Year 12 Mixed Exam Questions ABP - Hinchley Wood School

Year 12 – Mixed exam questions – ABP - Hinchley Wood School

Q1.(a)Name the constituent of an atom which

(i)has zero charge,

......

(1)

(ii)has the largest specific charge,

......

(1)

(iii)when removed leaves a different isotope of the element.

(1)

......

(b) The equation

represents the decay of technetium−99 by the emission of a β− particle.

(i)Identify the particle X.

......

(1)

(ii)Determine the values of A and Z.

A =......

Z = ......

(2)

(iii)Calculate the specific charge of the technetium−99 nucleus. State an appropriate unit for your answer.

specific charge = ...... unit ......

(4)

(Total 10 marks)

Q2. (a) (i) Determine the charge, in C, of a nucleus.

......

(ii) A positive ion with a nucleus has a charge of 4.80 × 10–19 C.
Determine how many electrons are in this ion.

......

(4)

(b) A nucleus may decay by emitting two β– particles to form a plutonium nucleus . State what X and Y represent and give the numerical value of each.

X ...... …...... ………….

......

Y ...... …...... ………….

......

(4)

(Total 8 marks)

Q3. (a) (i) Name a force which acts between an up quark, u, and an electron. Explain, with reference to an exchange particle, how this force operates.

You may be awarded marks for the quality of written communication in your answer.

......

(ii) With what particle must a proton collide to be annihilated?

......

(4)

(b) A sigma plus particle, Σ+, is a baryon.

(i) How many quarks does the Σ+ contain?

......

(ii) If one of these quarks is an s quark, by what interaction will it decay?

......

(iii) Which baryon will the Σ+ eventually decay into?

......

(3)

(Total 7marks)

Q4. (a) Give the number of nucleons and the number of electrons in an atom of Na.

nucleons ......

electrons ......

(2)

(b) The isotope Na is a positron emitter. In positron emission an up quark undergoes the following change,

u → d + β+ +ve.

Show that charge, lepton number and baryon number are conserved in this decay.

charge ......

lepton number ......

baryon number ......

(3)

......

(2)

(Total 7 marks)

Q5. (a) (i) Explain the meaning of the term work function of a metal.

......

(ii) State what you would need to change in an experiment to investigate the effect of the work function on the photoelectric effect.

......

(3)

(b) Experiments based on the photoelectric effect support the particle theory of light. State one conclusion drawn from these experiments and explain how it supports the particle theory.

You may be awarded marks for the quality of written communication in your answer.

......

(2)

(c) Monochromatic light of wavelength 4.80 × 10–7 m falls onto a metal surface which has a work function of 1.20 × 10–19 J.

Calculate

(i) the energy, in J, of a single photon of this light,

......

(ii) the maximum kinetic energy, in J, of an electron emitted from the surface.

......

(5)

(Total 10 marks)

Q6. (a) Line spectra were observed before they could be explained by theory. We now know that photons of characteristic frequency are emitted when the vapour of an element is bombarded by energetic electrons. The spectrum of the light emitted contains lines, each of a definite wavelength.

Explain how

• the bombarding electrons cause the atoms of the vapour to emit photons

• the existence of a spectrum consisting of lines of a definite frequency supports the view that atoms have discrete energy levels.

The quality of your written communication will be assessed in this question.

......

(6)

(b) The ionisation energy of a hydrogen atom is 13.6eV.

(i) State what is meant by the ionisation energy of hydrogen.

......

(2)

(ii) Express the ionisation energy of hydrogen in joules, giving your answer to an appropriate number of significant figures.

answer = ...... J

(3)

(Total 11 marks)

Q7. (a) When determining the Young modulus for the material of a wire, a tensile stress is appliedtothe wire and the tensile strain is measured.

(i) State the meaning of

tensile stress ......

......

tensile strain ......

......

(ii) Define the Young modulus ......

......

