AS Unit 1 –January 2010 –Solutions and feedback
Q1
(a)
(b)
(c)
Hadrons are a group of particles composed of quarks. Hadrons can either be baryons or mesons.
(i) What property defines a hadron?
particles that experience the strong (nuclear) force/interaction(1 mark)
(ii) What is the quark structure of a baryon?
particles composed of three quarks(1 mark)
(iii) What is the quark structure of a meson?
particles composed of a quark and an antiquark (1 mark)
State one similarity and one difference between a particle and its antiparticle.
similarity: the same (rest) mass or rest energy(1 mark)
difference: opposite quantum states eg charge(1 mark)
Complete the table below which lists properties of the antiproton.
antiproton
charge/C
-19
-1.6 × 10
baryon number
-1
it is the antiparticle
NOT -1 !!!!
(d)
quark structure
-1 for each mistake (2 marks)
The K- is an example of a meson with strangeness -1. The K- decays in the following way:
(i)
State, with a reason, what interaction is responsible for this decay.
The weak interaction (1 mark).
This can be deduced because strangeness is not conserved or there is a change/decay of quark (flavour) (1 mark)
(ii) State two properties, other than energy and momentum, that are conserved in this decay.
any two eg charge; baryon number; (muon) lepton number (1 mark for each)
(11 marks total)
Q2
(a)
A fluorescent tube is filled with mercury vapour at low pressure. In order to emit electromagnetic
radiation the mercury atoms must first be excited.
(i) What is meant by an excited atom?
An atom is excited when an electron is at a higher level than the ground state or an electron
has jumped/moved up to higher energy level (1 mark)
(ii) Describe the process by which mercury atoms become excited in a fluorescent tube.
The tube is filled with mercury vapour at low pressure –mercury is a metal –free electrons –these are accelerated
when a p.d. is applied to the tube.
Electrons (or an electric current) flow through the tube (1 mark)
and collide with orbiting atomic electrons within the mercury atoms (1 mark)
givin their energy to them, raising the electrons to a higher level (within the mercury atoms) (1 mark)
(3 marks)
(iii) What is the purpose of the coating on the inside surface of the glass in a fluorescent tube?
Photons emitted from mercury atoms as they de-excite are part of the ultra violet (spectrum) or
are UV photons (1 mark)
These UV photons are absorbed by electrons in the atoms of the powder (1 mark)
and the powder emits photons in the visible spectrum as the atoms de-excite (1 mark)
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LOJ 2010
AS Unit 1 –January 2010 –Solutions and feedback
The visible photons have a variety of different wavelengths and these give white light from the tube (1 mark)
(3 marks MAX)
(b)
The lowest energy levels of a mercury atom are shown in the diagram below. The diagram is not to scale.
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Interval 3 to 2 looks the smallest but do the math to be sure it is!
(i) Calculate the frequency of an emitted photon due to the transition level n = 4 to level n = 3.
E = (-0.26 x 10-18) –(-0.59 x 10-18) = 0.33 x 10-18J (1 mark)
Always look carefully at the diagram –many missed out the 10-18
E = hf = 6.63 × 10-34 × f (1 mark) = 0.33 x 10-18J
f = 0.33 × 10-18/(6.63 × 10-34) = 5.0 × 1014 (1 mark)
answer = 5.0 × 1014 Hz
(3 marks)
(ii) Draw an arrow on the diagram above to show a transition which emits a photon of a longer wavelength
than that emitted in the transition from level n = 4 to level n = 3.
For an emitted photon the electron must be falling down the levels –towards the ground state
A longer wavelength means a shorter frequency –a lower energy. Therefore the jump must be smaller!
Arrow from 3 to 2 (1 mark for the interval) (1 mark for the direction)
(2 marks)
(12 marks total)
Q3.
(a)
An unstable nucleus, A X, can decay by emitting a particle,
Z
What part of the atom is the same as a particle?
An (orbital) electron (1 mark)
(ii) State the changes, if any, in A and Z when X decays.
change in A: A is the mass number –in beta decay a neutron changes to a proton –so there is no change to A
change in Z: Z is the atomic number (the proton number) –a neutron changes to a proton so Z increases by 1
(i)
(b)
In the process of decay an anti-neutrino is also released,
(i) Give an equation for this decay.
(2 marks)
The neutrino is an electron anti-neutrino –don’
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!
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LOJ 2010
AS Unit 1 –January 2010 –Solutions and feedback
(1 mark)
State and explain which conservation law may be used to show that it is an anti-neutrino rather
than a neutrino that is released.
Here they already told you a lepton was in question. May of you checked out charge and baryons too –
that was unnecessary!
