A2 Wave Particle Duality

A2 Wave Particle Duality

155

minutes

141

marks

Q1.

The diagram below shows a Transmission Electron Microscope. Electrons from the electron gun pass through a thin sample and then through two magnetic lenses A and B on to a fluorescent screen. An enlarged image of the sample is formed on the screen.

(a) (i) Sketch the path of an electron that reaches point Q on the screen after passing through the sample at point P and through the two magnetic lenses A and B.

(ii) State the function of magnetic lens A and the function of magnetic lens B. magnetic lens A ..................................................................................

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............................................................................................................. magnetic lens B ...................................................................................

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(4)

(b) Explain why greater image detail is seen when the anode voltage is increased.

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(3)

(Total 7 marks)

Q2.

Light from a point source was passed through two closely spaced parallel slits, as shown in the diagram. A pattern of alternate bright and dark fringes was observed on the screen.

(a) Use Huygens’ wave theory of light to explain the formation of these fringes by the double slits. You may be awarded marks for the quality of written communication provided in your answer.

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(4)

(b) (i) Explain what Newton’s theory of light would predict for the same experimental arrangement.

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(ii) Give one reason why Huygens’ wave theory of light did not replace Newton’s theory of light when the fringe pattern due to the double slits was first observed.

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(3)

(Total 7 marks)

Q3.

(a) Describe one piece of evidence that shows that matter has

(i) a wave-like nature,

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(ii) a particle-like nature.

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(b) For a proton of kinetic energy 5.0 MeV,

(i) show that its speed is 3.1 × 10

7

m s –1 ,

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(ii) calculate its de Broglie wavelength.

(3)

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(4)

(Total 7 marks)

Q4.

The diagram below shows the probe tip of a scanning tunnelling microscope (STM) above a metal surface. The probe tip is at a constant negative potential relative to the metal surface.

(a) Explain why electrons can cross the gap between the probe tip and the surface, provided the gap is sufficiently narrow.

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

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(4)

(b) Describe one way in which an STM is used to investigate a surface.

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(3)

(Total 7 marks)

Q5.

(a) The diagram below shows the path followed by a light ray travelling from air into glass.

Use Newton’s theory of light to explain the refraction of the light ray at the air/glass boundary.

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(3)

(b) Newton’s theory of light was eventually abandoned in favour of Huygens’ wave theory which correctly predicted the speed of light in glass in comparison with the speed of light in air.

(i) What did each theory predict about the speed of light in glass in comparison with the speed of light in air?

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(ii) Describe one further piece of evidence that supports Huygens’ wave theory.

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(3)

(Total 6 marks)

Q6.

Hertz discovered how to produce and detect radio waves. He measured the wavelength of radio waves produced at a constant frequency using the arrangement shown in the diagram below.

(i) Explain why the strength of the detector signal varied repeatedly between a minimum and a maximum as the detector was moved slowly away from the transmitter along the dotted line.


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(ii) Hertz found that a minimum was detected each time the detector was moved a further

1.5 m away from the transmitter.

Calculate the frequency of the radio waves.

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(Total 5 marks)

Q7.

In an experiment to demonstrate the wave nature of light, a parallel beam of monochromatic light was directed at two closely spaced slits, as shown in Figure 1 . A pattern of bright and dark fringes due to this light passing through the slits was seen on the screen.

Figure 1

(a) Explain why this fringe pattern was formed.


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(4)

(b) Discuss why this fringe pattern cannot be explained using Newton’s corpuscular theory of light.

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(2)

(Total 6 marks)

Q8.

Photoelectric emission occurs from a certain metal plate when the plate is illuminated by blue light but not by red light.

(a) Explain why photoelectric emission occurs from this plate using blue light but not using red light.

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(4)

(b) Outline why Huygens’ wave theory of light fails to explain the fact that blue light causes photoelectric emission from this plate but red light does not.

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(2)

(Total 6 marks)

Q9.

In a transmission electron microscope, electrons from a heated filament are accelerated through a certain potential difference and then directed in a beam through a thin sample. The electrons scattered by the sample are focused by magnetic lenses onto a fluorescent screen where an image of the sample is formed, as shown in the figure below.

(a) State and explain one reason why it is important that the electrons in the beam have thesame speed.

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(2)

(b) When the potential difference is increased, a more detailed image is seen. Explain why this change happens.

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(3)

(Total 5 marks)

Q10.

In a scanning tunnelling microscope (STM), a metal probe with a sharp tip is scanned across a surface, as shown in the figure below.

(a) Explain why electrons transfer between the tip of the probe and the surface when the gap between the tip and the surface is very narrow and a pd is applied across it.

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(3)

(b) Describe how an STM is used to obtain an image of a surface.

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(3)

(Total 6 marks)

Q11.

(a) Describe, in terms of electric and magnetic fields, the nature of electromagnetic waves travelling in a vacuum. You may wish to draw a labelled diagram.

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(b) Electrons are emitted from a metal plate when monochromatic light is incident on it, provided that the frequency of the light is greater than or equal to a threshold value.

You may be awarded additional marks to those shown in brackets for the quality of written communication in your answer.

(i) How did Einstein explain this effect?

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(3)

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(ii) Discuss the significance of Einstein’s explanation.

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(4)

(Total 7 marks)

Q12.

(a) The discovery of photoelectricity and subsequent investigations led to the wave theory of light being replaced by the photon theory. State one feature of photoelectricity that could not be explained using the wave theory of light and describe how it is explained using photon theory.The quality of your written answer will be assessed in this question.

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(6)

(b) A certain metal has a work function of 2.2 eV.

(i) Explain what is meant by this statement.

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(ii) The surface of the metal is illuminated with light of wavelength 520 nm.