(3)

(b) The diagram below shows two wires, one made of steel and the other of brass, firmly clamped together at their ends. The wires have the same unstretched length and the same cross-sectional area.
One of the clamped ends is fixed to a horizontal support and a mass M is suspended from theother end, so that the wires hang vertically.

(i) Since the wires are clamped together the extension of each wire will be the same.
If ES is the Young modulus for steel and EB the Young modulus for brass, show that

where FS and FB are the respective forces in the steel and brass wire.

......

(ii) The mass M produces a total force of 15 N. Show that the magnitude of the force
FS = 10 N.

the Young modulus for steel = 2.0 × 1011 Pa
the Young modulus for brass = 1.0 × 1011 Pa

......

(iii) The cross-sectional area of each wire is 1.4 × 10–6 m2 and the unstretched length is 1.5 m. Determine the extension produced in either wire.

......

(6)

(Total 9 marks)

Q8. (a) The graph shows the variation of tensile stress with tensile strain for two wires X and Y, having the same dimensions, but made of different materials. The materials fracture at the points FX and FY respectively.

You may be awarded marks for the quality of written communication provided in your answer to the following questions.

State, with a reason for each, which material, X or Y,

(i) obeys Hooke’s law up to the point of fracture,

......

(ii) is the weaker material,

......

(iii) is ductile,

......

(iv) has the greater elastic strain energy for a given tensile stress.

......

(8)

(b) An elastic cord of unstretched length 160 mm has a cross-sectional area of 0.64 mm2. The cord is stretched to a length of 190 mm. Assume that Hooke’s law is obeyed for this range and that the cross-sectional area remains constant.

the Young modulus for the material of the cord = 2.0 × 107 Pa

(i) Calculate the tension in the cord at this extension.

......

(ii) Calculate the energy stored in the cord at this extension.

......

(5)

(Total 13 marks)

Q9. An aerial system consists of a horizontal copper wire of length 38 m supported between two masts, as shown in the figure below. The wire transmits electromagnetic waves when an alternating potential is applied to it at one end.

(a) The wavelength of the radiation transmitted from the wire is twice the length of the copper wire. Calculate the frequency of the transmitted radiation.

......

(1)

(b) The ends of the copper wire are fixed to masts of height 12.0 m. The masts are held in a vertical position by cables, labelled P and Q, as shown in the figure above.

(i)P has a length of 14.0 m and the tension in it is 110 N. Calculate the tension in the copper wire.

......

(ii) The copper wire has a diameter of 4.0 mm. Calculate the stress in the copper wire.

......

(iii) Discuss whether the wire is in danger of breaking if it is stretched further due to movement of the top of the masts in strong winds.

breaking stress of copper = 3.0 × 108 Pa

......

(7)

(Total 8 marks)

Q10.(a) The diagram below shows an arrangement used to investigate the properties of microwaves.

When the transmitter T was rotated through 90° about the straight line XY, the receiver signal decreased to zero. Explain why this happened and state the property of microwaves responsible for this effect.

You may be awarded marks for the quality of written communication in your answer.

......

(3)

(b) A microwave oven produces microwaves of wavelength 0.12 m in air.

(i) Calculate the frequency of these microwaves.

......

(ii) In a certain oven, explain why food heated in a fixed position in this oven would be uncooked at certain points if stationary waves were allowed to form.

......

(3)

(Total 6 marks)

Q11. Just over two hundred years ago Thomas Young demonstrated the interference of light by illuminating two closely spaced narrow slits with light from a single light source.

(a) What did this suggest to Young about the nature of light?

......

(1)

(b) The demonstration can be carried out more conveniently with a laser. A laser produces coherent, monochromatic light.

(i) State what is meant by monochromatic.

......

(ii) State what is meant by coherent.

......

(2)

(iii) State one safety precaution that should be taken while using a laser.

......

(1)

(c) The diagram belowshows the maxima of a two slit interference pattern produced on a screen when a laser was used as a monochromatic light source.

The slit spacing = 0.30 mm.
The distance from the slits to the screen = 10.0 m.