Lepton number must be conserved (1 mark)
Total lepton number before decay equals zero so after decay you need a particle with a negative lepton
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onnumberof+
1–therefore you have to have an anti-neutrino
(1 mark)
(2 marks)
(iii) What must be done to validate the predictions of an unconfirmed scientific theory?
The hypothesis needs to be tested by experiment (1 mark).The experiment must be repeated by other
scientists so that the hypothesis can be confirmed as valid or rejected by the scientific community. (1
mark)
(2 marks)
(8 marks total)
(ii)
Q4.
(a)
Experiments based on the photoelectric effect support the particle nature of light. In such experiments
light is directed at a metal surface.
(i) State what is meant by the threshold frequency of the incident light.
The threshold frequency is the minimum frequency of the incident light required for electrons
to be emitted from a metal surface by the photoelectric effect (1 mark) Light of lower
frequencies will not release electrons no matter how great the intensity of it.
(1 mark)
(ii) Explain why the photoelectric effect is not observed below the threshold frequency.
The energy of a photon depends on its frequency (E = hf). (1 mark) Photons of the threshold
frequency have the minimum energy required to overcome the work function of the metal (1
mark)
(2 marks)
(b) Monochromatic light of wavelength 5.40 x 10-7 m is incident on a metal surface which has a work function
of 1.40 x 10-19J.
(i) Calculate the energy of a single photon of this light.
E = hf = hc/λ
E = 6.63 × 10-34 × 3.00 × 108/5.40 × 10-7 (1 mark)
answer = 3.68 × 10-19 (1 mark) J
(2 marks)
(ii) Calculate the maximum kinetic energy of an electron emitted from the surface.
The photon gives all of its enegy to the electron. The electron uses part of that to free itself and the rest becomes
kinetic energy
hf = Ek + 
-19
3.68 × 10 = Ek + 1.40 x 10-19J (1 mark)
answer = 2.28 x 10-19 (1 mark) J
(2 marks)
(iii) Calculate the maximum speed of the emitted electron.
An electron is matter not a photon.
Ek = ½ mv2
2.28 x 10-19 = ½ × 9.11 × 10-31 × v2(1 mark)
v2 = 2 × 2.28 × 10-19/9.11 × 10-31 = 5.0 × 1011
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AS Unit 1 –January 2010 –Solutions and feedback
v = 7.1 × 105
answer - 7.1 × 105 (1 mark) ms-1
(iv)
Calculate the de Broglie wavelength of the fastest electrons.
dB = h/p = h/mv
-34
= 6.63 × 10 (9.11 × 10-31 × 7.1 × 105) (1 mark)
answer = 1.03 × 10-9 (1 mark) m
(2 marks)
(2 marks)
(11 marks total)
Q5.
(a) A sample of conducting putty is rolled into a cylinder which is 6.0 x 10 -2m (6.0 cm) long and has a radius
of 1.2 x 10-2m. (1.2cm)
resistivity of the putty = 4.0 x 10-3 m.
(i) Calculate the resistance between the ends of the cylinder of conducting putty. Your answer should
be given to an appropriate number of significant figures.
In this question ALL figures are given to 2sf –therefore you have to give the answer to 2sf. There is no mark for sf
unless you are spot on!
L = 6.0 x 10 -2m (1 mark)
r = 1.2 x 10-2m (1 mark)
R = ρL /A
-3
R = 4.0 × 10 × 0.060 /(π× 0.0122)
2 significant figures (1 mark)
answer = 0.53 (1 mark) 
(4 marks)
(ii) The putty is now reshaped into a cylinder with half the radius and a length which is four times as
great. Determine how many times greater the resistance now is.
Halving the diameter will increase resistance by factor of 4 and increasing the length by a factor of 4 will increase
resistance by factor of 4 (1 mark)
(hence) resistance will be 16 times greater (1 mark)
(2 marks)
(b)
Given the original cylinder of the conducting putty described in part (a), describe how you would use a
voltmeter, ammeter and other standard laboratory equipment to determine a value for the resistivity
of the putty.
Your description should include
•alabelled circuit diagram,
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AS Unit 1 –January 2010 –Solutions and feedback
•details of the measurements you would make,
- length L –using a ruler marked in mm or calipers that measure to 10cm –there are
lots of types of them –most would not be used for more than 4 cm! (but be careful not
to squeeze it when measuring)
- diameter of the cylinder –using calipers (but be careful not to squeeze it when
measuring). A micrometer would not really be suitable as you would lose the edge of its
sensitivity in avoiding squishing the cylinder –but this was accepted by the board!
A series of voltage V and current I readings would be taken using the ammeter to
measure the current and the voltmeter to measure the voltage.
•anaccountofhow you would use your measurements to determine the result,
The diameter would be halved and then inserted into the equation A = r2 to work out
the cross sectional area A of the sample and a graph would be plotted of V against I
to determine the resistance (V = IR). R = V/I –resistance can be read off the graph
once a best fit line had been plotted.