Calculate the maximum kinetic energy of electrons emitted from the surface.

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(5)

(Total 11 marks)

Q13.

The diagram below shows the path followed by a ray light which is incident at non-normal incidence on a glass block in air.

(a) Use Newton’s theory of light to explain the path of the light ray shown in the diagram above.

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(4)

(b) Newton’s theory of light was eventually abandoned by the scientific community in favour of Huygen’s theory of light. State one piece of evidence that supports Huygen’s theory and explain why it supports Huygen’s theory.

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(3)

(Total 7 marks)

Q14.

When light of wavelength 590 nm is directed at an uncharged surface of a certain metal X, electrons are emitted from the metal surface causing a photoelectric current.

(a) When the metal surface is charged positively, the photoelectric current decreases and becomes zero when the potential of the surface is +0.35 V.

(i) Calculate the maximum kinetic energy of a photoelectron emitted from the surface when the metal surface is uncharged. answer = ...................................... J

(2)

(ii) Calculate the work function of the metal surface, in J.

answer = ...................................... J

(3)

(b) When the experiment was repeated using a different metal, Y, illuminated by light of the same wavelength, there was no photoelectric emission when the metal surface was uncharged.

(i) Explain this observation.

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(2)

(ii) How did this observation contribute to the failure of the wave theory of light?

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(2)

(Total 9 marks)

Q15.

(a) Newton suggested a theory that light is composed of corpuscles. He used his theory to explain the refraction of a light ray travelling from air to glass, as shown in Figure 1 .

Huygens explained the refraction of light using his own theory that light consists of waves.

Figure 1

(i) State one reason why Huygens’ theory of light was rejected for many years after itwas first proposed, in favour of Newton’s corpuscular theory of light.

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(ii) Explain why the eventual measurement of the speed of light in water led to thedefinite conclusion that light consists of waves and not corpuscles.

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(b) Young demonstrated that a pattern of alternate bright and dark fringes was observedwhen light from a narrow single slit passed through double slits, as shown in Figure 2 .

Figure 2

(1)

(2)

Newton’s corpuscular theory predicted incorrectly that just two bright fringes would be

formed in this pattern. Use Huygens’ theory of light to explain why more than two bright fringes are formed in this pattern.

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

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(6)

(Total 9 marks)

Q16.

(a) Light has a dual wave-particle nature. State and outline a piece of evidence for the wave nature of light and a piece of evidence for its particle nature. For each piece of evidence, outline a characteristic feature that has been observed or measured and give a short explanation of its relevance to your answer. Details of experiments are not required.

The quality of your written communication will be assessed in your answer.

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(6)

(b) An electron is travelling at a speed of 0.890 c where c is the speed of light in free space.

(i) Show that the electron has a de Broglie wavelength of 1.24 × 10

–12 m.

(2)

(ii) Calculate the energy of a photon of wavelength 1.24 × 10

–12 m.

answer = ...................................... J

(1)

(iii) Calculate the kinetic energy of an electron with a de Broglie wavelength of 1.24 × 10

–

12

m.Give your answer to an appropriate number of significant figures.

answer = ...................................... J

(2)

(Total 11 marks)

Q17.

(a) Describe, in terms of electric and magnetic fields, a plane polarised electromagnetic wave travelling in a vacuum. You may wish to draw a labelled diagram.

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(3)

(b) In his theory of electromagnetic waves, Maxwell predicted that the speed of all electromagnetic waves travelling through free space is given by where μ o

is the permeability of free space and ε o

is the permittivity of free space.

Explain why this prediction led to the conclusion that light waves are electromagnetic waves.

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(c) Hertz discovered how to produce and detect radio waves. The figure below shows a transmitter of radio waves, T , and a detector D . The detector loop and the transmitter aerial are in the same vertical plane.

(2)

(i) Explain why an alternating emf is induced in the loop when it is in this position.

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(3)

(ii) Explain why an alternating emf cannot be detected if the detector loop is turned through 90° about the axis XY .

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(1)

(Total 9 marks)

Q18.

In his investigation of radio waves, Hertz created stationary waves by using a large flat metal sheet to reflect radio waves as shown in the diagram below.

(a) Explain why stationary waves are formed in this arrangement and describe how the wavelength of the radio waves can be determined by moving a suitable detector along XY .

The quality of your written communication will be assessed in your answer.

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(6)

(b) Hertz knew the frequency of the radio waves from the electrical characteristics of the transmitter. He found the wavelength from the investigation described in part (a) and was then able to calculate the speed of the radio waves. Explain the significance of the result of this calculation.

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(2)

(Total 8 marks)

Q19.

(a) State de Broglie’s hypothesis.

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(2)

(b) Neutrons in a narrow beam can be diffracted by crystals thereby exhibiting wave behaviour. Calculate the de Broglie wavelength of a neutron of kinetic energy 0.021 eV.

Give your answer to an appropriate number of significant figures. de Broglie wavelength .......................................... m

(4)

(c) Explain why an electron of the same de Broglie wavelength as the neutron in part (b) has much more kinetic energy than 0.021 eV. Assume relativistic effects are negligible.

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(2)

(Total 8 marks)

M1.

(a) (i) straight paths outside the lenses (1) correct direction of deflection on passing through A (1) path through B correct for path drawn through A (1) for examples

(only one required)

(ii) lens A: magnifies (or forms an intermediate image before B) (1) lens B: magnifies and focuses (or forms an enlarged image on the screen) (1) max 4

(b) increase of voltage causes increase of speed (of the electrons) (1) hence a reduced de

Broglie wavelength (1) less diffraction for reduced wavelength (1) better resolution if less diffraction (1) max 3

M2.