Use the diagram aboveto calculate the wavelength of the light that produced the pattern.

answer = ...... m

(3)

(d) The laser is replaced by another laser emitting visible light with a shorter wavelength.
State and explain how this will affect the spacing of the maxima on the screen.

......

(2)

(Total 9 marks)

Q12. The diagram below shows a cross-section through a step index optical fibre.

(a) (i) Name the parts A and B of the fibre.

A
B

(1)

(ii) On the diagram above, draw the path of the ray of light through the fibre.
Assume the light ray undergoes total internal reflection at the boundary between Aand B.

(2)

(b) Calculate the critical angle for the boundary between A and B.
Give your answer to an appropriate number of significant figures.

The refractive index of part A = 1.46
The refractive index of part B = 1.48

answer = ...... degrees

(2)

......

(2)

(d) State one application of optical fibres and explain how this has benefited society.

Application

......

Benefit

......

(2)

(Total 9 marks)

Q13. A cyclist pedals downhill on a road, as shown in the diagram below, from rest at the top of the hill and reaches a horizontal section of the road at a speed of 16 m s–1. The total mass of the cyclist and the cycle is 68 kg.

(a) (i) Calculate the total kinetic energy of the cyclist and the cycle on reaching the horizontal section of the road.

answer ...... J

(2)

(ii) The height difference between the top of the hill and the horizontal section of road is 12 m.
Calculate the loss of gravitational potential energy of the cyclist and the cycle.

answer ...... J

(2)

(iii) The work done by the cyclist when pedalling downhill is 2400 J. Account for the difference between the loss of gravitational potential energy and the gain of kinetic energy of the cyclist and the cycle.

......

(3)

(b) The cyclist stops pedalling on reaching the horizontal section of the road and slows to a standstill 160 m further along this section of the road. Assume the deceleration is uniform.

(i) Calculate the time taken by the cyclist to travel this distance.

answer...... s

(3)

(ii) Calculate the average horizontal force on the cyclist and the cycle during this time.

answer ...... N

(3)

(Total 13 marks)

Q14. 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)

(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)

M1.(a) (i)neutron

accept symbols
symbols e.g. n

(ii)electron

accept symbols

(iii)neutron

accept symbols

(b) (i)antineutrino

V(e)

(ii)A=99
Z= 44

(iii)specific charge = 43 × 1.6 × 10−19 / 99 × 1.66 × 10−27
specific charge = 4.2 × 107C kg−1

Correct answer no working −1
If include mass of electrons lose 2 and 3 mark

[10]

M2. (a) (i) (charge) = 92 × 1.60 × 10–19
= 1.47 × 10–17 (C) (1)

(ii) (magnitude of ion charge) = 3(e) (1)
number of electrons (= 92 – 3) = 89 (1)

(b) X: number of nucleons [or number of neutrons plus protons or
mass number] (1)
239 (1)
Y: number of protons [or atomic number] (1)
94 (1)

[8]

M3. (a) (i) (named force) from weak (nuclear), electromagnetic or gravity (1)
uses a mediating/exchange particle, named particle from W(±) (boson),
() photon or graviton (1)
to transfer energy/momentum (1)
when electron emits/receives exchange particle,
disappearance/creation of new particle occurs (1)

QWC 1

(ii) anti proton (1)

max 4

(b) (i) 3 (quarks) (1)

(ii) weak (nuclear) (1)

(iii) proton (1)

[7]

M4. (a) 22 (nucleons) (1)
11 (electrons) (1)

(b) charge: += –+ 1 + 0 (1)

lepton number: 0 = 0 – 1 + 1 (1)

baryon number: = + 0 + 0 (1)

(c) the electron and the positron are annihilated (1)
photon(s)/ ray(s) are produced (1)
specifying two () photons/rays (1)
masses converted into energy (1)

max 2

[7]

M5. (a) (i) the minimum energy (1)
energy required to eject a (photo)electron
(from the metal surface) (1)

