Chapter 5: Particle model of light
solutions
1.
[2 marks]
(2001 Q4)
The work function is given by either
W = hf0 or by extrapolating the graph back
to the vertical axis.
3.
[3 marks]
(VCE 2002 Q6)
Using KEmax = hf – W.
The cut off potential will correspond
to the energy of the electron, if it is
expressed in eV. W is given by hf0.
(Graph property of straight lines)
V0 = hf - hf0
= h(f - f0 )
= 4.14 × 10-15 (6.25× 1014 - 5.50 × 10 14 )
= 0.31V (ANS)
4.
[2 marks]
(VCE 2002 Q7)
current
Using W = hf0 gives
W =6.63  10-34  5.5  1014
= 3.65  10-19
= 3.7  10-19J (ANS)
Blue
Ultraviolet
Extrapolating the graph, gives an answer of
–3.7 or –3.8.
This would give a work function of either
3.7  10-19J or 3.8  10-19J. As long as you
had shown your working, by drawing the
extrapolation on the graph, either would
have been correct. But you do need to be
careful when doing this.
The examiners accepted values in the range
of (3.5 – 3.9)  10-19J.
2.
[2 marks]
(2001 Q5)
The cut-off frequency, from graph, was
5.5  1014 Hz.
The wavelength is given by c = f.
Where c = 3.0 108.
c
So  =
f
3× 108
=
5.5× 1014
= 5.45  10-7
 545 nm
 Sodium (ANS)
potential
difference
The violet line has:
a lower plateau corresponding to the
lower intensity
a larger Vo value due to the higher
energy of each photon
5.
[2 marks]
(2003 Q5)
You need to be very careful reading the
units on this graph. The vertical axis is in
Volts, not the expected “Units of Energy”.
To convert Volts to ‘Energy’ you need to
multiply by ‘e’, the charge on an electron.
The work function (or binding energy) is
the ‘vertical intercept’.
Extrapolating the graph gives a value of
1.9V.
This becomes 1.9eV.
As this is slightly imprecise, the examiners
would accept values in the range of
1.6  2.2 eV (ANS)
6.
[3 marks]
(2003 Q6)
The maximum KE that the photoelectron
can have is the initial energy of the light
minus the work function.
 hf – W = KE
 4.1  10-15  1.93  1016 – 1.9
= 77.2eV (ANS)
7.
The results for Red and Yellow cannot be
shown on this graph. The stopping voltage
of 0 shows that photo-electrons were not
emitted.
[2 marks]
11.
[3 marks]
(2004 Sample Q6)
Planck’s constant is the gradient of the line.
(2004 Q1)
hc
λ
4.14 × 10 -15 × 3.00 × 108
=
550 × 10 -9
= 2.26 eV (ANS)
E=
8.
[2 marks]
(2004 Q2)
EK max = hf - W
EK max = photon energy – work function
= 2.26 – 2.10
= 0.16 eV (ANS)
9.
[3 marks]
(2004 Q3)
EK max (joules) = 21 mv2 ,
2.80 × 1.6 × 10-19 = 21 × 9.1× 10-31 v2
 v2 =
1.9 - 0.1
(10 - 5.8)  10 14
1.8
=
4.2  1014
= 4.3  10-15 eVs (ANS)
The gradient =
2 × 2.80 × 1.6 × 10-19
9.1× 10-31
12.
[2 marks]
KEmax = hf – W
KEmax = Ephoton – W
W = Eph – Ekmax
= 3.0 – 0.9
= 2.1 eV (ANS)(for blue, same for UV)
 v = 9.9× 105 ms-1. (ANS)
13.
10.
Eph = hf
= 4.14x10-15 x 5.4x1014
= 2.24 eV (ANS)
[3 marks]
(2004 Sample Q5)
(2004 Pilot Q1)
14.
[2 marks]
[2 marks]
(2004 Pilot Q2)
(2004 Pilot Q3)
Current only depends on intensity, given
that photo-electrons are being emitted.
 A (ANS)
15a [1 mark]
(2004 Pilot Q4a)
6.2
14×1014
= 4.4  10-15 eVs (ANS)
=
The question required you to show
working, so just quoting Planck’s constant
as 4.14  10-15 eV s did not get any marks.
Showing your working means; drawing the
line of best it AND showing your
calculation.
17.
[4 marks]
Graph should show three points giving
straight line from (3.0,-1.0) to (12,3.0)
15b. [2 marks]
Y
(2004 Pilot Q4b)
Y
Red is N/A as no electrons emitted.
UV2 is 2.9 eV (ANS)
16.
[2 marks]
17.
N
N
(2005 Q3)
The work function is given by the
intersection of the graph with the vertical
axis.
The line of best fit, gave ~2.7 V.
 Sodium. (ANS)
[3 marks]
(2005 Q4)
Planck’s constant is the gradient of the line.
Using the line of best fit, and the widest
range of values possible, calculate the rise
over the run.
(2006 Q2)
19.
[2 marks]
(2006 Q3)
The minimum energy required is given by
the intersection of the graph with the y-axis.
 1.8 eV
This is the energy required to release an
electron from the surface of the metal.
The minimum energy required can also be
found by E = hfcut-off.
 E = 4.14 x 10-15 x 4.4 x 1014
= 1.8(2) eV (ANS)
20.
[2 marks]
(2006 Q4)
The gradient of the lines had to be the same.
They both give Planck’s constant. The work
function for silver is higher, then the graph
for silver needs a more negative ‘y-intersect’
value.
Alternatively it will take more energy to
release an electron, and so the cut-off
frequency will be higher.
 B (ANS)
The gradient is
3.5 - (-2.7)
14×1014 - 0
21.
[2 marks]
(2007 Q1)
The minimum energy required to remove
photoelectrons (work function) is given by
either W = hf0
or by extrapolating to the vertical axis.
23.
[3 marks]
(2007 Q3)
The wave model predicts that if you
increase the intensity of the light, then this
increase in energy will emit photoelectrons.
This is not the case.
Einstein’s equation for the photoelectric
effect
KE = hf – W
predicts that as you increase the frequency
(not the intensity) of the incident photons
then the emitted photoelectrons will have
more energy. This is a prediction from the
particle-like theory of light.
frequency  1014 Hz
Using W = hf0 gives
W =4.14x10-15  5.1  1014
= 2.1 eV (ANS)
or vertical intercept
= 2.0 eV (ANS)
22.
[2 marks]
Using f =
[3 marks]
Stopping voltage (V)
24.
(VCE 2007 Q6)
6
7
8
Frequency
×1014 Hz
(2007 Q2)
c
λ
3× 108
400 × 10 -9
= 7.5  1014 Hz
=
25.
frequency  1014 Hz
[3 marks]
(VCE 2007 Q7)
From the gradient of the graph will give
you Planck’s constant in eVs. The gradient
is:
rise
h
run
1.5  0
h
(8  5)  1014
h = 5.0 × 10 -15 eVs (ANS)
From the graph,
KE = 1.0 eV (ANS)
26.
[2 marks]
(VCE 2007 Q8)
Use the graph again and find the y –
intercept, this was around -2.5 Volts. This
means the work function of the metal is:
2.5 eV. (ANS)