(a) waves/ wavelets are emitted by each slit (1) each slit diffracts light (1) the two slits are coherent emitters / sources of light waves (1) bright fringes formed where light from one slit reinforces light from the other slit

[or dark fringes formed where light from one slit cancels light from the other slit] (1) path difference to a bright fringe = whole number of wavelengths

[or path difference to a dark fringe =

(whole number + half) wavelengths)] (1) max 4

[7]

QWC 2

(b) (i) light consists of corpuscles/particles (1) corpuscles would not be diffracted

[or pass straight through] (1) only two bright fringes would be seen (1)

(ii) Newton’s scientific pre-eminence

[or there was no evidence that light travelled slower in water as predicted by Huygens’ theory]

[or Huygens’ theory considered light waves as longitudinal and therefore could not explain polarisation] (1) max 3

M3.

(a) (i) suitable description and outline detail (1) for an appropriate named particle (1)

(e.g. electron diffraction of a beam of electrons by a thin metal sample or tunnelling in the STM across a gap by electrons)

(ii) suitable description and outline detail (1) for an appropriate named particle (1)

(e.g. a beam of electrons deflected by an electric or magnetic field or collision/impact on a screen of electrons/ions)

(b) (i) E k

= 5.0 × 10

6

× 1.6 × 10

–19 (J) (1) max 3

(use of E k

= ½ mv

2

gives) v ( =

(1) =

(= 3.1 × 10

–7 ms

–1

)

(ii) (use of λ =

= 1.3 × 10

–14 m

gives) λ =

)

(1)

[or alternatively

= 1.3 × 10

–14 m] (1)

[7]

4

[7]

M4.

(a) electrons can behave as waves

[or electrons can tunnel across gap] (1) waves can cross narrow gaps

[or non-zero probability of crossing gap] (1) electron waves would be attenuated too much by large gap

[or probability of transfer negligible if gap too wide]

[or the narrower the gap, the greater the probability] (1) electron transfer is from – to + (1)

(b) constant height mode: tip height constant (1) current varies as gap width changes (1) image built up as tip moves across surface

[or as tip moves across, a decrease (or increase) of current means the gap widens (or narrows)] (1)

[or constant current mode: tip height altered (1) to keep current constant (1) image built up as above or as tip moves across, the tip height rises (or falls) if the surface rises or falls (1) ]

M5.

(a) particles of light/corpuscles (1) attracted towards glass surface (on entry into glass) (1)

4

QWC 2

3

[7]

velocity/momentum normal to surface increased (1) velocity/momentum parallel to surface unchanged (1)

(b) (i) Newton predicted speed glass

> speed air and Huygens predicted speed glass

< speed air

(1)

(ii) named experiment (1) relevance explained (1)

(e.g. Young’s double slit (1) give rise to fringes/interference which is a wave property (1) or diffraction of light (1) which is a wave property (1) ) max 3

3

[6]

M6.

(i) reflected waves and incident waves form a stationary/standing wave pattern or interfere/reinforce/cancel (1) nodes formed where signal is a minimum (1)

(ii) λ /2 = 1.5 (m) [or λ = 3 (m)]

[or nodes formed at half

–wavelength separation]

(1)

(use of c = f λ gives)

= 100 MHz (1)

(1)

M7.

(a) light, passing through each slit, is diffracted (1) diffracted light from one slit overlaps with (diffracted) light from theother slit (1) bright fringes formed where light waves from each slit reinforce(or in phase)(or interfere constructively) (1) dark fringes formed where light waves (from the two slits) cancel (1) (or out of phase by 180°) (1) path difference = whole number of wavelengths for a bright fringe [or whole number + ½ wavelength for a dark fringe] (1) max 4

QWC 1

(b) corpuscles passing through a slit form a bright fringe (1) two slits produce only two bright fringes according to corpuscular theory (1) more than two fringes are observed (1) dark fringes (or cancellation) cannot happen with corpuscles (1) max 2

[5]

[6]

M8.

(a) light consists of photons (1) an electron in the metal absorbs a photon (1) an electron needs a minimum amount of energy to escape (1) a blue photon has more energy than a red photon (1) hf > for blue photon , < for red photon (1)

Max 4

(b) every electron would gain sufficient energy from the waves in time (1) no matter what the frequency/colour/wavelength of the light is (1)

2

M9.

(a) force on an electron in a magnetic field depends on speed (1) electrons at different speeds would be focussed differently so image would be blurred (1)

[or electrons at different speeds would have different (de Broglie) wavelengths therefore resolution would be reduced]

2

(b) increase in pd increases speed (1) increase in speed/momentum/ E k

causes reduction of

(de Broglie)wavelength (1) reduced (de Broglie) wavelength gives better resolution (1)

3

M10.

(a) electrons have a wave-like nature (1) there is a (small) probability that an electron can cross the gap

[or an electron can tunnel across the gap] (1) transfer is from - to + only (1)

3

(b) constant height mode: gap width varies as tip scans across at constant height (1) current due to electron transfer is measured (1) current decreases as gap width increases (or vice versa) (1) variation of current with time is used to map surface (1)

[6]

[5]

[or constant current mode: current due to electron transfer is measured (1) feedback used to keep current constant by changing height of probe tip (1) height of probe tip changed to keep gap width constant (1) variation of height of probe tip with time used to map surface (1) ]

3

M11.