(ii) changing the metal/cathode (1)

(b) conclusion: light below athreshold frequency does not
release electrons (1)
explanation: photons carry quanta of energy (1)

[or conclusion: electrons are emitted immediately the
light hits the metal surface (1)
explanation: photons carry quanta of energy (1)]

4.14 × 10–19 (J)

(ii)hf = + Ek(1)
Ek = 4.14 × 10–19 – 1.20 × 10–19(1)
= 2.94 × 10–19 (J) (1)
(allow C.E for value of E from (i))

[10]

M6. (a)The candidate’s writing should be legible and the spelling,
punctuation and grammar should be sufficiently accurate
for the meaning to be clear.

The candidate’s answer will be assessed holistically. The
answer will be assigned to one of three levels according to the
following criteria.

High Level (Good to excellent): 5 or 6 marks

The information conveyed by the answer is clearly organised,
logical and coherent, using appropriate specialist vocabulary
correctly. The form and style of writing is appropriate to answer
the question.

The candidate provides a comprehensive and coherent
description which includes a clear explanation of constant
energy level differences and how electrons can be excited
by electron collisions. The link between the energy of a photon
and its frequency should be clear. The description should
include a clear explanation of the reason atoms of a given
element emit photons of a characteristic frequency or there
is a clear link between constant energy differences and photon
frequency/wavelength (eg E=hf).
The candidate should relate the energy difference between
levels to the energy of emitted photons and state the energy
difference is fixed/constant.

Intermediate Level (Modest to adequate): 3 or 4 marks

The information conveyed by the answer may be less well
organised and not fully coherent. There is less use of specialist
vocabulary, or specialist vocabulary may be used incorrectly.
The form and style of writing is less appropriate.

The candidate provides an explanation of energy levels and
how excitation takes place by electron collision with
atomic/orbital electrons. The candidate explains how an
orbital/atomic electron loses energy by emitting a photon.

Low Level (Poor to limited): 1 or 2 marks

The information conveyed by the answer is poorly organised
and may not be relevant or coherent. There is little correct
use of specialist vocabulary.

The form and style of writing may be only partly appropriate.
Some mention of energy levels and the idea of excitation of
electron. Talk about excitation of atom instead of electron limits
the mark to 1.

Incorrect, inappropriate of no response: 0 marks

No answer or answer refers to unrelated, incorrect or inappropriate physics.

The explanation expected in a competent answer should include a
coherent account of the significance of discrete energy levels
and how the bombardment of atoms by electrons can lead to
excitation and the subsequent emission of photons of a characteristic
frequency.

electrons bombard atoms of vapour and give energy to electrons in atom

electrons move to a higher energy level

electrons are excited

excited electrons move down to lower energy levels losing energy by
emitting photons

photons have energy hf

photons of characteristic frequencies emitted from atoms of a
particular element

this is because atoms have discrete energy levels which are
associated with particular energy values

max 6

(b) (i) energy required to (completely) remove an electron
from atom/hydrogen

ground state/lowest energy level

(ii) 13.6 × 1.6 × 10–19= 2.18 × 10–18 (J) 3 sfs

[11]

M7. (a) tensile stress: force/tension per unit cross-sectional area or

with F and A defined (1)
tensile strain: extension per unit length or with e and l defined (1)

the Young modulus: (1)

(b) (i)

(ii)= 2 (1)

F = 2FB (1)

FS + FB = 15 N (1) gives FS = 10 N

[or any alternative method]

(iii)e = = (1)

= 5.36 ×10–5m (1)

[9]

M8. (a) (i) X (1)
stress (force) strain (extension) for the whole length (1)

(ii) Y (1)
has lower breaking stress (or force/unit area is less) (1)

(iii) Y (1)
exhibits plastic behaviour (1)

(iv) Y (1)
for given stress, Y has greater extension
[or greater area under graph] (1)

QWC 2

(b) (i) (use of E =gives)

=(1)

