low (hence the derived
frequencies are too high, shifting the graph to the right). Or the measured values of p.d. could be
systematically too low (pulling the graph downwards). Or a combination of both!
5. Wavelengths too high: No, this would give lower frequencies, pushing the line of best fit further to
the left and towards the origin.
Zero error in voltmeter: Yes, if the voltmeter read consistently low this could account for the
negative y intercept of the best-fit line at –9.9 × 10–20 J or –0.62 V. The zero error would need to
be V = –9.9 × 10–20 J / 1.6 × 10–19 C = –0.62 V and this could be checked.
Incorrect wiring: No, the voltmeter would then be reading too high and would have to be corrected
downwards, which would lower the graph further and put the y intercept further from the origin.
Frequency spread: Yes, the threshold p.d. to turn on the diode could relate to a lower frequency
than the peak emission frequency. This would lower the value of the frequencies at turn on, and
shift the graph to the left closer to intercepting the origin. The graph would need to shift left by the
value of the x-axis intercept, given by the x-intercept ( when y = 0 ) in y = 7.3 × 10–34 x – 9.9 ×
10-20 so that x = 9.9 × 10-20 J / 7.3 × 10–34 J s = 1.4 × 1014 Hz. This is about 40% of the average
infrared frequency, 30% of the red light frequency and 20% of the blue light frequency.
Non-linear potential divider: No, all that is required of the divider is that it varies the p.d., which is
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then measured by the voltmeter. A non-linear divider might be less convenient to use to set the
threshold voltage, but should not have a systematic effect on the results.
Open the Excel Worksheet
Photons streaming from a lamp
Question 20E: Estimate
1. P = 40 W
2.  = 5  10–7 m
3. Calculate the frequency of the photons corresponding to this wavelength:
c
f 


3  10 8 m s
5  10  7 m
 6  10 14 Hz .
Now calculate the energy of each photon:
E  hf
 6  10  34 J s 1  6  10 14 Hz
 4  10 19 J.
4. Energy per second = 40 J s–1
Energy per photon = 4  10–19 J.
energy per second
photons per second 
energy per photon

40 J s -1
4  10 19 J
 1 10 20.
Spacing a grating
Question 60X: Explanation–Exposition
1. A good answer would explain the links between the geometry, differences in trip times, and the
arrows from adjacent paths lining up.
2. A good answer would relate the differences in trip times to the changes in slit separation, noting
that there must be more than one extra rotation between the paths; two extra rotations is a
possibility.
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CD:
Mirror and grating?
Question 70X: Explanation–Exposition
1. A good answer would explain the conditions for lots of green photons to be found in one place,
where no red photons are likely to be found and vice versa. This explanation hangs on the
differences in trip times between adjacent paths being the time for one (or two, for second order
etc) complete rotation of the arrow for one frequency, but not for another. This relies on the
geometry, since all paths are explored at the same rate. Conditions for places where lots of
photons and a few photons are found should both be covered.
2. A good answer would build on question 1, using the very much shorter differences in trip times to
suggest that significant differences in the alignment of the red and green arrows will not arise.
Thus the amplitudes for photons of all colours to appear at any place is approximately constant.
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