(a) diagram/description of electric wave and magnetic wave in phase (1) diagram/description/statement that electric wave is at 90° to the magnetic wave (1) diagram/description/statement that direction of propagation/travel is perpendicular to both waves (1)

3

(b) (i) (conduction) electron (in the metal) absorbs a photon and gains energy hf (1) work function of a metal is the minimum energy needed by an electron to escape from the metal (surface) (1) an electron can only escape if hf > work function (1)

any two (1)(1)

(ii) the photon is the quantum of em radiation/light (1) classical wave theory could not explain threshold frequency (1) classical wave theory was replaced by the photon theory (1) [or photons can behave as waves or particles][or photons have a dual wave/particle nature]

any two (1)(1)

4

M12.

(a) one feature (1 mark for one of the following)

• there is a threshold (minimum) frequency (of light) for photoelectric emission from a given metal

• photoelectric emission is instant explanation

[6]

[7]

• light consists of photons (or wavepackets) (1)

• energy of a photon = hf where f is the light frequency (1)

• work function of metal is the minimum amount of energy it needs to escape (1)

• 1 electron absorbs 1 photon and gains energy hf (1)

• electron can escape if energy gained hf > (1)

(b) (i) an electron requires 2.2 eV of energy to escape from the metal surface (1)

6

(ii) photon frequency, f (= c /

λ

=

) = 5.77 × 10

–19 J (1)

photon frequency (= hf ) = 6.63 × 10

–34

× 5.77 × 10

14

= 3.83 × 10

–19 J (1)

E

K

max (= hf

– ) = 3.83 × 10

–19

– (2.2 × 1.6 × 10

–19 ) (1)

= 3.1 × 10

–20 J (1)

5

M13.

(a) particles of light (or corpuscles) (1) attracted towards glass surface (1) ( on entry to glass (or leaving glass) ) velocity (or momentum) parallel to surface unchanged (1) velocity (or momentum) perpendicular to surface increased (ordecreased on leaving) (1) direction (or velocity or momentum) same after leaving glass asbefore entry to glass (1) max 4

(b) named experiment (1) observational evidence (1) how it supports

Huygens’ theory (1)

(e.g. Young’s double slits

(1) shows interference (1) which is a wave property (1) or measurement of the speed of light (1) speed of light is less than in air (1) as predicted by wave theory (1) ) max 3

[11]

[7]

M14.

(a) (i) work done (due to stopping potential V ) = eV (1)

E

Kmax

= work done due to stopping potential

= (1.6 × 10

–19

× 0.35) = 5.6 × 10

–20 J (1)

(ii) (rearranging hf = + E

Kmax

) gives = hf – E

Kmax

(1)

2 photon energy (= hf =

= 3.37 × 10

–19 J (1)

)

= hf – E

Kmax

= 3.37 × 10

–19

– 5.6 × 10

–20

= 2.8(1) × 10

–19 J (1)

3

(b) (i) photons have the same energy (as in a)) (1) when a (conduction) electron in the metal absorbs a photon, it gains all the energy of the photon (1) work function (of Y) is the minimum energy needed by an electron to escape (1) work function of Y is greater than the energy gained by an electron (so electron cannot escape) (1) max 2

(ii) wave theory predicts that incident light (of any frequency)would cause photoelectric emission (from any metal) (1) and any one of the following points wave theory could not explain why light below a certain frequency(or below a threshold frequency) could not cause photoelectricemission (1) or this (threshold) frequency is characteristic of the metal(or depends on the metal) (1) or wave theory could not explain the instantaneous emission ofphotoelectrons (1)

2

[9]

M15.

(a) (i) Newton’s other theories were successful (or Newton was more eminent so Newton’s view was accepted)

alternatives , Huygens’ theory was based on longitudinal waves which cannot explain polarisation or

Huygens’ theory could not explain sharp shadows

1

(ii) either

Newton predicted that light travels faster in glass than in air, Huygens predicted the opposite or there was no evidence (for many years) that light travels slower or faster in glass than in air the speed of light in water (or glass) was (eventually) found to be less than the speed of light in air diffraction/interference observations not conclusive max 2

(b) 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 vocabularycorrectly. The form and style of writing is appropriate to answerthe question.

The candidate provides a comprehensive, coherent and logicalexplanation which recognises that the pattern is due to interferenceof light which is a wave property. They should know that at a brightfringe, the waves from the two slits are in phase and thereforereinforce each other and this can happen at positions where thepath difference is zero or a whole number of wavelengths.They may not refer to the need for the waves to be coherent.Their answer should be well-presented in terms of spelling,punctuation and grammar.

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 a logical explanation which recognises that interference of light is a wave property. They should know either a

bright fringe is where the waves from the two slits are in phase or a dark fringe is where they are out of phase by 180° and be aware there are different positions where these conditions apply. They may know the general condition for the path difference for a bright fringe or a dark fringe although they may not recognise that this condition explains why there are more than two bright fringes.

Their answer should be adequately or well-presented in terms of spelling, punctuation and grammar.

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.

The candidate recognises that interference of light is a wave property and that the waves from the two slits reinforce at a bright fringe or cancel at a dark fringe. They may confuse path difference and phase difference and their explanation of why there are more than two bright fringes may be vague or absent.

Their answer may lack coherence and may contain a significant number of errors in terms of spelling and punctuation.

Incorrect, inappropriate of no response: 0 marks

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

Statements expected in a competent answer should includesome of the following marking points.

the pattern is due to interference of light from the two slits interference is a wave property light from the two slits is in phase at a bright fringe and thereforereinforces the path difference (from the central bright fringe to the two slits)is zero either bright fringes are formed away from the centre wherever thepath difference is a whole number of wavelengths or dark fringesare formed away from the centre wherever the path difference isa whole number of wavelengths + a half wavelength the path difference for the m th

bright fringe from the centre is m wavelengths where m is any whole number since m is any whole number, more than two bright fringes areobserved max 6

[9]

M16.