(1) for data into correct equation, (1) for correct area

= 2.4 N (1)

(allow C.E. for incorrect area conversion)

(ii) (use of energy stored = ½Fe gives) energy = (1)

= 36 × 10–3 J (1)

(allow C.E. for value of F from (i))

[13]

M9. (a)λ(=2 × 38) = 76(m)

MHz (1)

(b) (i) angle between cable and horizontal = (1)

T= 110 cos59° = 57N • (56.7N) (1)

(allow C.E. for value of angle)

(ii) cross-sectional area (= (2.0 × 10–3)2)

=1.3 × 10–5(m2) (1)

(1.26 × 10–5(m2))

stress (1)

= 4.4 × 106Pa (1)

(4.38 × 106Pa)
(use of 56.7 and 1.26 gives 4.5 × 106 Pa)
(allow C.E. for values of T and area)

(iii) breaking stress is 65 × stress
copper is ductile
copper wire could extend much more before breaking
because of plastic deformation
extension to breaking point unlikely

any three (1)(1)(1)

[8]

M10. (a) microwaves from transmitter are polarised
[or vibrate in certain plane or direction] (1)
rotating transmitter through 90° rotates plane of
vibration/polarisation of the microwaves (1)
receiver signal becomes zero when receiver is perpendicular
to plane/direction of vibration/polarisation of the microwaves (1)

QWC 2

(b) (i) (use of c = fλ gives) = 2.5 × 109 Hz (1)

(ii) no energy/amplitude/intensity/vibrations at nodes (1)
food at nodes would not be heated (1)

[6]

M11. (a) showed that light was a wave (rather than a particle)/wave nature
(of light) (1)

(b) (i) single wavelength (or frequency) (1)

(ii) (waves/source(s) have) constant phase difference (1)

(iii) any sensible precaution, eg do not look into laser/do not point
the laser at others/do not let (regular) reflections enter the
eye/safety signs/suitable safety goggles (1)

= (1) ecf from calculation of fringe spacing

= 6.0 × 10–7 m (1) (= 600 nm) ecf from calculation of fringe spacing

(d) maxima closer together (1)

(quotes equation and states that) spacing is proportional to wavelength/
D and s are constant therefore as λ decreases so ω decreases (1)

or links smaller wavelength to smaller path difference (1)

[9]

M12. (a) (i) A: cladding + B: core (1)

(ii)

refraction towards the normal line (1)

continuous lines + strikes boundary + TIR correct angles by
eye + maximum 2 TIRs (1)

(b)or = 0.9865 (1)

80.6 or 80.8 or 81 (°) only (1)

(which would cause) light travelling at different angles to arrive at
different times/pulse broadening/merging of adjacent pulses/’smearing’/
poor resolution/lower transmission rate/lower bandwidth/less distance
between regenerators (1)

or to prevent light/data/signal loss (from core or fibre) (1)

(which would cause) signal to get weaker/attenuation/crossover/data
to be less secure (1)

(d) correct application (1) (endoscope, cytoscope, arthroscope etc,
communications etc)

linked significant benefit stated eg improve medical diagnosis/improve
transmission of data/high speed internet (1)

[9]

M13. (a) (i) (EK = ½ mv2 =) 0.5 × 68 × 162 (1) = 8700 or 8704(J) (1)

(ii) (ΔEP = mgΔh =) 68 × 9.8(1) × 12 (1) = 8000 or 8005 (J) (1)

(iii) any three from

gain of kinetic energy loss of potential energy (1)

(because) cyclist does work (1)

energy is wasted (on the cyclist and cycle) due to air resistance
or friction or transferred to thermal/heat (1)

KE = GPE + W – energy ‘loss’ (1) (owtte)

energy wasted (= 8000 + 2400 - 8700) = 1700(J) (1)

(b) (i) (u = 16 m s–1, s = 160 m, v = 0, rearranging s = ½ (u+v) t gives)

160 = ½ × 16 × t or t = or correct alternative