ρ= RA/L would be used to determine the resistivity of the putty.
•det
ai
l
sofhow to improve the precision of your measurements.
Length and diameter would be taken at three orientations of the cylinder and averaged.
Current and potential difference would be taken over a range of values –at least 8
readings would be taken and a graph would be plotted of V against I to determine the
resistance of the putty. This would help eliminate anomalies.
You would need to mention how precision could be improved, eg use of vernier calipers
rather than a ruler or full scale readings on the meters for V and I
The use of flat metal electrodes at each end of the cylinder would improve electrical
connection
Or you could vary the length of the putty –butt
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mean big errors in the length measurements.
The quality of your written communication will be assessed in this question.
(8 marks)
The mark scheme for this part of the question includes an overall assessment for the Quality of Written
Communication
The circuit diagram must include:
 voltmeter and ammeter connected correctly - the voltmeter in parallel with the putty, the
ammeter in series with it (1 mark)
 power supply with means of varying current –either a battery with a rheostat or a potential divider
set up or a variable lab pack (1 mark)
The other six marks went on your communication skills:
good –excellent (5/6 marks)
(i) Uses accurately appropriate grammar, spelling, punctuation and legibility.
(ii) Uses the most appropriate form and style of writing to give an explanation or to present an argument in a
well structured piece of extended writing [may include bullet points and/or formulae or equations].
An excellent candidate will have a working circuit diagram with correct description of measurements
(including range of results) and processing. An excellent candidate uses a range of results and finds a mean
value or uses a graphical method, eg I-V characteristics. They also mention precision eg use of vernier
callipers.
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AS Unit 1 –January 2010 –Solutions and feedback
modest –adequate (3/4 marks)
(i) Only a few errors.
(ii) Some structure to answer, style acceptable, arguments or explanations partially supported by evidence or
examples.
An adequate candidate will have a working circuit and a description with only a few errors, eg do not consider
precision. They have not taken a range of results or fail to realise that the diameter needs to be measured
in several places.
poor –limited (1/2 marks)
(i) Several significant errors.
(ii) Answer lacking structure, arguments not supported by evidence and contains limited information.
Several significant errors, eg important measurement missed, incorrect circuit, no awareness of how to
calculate resistivity.
incorrect, inappropriate or no response (0 marks)
(14 marks total)
Q6.
The circuit below contains four identical new cells, A, B, C and D, each of emf 1.5V and negligible internal
resistance.
(a)
The resistance of each resistor is 4.0.
(i) Calculate the total resistance of the circuit.
Parallel arrangement = 2.0…t
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ha4resistor so the total is 6.0
(1 mark)
(ii)
Calculate the total emf of the combination of cells.
A and B have combined EMF of 1.5V –EMF of a single cell
The parallel arrangement is in series with C and D –cells in series have EMFs added to make a sum of their values = 1.5
+ 1.5 + 1.5 = 4.5V
answer = 4.5 V (1 mark)
(iii) Calculate the current passing through cell A.
For total circuit: =IR(
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I = 4.5/6.0 = 0.75A (1 mark)
This splits at the parallel cell arrangement so cell A gets 0.375A
answer = 0.38 A (1 mark) (2 marks)
(iv) Calculate the charge passing through cell A in five minutes, stating an appropriate unit.
5 minutes = 5 x 60 = 300 s
charge = Qt = 0.375 × 300
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AS Unit 1 –January 2010 –Solutions and feedback
(b)
answer = 112 (1 mark) C (1 mark) (2 marks)
Each of the cells can provide the same amount of electrical energy before going flat. State and explain
which two cells in this circuit you would expect to go flat first.
Cells C and D will go flat first (1 mark)
The current (or charge per second) passing through cells C and D is double/more than that passing
through A or B (1 mark).
E/t = P = I2R so the energy given to charge passing through cells per second is double or more
than in cells C and D (1 mark) or in terms of power (3 marks)
(14 marks total)
Q7.
An alternating current (ac) source is connected to a resistor to form a complete circuit. The trace obtained on an
oscilloscope connected across the resistor is shown below.
The oscilloscope settings are: Y gain 5.0V per division
time base 2.0ms per division.
(i) Calculate the peak voltage of the ac source.
answer = 10.0 V (1 mark)
(ii) Calculate the rms voltage.
RMS = Peak /√2 =10.0/√2 = 7.1
answer = 7.1 V (1 mark)
(iii) Calculate the time period of the ac signal.
answer = 6.0 ms (1 mark)
(iv) Calculate the frequency of the ac signal.
for this T must be in seconds! (1 mark)
f = 1/T = 1/0.0060
answer = 167 (1 mark) Hz (2 marks)
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(14 marks total)
LOJ 2010