(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 answer that includes a stated property of light such as interference or diffraction that can only be explained in terms of the wave nature of light and a stated property such as photoelectricity that can only be explained in terms of the particle nature of light. In each case, a relevant specificobservational feature should be referred to and should be accompanied by a coherent explanation of the observation. Both explanations should be relevant and logical.

For full marks, the candidate may show some appreciation as to why the specific feature of either the named wave property cannot be explained using the particle nature of light or the named particle property cannot be explained using the wave nature of light.

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 a logical and coherent explanation that includes a stated property of light such as interference or diffraction that can only be explained in terms of the wave nature of light and a stated property such as photoelectricity that can only be explained in terms of the particle nature of light.

For 4 marks, the candidate should be able to refer to a relevant specific observational feature of each property, at least one of which should be followed by an adequate explanation of the observation. Candidates who fail to refer to a relevant specific observational feature for one of the properties may be able to score 3 marks by providing an adequate explanation of the observational feature referred to.

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.

The candidate provides some relevant information relating to two relevant stated properties for 1 mark. Their answer may lack coherence and may well introduce irrelevant or incorrect physics ideas in their explanation.

Points that can be used to support the explanation:

Wave-like nature property

• property is either interference or diffraction

• observational feature is either the bright and dark fringes of a double slit interference pattern or of the single slit diffraction pattern (or the spectra of a diffraction grating)

• explanation of bright or dark fringes (or explanation of diffraction grating spectra) in terms of path or phase difference

• particle/corpuscular theory predicts two bright fringes for double slits or a single bright fringe for single slit or no diffraction for a diffraction grating

Particle-like nature

• property is photoelectricity

• observational feature is the existence of the threshold frequency for the incident light or instant emission of electrons from the metal surface

• explanation of above using the photon theory including reference to photon energy hf , the work function of the metal and ‘1 photon being absorbed by 1 electron’

• wave theory predicts emission at all light frequencies or delayed emission for (very) low intensity

6

(b) (i) m (= m o

(1 - v

2

/ c

2

) –0.5

= 9.11 × 10

–31 (1 - 0.890

2

) –0.5

)

(= 1.998 × 10

–30 kg) = 2.0(00) × 10

–30 kg

(= 1.2(4) × 10

–12 m)

2

(ii)

(iii) E

K

= ( m - m o

) c

2

= (1.998 × 10

–30

– 9.11 × 10

–31

) × (3.0 × 10

8

)

2

= 9.78 × 10

–14 J 3 sf only

= 1.6(0)× 10

–13 J

1

2

[11]

M17.

(a) (vibrations of) the electric wave and magnetic wave; perpendicular to each other perpendicular to direction of propagation in phase with each other

3

(b) μ o

and ε o

determined experimentally (or μ o

and ε o

values were known)

(substitution of values of μ o

and ε o into) predicted equation gives 3(.0) × 10

8

m s –1

(or the speed of light) which is the speed of light (or 3(.0) × 10

8

m s –1 )

2

(c) (i) magnetic wave vibrations perpendicular to (plane of) loop

(magnetic wave) causes alternating (or changing) magnetic flux (linkageor cutting) through the loop alternating magnetic flux (or field) induces an alternating (or changing) emf(or pd) in the loop

[ or equivalent E-field statements

E-wave (or field) vibrations parallel to loop

E-wave (or field) induces emf (or pd) in wire of loop

E-wave (or field) alternates so induced emf is alternating ]

3

(ii) no magnetic flux (linkage or cutting) through the loop (as loop is now parallel to magnetic wave vibrations) so no induced emf (or pd)

( or electric field perpendicular to loop so no induced emf (or pd) )

1

[9]

M18.

(a) Quality of written communication:

Good – Excellent

The candidate provides a comprehensive, coherent and logical explanation which recognises what a stationary wave is and that the conditions for the formation of a stationary wave are present. They should know that nodes and antinodes are formed at alternate positions along XY which are equally spaced with nodes every half wavelength. They should know how the detector is used to locate the position of each node or antinode and how the wavelength is determined from the distance between two such positions. They may know that the nodes can be located more accurately than the antinodes and that their chosen two positions should be as far apart as possible.

Their answer should be well-presented in terms of spelling, punctuation and grammar.

For top band, explanation = at least b and e description = at least f, g,h

(5-6 marks)

Modest – Adequate

The candidate provides a logical explanation which recognises what a stationary wave is and what some of the conditions for the formation of a stationary wave are. They may know that nodes and antinodes are formed at alternate positions along XY with nodes every half- wavelength. They may know how the detector is used to locate the position of each node or antinode and how the wavelength is determined from the distance between two such positions.

They may know that the nodes can be located more accurately than the antinodes and that their chosen two positions should be as far apart as possible.Their answer should be well-presented in terms of spelling, punctuation and grammar.

For middle band , explanation = at least any two of a-e description = at least any two of f-i

(3-4 marks)

Poor to Limited

The candidate may recognise that the reflector reflects radio waves which then form a stationary wave pattern with the incident waves. They may be unaware what the conditions for the formation of a stationary wave are and their understanding of nodes and antinodes may be poor. They may have some awareness that the stationary wave causes the detector signal to vary with position along XY and that the wavelength can be determined from this variation although they might not be able to link the wavelength to the changes of detector position correctly.

Their answer may lack coherence and may contain a significant number of errors in terms of spelling and punctuation.

For lowest band,

Any 2 points ,must be 1 of each for 2 marks

The explanations expected in a good answer should include most of the following physics ideas

Explanation of stationary wave formation;- a. radio waves from the transmitter are reflected back towards the transmitter b. reflected and incident waves pass through eachother

c. both waves have same frequency (and speed) andamplitude d. superposition (of reflected and incident waves)occurs to form a stationary wave (as above) e. equally spaced nodes and antinodes formed along XY

Description of measurement of wavelength;- f. Detector signal is zero ( or least) along XY atnodes g. distance between adjacent nodes is ½λ h. move detector along XY to measure distancebetween adjacent nodes and double to give thewavelength i. measure distance over n nodes and divide by n-1to give distance between adjacent nodes

6

(1-2 marks)

(b) Speed of radio waves (obtained by Hertz ) is the same as the speed of light

Speed of electromagnetic waves (calculated or predicted by Maxwell) is the same as the speed of light ( or of radio waves) so radio waves are electromagnetic waves

2

(Total 8 marks)

M19.

(a) (Matter) particles have wave-like properties (owtte)

Accept mv or mass x velocity in place of p and an associated wavelength = h / p where p is the momentum of the particles .

Accept ‘inversely proportional to momentum ( or mv)’ after

‘wavelength’

2

(b) E

K

(= 0.021 eV ) = 0.021 × 1.60 × 10

−19 or 3.36 × 10

−21 J

(Using E

K

= ½ m v

2

gives )

For 2 nd

mark, allow individual values of e and V in place of E

K value in data substitution mv =( 2 m E

K

)

1/2

= ( 2 × 1.67(5)×10

−27

× 3.36 × 10

−21 )

1/2

( = 3.35 × 10

−24 kg m s −1 )

For 3 rd

mark, allow individual values of m and v in denominator

[OR v =(2 E

K

/ m)

1/2

= ( 2 × 3.36 × 10

−21

/ × 1.67(5)×10

−27 )

1/2

( = 2.0 × 10

3

m s −1 ) mv = (1.67(5)×10

−27

× 2.0 × 10

3

( = 3.35 × 10

−24 kg m s −1 ) ]

= 2.0 × 10

−10 m to 2 sf

Alternative;

Correct use of 0.021 eV in λ= h / (2meV)

1/2

= 1.88 × 10

−10 m = 2.0 × 10

−10 m to 2 sf

Final sf mark - need to see some valid working

4

(c) electron’s momentum (p) is the same (as that of the neutron) and its mass is (much) smaller than neutron mass kinetic energy = p

2

/ 2m so kinetic energy of electron is (much) greater

Alternative for 2 nd

mark;- (so) electron’s speed is (much) greater and as kinetic energy = ½ mv

2

, the electron’s kinetic energy is (much) greater as v

2

is more significant than m

(here)(owtte)

2 nd

alternative for 2 nd mark using λ = h / (2 mE

K

)

1/2

λ = h / (2 mE

K

)

1/2 so (same λ means) mE

K

( in equation) is the same for electron as for the neutron). So E

K

is (much) greater as electron mass is (much) smaller than neutron mass (owtte)

Note; allow use of eV in place of E

K

if eV is identified as E

K

.

2

[8]

E1.

Most candidates scored at least two marks on the diagram in part (a)(i). They were clearly aware that the path changed direction at each lens and needed to end at Q. In part (ii), many candidates did not realise that the second lens focused the image on the screen as well as providing further magnification.

In part (b) most candidates knew that the speed of the electrons increased with anode voltage and therefore that the wavelength decreased. However, few candidates failed to mention that the reduced wavelength meant less diffraction and therefore better resolution.

E2.

In part (a) few candidates referred to wavelets being emitted by the slits or to the slits as coherent sources of light. Most candidates scored a mark as a result of explaining why a bright or a dark fringe was formed. Very few candidates referred to the criteria for reinforcement or cancellation, either in terms of path difference or phase difference.

In part (b)(i) most candidates knew that Newton considered light as corpuscles and that they would pass straight through the slits. On the other hand, few candidates were aware that

Newton’s theory predicted that just two fringes would be seen. In part (iii), many candidates

failed to refer to Newton’s pre-eminence in science as the reason why his theory was preferred.

Very few candidates knew that Newton predicted that light travels faster in glass than in air, whereas Huyghens gave the opposite prediction.

E3.

Candidates who scored well in part (a)(i) usually referred to electron diffraction and were able to provide a description of a piece of relevant evidence. Some candidates however, did lose a mark through vague references to diffraction (of electrons) at a slit. Many candidates provided irrelevant answers to both part (i) and part (ii) as a result of not realising the question asked about matter.

The few good answers seen in part (a)(ii) usually referred to the deflection of a beam of electrons or protons by an electric or a magnetic field. Some candidates lost a mark by referring to the deflection of such a beam by a gravitational field. Some candidates scored well by describing ionisation by collision or scattering of alpha particles by nuclei. A significant number of candidates lost marks by describing evidence of the particle-like nature of light.

In the calculations in part (b), many candidates scored all possible marks with clearly explained calculations. Those who attempted to calculate the de Broglie wavelength using the voltage formula were often confused between the speed and the voltage, or else failed to include e in their calculation. Almost all candidates who chose to calculate the de Broglie wavelength directly from the speed did so correctly.

E4.

Although most candidates knew in part (a) that electrons crossed the gap in the scanning tunnelling microscope because of the wave nature of the electron, they usually referred to tunnelling and probability rather than attenuation of the wave. Many candidates knew that the probability of an electron crossing the gap was greater the narrower the gap. The best candidates were able to explain satisfactorily why a potential difference was necessary although many candidates considered the potential difference caused insulation breakdown between the tip and the surface.

Very few good answers were seen in part (b). Most candidates knew that the tunnelling current varied according to the gap width but few gave a clear answer in terms of a constant current or the tip being at a constant height. Many candidates switched from constant current to constant height, often as a result of confusion about feedback giving a constant gap width. Some candidates failed to mention that the current or the tip height was monitored as the tip was scanned across the surface.

E5.

Relatively few good answers were seen in part (a). Most candidates mentioned that the theory required an increase of speed of the corpuscles on entry to glass. Some candidates did refer to an attractive force acting on the corpuscles. However, very few candidates were able to give a satisfactory account of the effect on the velocity or momentum components of the

corpuscles. Those candidates who did provide a satisfactory account usually gave an appropriate diagram to support their account.

In part (b)(i), a significant number of candidates often failed to state what each theory predicted.

In part (ii), most candidates scored both marks by giving an account in appropriate terms of y oung’s double slits experiment.

E6.

In part (i), the majority of candidates realised that the reflected waves and direct waves formed a stationary wave pattern and that the signal was a minimum at the nodes. Some candidates referred only to interference, without demonstrating any awareness that stationary waves were formed. In part (ii), a significant number of candidates thought that the wavelength was 1.5 m although most candidates knew how to calculate the frequency of the radio waves from the wavelength.

E7.

In part (a) the majority of candidates realised that the light was diffracted by each slit but not all were aware that interference took place where the diffracted waves overlap. Explanations given for the formation of bright fringes were usually adequate, but satisfactory explanations of the formation of dark fringes were less common, usually because candidates did not appreciate that ‘out of phase’ does not mean a phase difference of 180° when referring to the superposition of two waves. In addition, many candidates referred incorrectly to path difference when they meant phase difference or vice versa. Few candidates were able to give the correct condition in terms of path difference for the formation of a bright or a dark fringe.

Most candidates, in part (b) knew that the corpuscular theory predicted only two bright fringes but very few candidates recognised that each fringe was also predicted as a result of corpuscles passing through each slit. No more than a minority of candidates knew that the corpuscular theory could not explain why more than two fringes are formed or that the theory could not explain why dark fringes are seen.

E8.

Although many candidates were able to score at least two marks for part (a), very few candidates were able to provide a coherent explanation. Weak candidates often failed to mention that light consists of photons and they sometimes considered photons, not electrons, being emitted from the plate. Few candidates stated the photon energy equation or mentioned that an electron absorbed a single photon or realised that an electron needed a minimum amount of energy to escape from the plate. Many candidates wrote vaguely, without any reference to photons, that blue light has more energy than red light. The work function of the metal was often referred to as a property of light and incorrect phrases such as ‘the frequency of light must be greater than the work function’ were often seen.

In part (b), many candidates realised that wave theory predicted that all electrons would gain some energy from the incident light, but fewer stated that wave theory also predicted that light

of any frequency (or colour or wavelength) would have this effect.

E9.

In part (a), many candidates were aware that electrons moving at different speeds would follow different paths, but few gave the reason for this, namely that the force on an electron moving in a magnetic field depends on the speed of the electron. Most candidates realised that different paths due to different speeds would produce a blurred image, but some candidates incorrectly thought that the blurred image was due to electrons having different speeds and arriving at the screen at different times.

Many candidates scored all three marks allocated to part (b), with an explanation that demonstrated not only a sound understanding of the physics principles involved, but also good awareness of the application of these principles. Some candidates lost marks as a result of not explaining why the de Broglie wavelength was reduced. A small number of candidates incorrectly considered the effect of the increased pd on the intensity of the image.

E10.

Most candidates were aware, in part (a), that electron transfer across the gap was due to the wave nature of the electron, and many candidates referred to tunnelling or to a non-zero probability of electrons crossing the gap. Few however, explained that a pd was necessary to ensure that electron transfer occurred in one direction only. A significant number of candidates considered incorrectly that electrostatic forces or secondary emission to be the cause of electron transfer.

In part (b) the majority of candidates were unable to give a coherent description of how an STM was used to obtain an image. Some failed to make it clear whether their description referred to

‘constant height’ or ‘constant current’ mode. Those who chose to describe the former frequently confused the constant height of the tip with the fixed gap between the tip and the surface needed for the latter mode. Those who described the ‘constant current’ mode failed to refer to the change of current over a gap or depression being used to raise or lower the tip in order to keep the gap width constant. Again, few candidates stated clearly what variable was used in their chosen mode to measure the surface height.

E11.

Many candidates scored full marks in part (a), either through a clearly labelled diagram or a detailed description accompanied by a sketch. Many knew that the electric wave and the magnetic wave were in phase and perpendicular to each other but they often failed to state or show that the direction of propagation was perpendicular to both waves.

In part (b) (i), few candidates were able to give a coherent account of Einstein’s explanation, usually as a result of failing to mention that the energy of a photon was gained by a single electron. Most candidates, however, knew that Einstein introduced the photon theory of light to explain photoelectricity, but few mentioned that the energy of a single photon was gained by a single electron. It was also well known that electrons required energy to escape from a metal

surface but only a minority of candidates stated that an electron needed a minimum amount of energy to do this. In part (ii), although a considerable number of candidates knew that light consisted of photons which were ‘packets of energy’, few referred to the photon as the ‘quantum of light’. A significant number were not aware that the wave theory was replaced by the photon theory because it could not explain the existence of the threshold frequency of light, although some candidates thought that Newton’s corpuscular theory provided the explanation.

E14.

In (a) the calculation was considered straightforward. Even though the photoelectric equation is written on the data sheet candidates need to be aware what each symbol represents. In (i) many tried to use the photoelectric equation before realising that this part involved the stopping potential only. Others worked out the work function here and again in (ii).

For the able candidates who correctly worked out the maximum ke of the photoelectrons in (i) they invariably scored full marks in (ii) as well.

In (b) many candidates knew the correct physics in this situation but the majority of answers lacked detail and a logical argument was not presented. Most candidates were content in (i) with just saying that the work function of Y was greater than the work function of X. In (b) (ii) very few candidates could clearly make a statement that the wave theory of light predicts that any incident light would cause the emission of electrons. Most picked up a mark for saying the wave theory of light could not predict the instantaneous emission of electrons.

E15.

Many candidates did not score the mark in (a)(i) because they gave general statements which could apply to a comparison of any two individuals rather than referring to Newton’s greater scientific reputation.

In part (a)(ii), most candidates knew that light travels slower in water than in air but many lost a mark because they only gave the correct prediction of one of the two theories.

Although considerable variation was seen in the depth of knowledge and understanding of candidates in part (b), many candidates were able to express their ideas adequately. Relatively few candidates were hampered by very poor quality of written communication. Most candidates gained some credit for knowing the light from the two slits produced an interference pattern and were able to give a simple explanation of why bright and dark fringes were formed. Many candidates knew that bright fringes were formed where the light waves were in phase but a significant number of candidates did not indicate the exact phase difference for the formation of a dark fringe and merely stated that the waves needed to be out of phase. Few candidates were able to state the correct path difference for a bright fringe or for a dark fringe, often referring to

‘phase’ difference in wavelengths or stating the path difference for a bright fringe as one wavelength instead of a whole number of wavelengths. The key explanation of why there are more than two bright fringes was often absent or too vague to gain any credit. The more able candidates wrote clearly that bright fringes are formed where the path difference is a whole number of wavelengths and that because the light is diffracted at each slit, there will be several positions where the path difference condition for a bright fringe is fulfilled.

E16.

In part (a) it was pleasing to see some well-written accounts that covered most if not all the relevant facts. Many students failed to support a reasonable or good account of one of the two properties with a similar account of the other property. Many students who were able to supply a reasonable ‘wave’ explanation of the double slits experiment often gave a limited account of photoelectricity as a particle property with little more than a statement of the meaning of the threshold frequency. Explanations often lacked depth as many students failed to link the threshold frequency to the work function and the photon energy equation. However, a significant number of students did provide a brief outline of why interference fringes could not be accounted for using corpuscular theory or why the threshold frequency could not be explained using wave theory.In part (b)(i), many students did not realise the relativistic mass needed to be calculated even though the speed of the electron was given in terms of the speed of light. In (ii) and (iii), whereas most students were able to calculate the photon energy in (ii), only the best students were able to calculate the kinetic energy in (iii). Frequent errors included the use of ½ mv

2

, some with the correct mass of the electron and some with its rest mass. A significant number of students did calculate the total energy correctly but then failed to subtract the rest energy.

E17.

In part (a) many students scored full marks by depicting as well as describing an electromagnetic wave. Those who limited their account to a description often failed to mention the fields were in phase or the direction of propagation was perpendicular to both fields. A significant number of students considered polarisation of an electromagnetic wave causes loss of the electric wave or the magnetic wave.

In part (b) many students did realise that the equation for the speed of electromagnetic waves gives a value equal to the speed of light but few students stated its value or appreciated the value of the speed of light with which the calculated value was compared was obtained by experiment.

For part (c) although most students in (i) chose to explain the induced emf by considering the magnetic wave variations, very few students provided a good explanation. Relatively few students answered in terms of the electric field variation. Many students knew the magnetic flux in the loop changed but failed to state the flux linkage changes continuously or that the magnetic field is perpendicular to the plane of the loop. Some students failed to score because they did not make clear whether their account was in terms of the magnetic wave or the electric wave. In (ii), very few students mentioned the significance of the wave being polarised although the best students did know that emf was zero at 90 degrees because the flux linkage was zero or that the electric field was then perpendicular to the loop.

E18.

(a) Most candidates made some progress in their description of how to measure the wavelength although some were unclear about the use of the detector. Quality of written communication was tested in this question and it is pleasing to see that many candidates

were able to write clearly and logically on this topic. However, candidates were often unable to recognise the key physics principles involved in the formation of stationary waves in this situation although they usually recognised the significance of nodes and antinodes.

Misunderstandings were often seen in their explanations such as nodes at maxima rather than minima or superposition occurring at antinodes only. Lengthy explanations that were about double slits interference rather than stationary waves were not uncommon.

(b) Many candidates knew that the speed of radio waves was found to be the same or very close to the speed of light but only a minority appreciated why it was concluded that radio waves and light are electromagnetic waves.

E19.

(a) Although many candidates were able to give the de Broglie equation, candidates frequently forgot to state clearly the basis of the theory, namely that matter particles exhibit wave like properties. A clear reference to ‘matter particles’ in general was expected rather than to specific particles or to vague forms such as ‘substance’.

(b) Candidates who calculated the speed or momentum from the given value of kinetic energy were generally able to complete the calculation successfully although some candidates used the wrong value of mass in the final stage of their work. Candidates who used h / (2 meV )

1/2

often substituted 0.021 as the value of eV , clearly unaware the given value of eV needed to be converted to joules before substitution in the formula. Those candidates who actively recognised eV in the formula as the kinetic energy generally went on to answer the question successfully.

(c) Explanations often lacked sufficient detail and candidates often failed to explain their reasoning. Most candidates worked out correctly from knowledge of their relative masses that electrons travel faster than neutrons with the same de Broglie wavelength but few recognised clearly their deduction in this context follows from the relationship between wavelength and momentum. Relatively few candidates explained that the greater speed was much more significant than the lesser mass because the speed is squared in the kinetic energy formula ½ mv

2

whereas the mass is not.

A2 Wave Particle Duality

A2 Wave Particle Duality

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