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IGCSE · 3 · extended response

Same light energy, different photon counts

Waves · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
6
Topics
1
Answer
Complete
Explore the question data

Photon-energy model

Supplied question data. Read the labelled axes and use the observations to test the physical model.

Vertical axis (y): Photon energy / 10⁻¹⁹ J

Photon-energy modelSupplied question data. Read the labelled axes and use the observations to test the physical model. Horizontal axis: Frequency / 10¹⁴ Hz. Vertical axis: Photon energy / 10⁻¹⁹ J. Exact values are available in the data table below.01.6573.3134.976.62602.557.510xy

Horizontal axis (x): Frequency / 10¹⁴ Hz

Reading 1Frequency / 10¹⁴ Hz: 0Photon energy / 10⁻¹⁹ J: 0
Read the exact data as a table
Photon-energy model — supplied values
Frequency / 10¹⁴ HzPhoton energy / 10⁻¹⁹ J
00
53.313
106.626

These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.

extended response6 marks

Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A source emits monochromatic pulses of equal energy 6.626×10⁻⁶ J at two frequencies: 5.0×10¹⁴ Hz and 1.0×10¹⁵ Hz. Use h = 6.626×10⁻³⁴ J s and c = 3.0×10⁸ m s⁻¹. The graph provides photon energy as a function of frequency. Supplied model: Use photon energy E = hf, wavelength λ = c/f and photon count N = Epulse/Ephoton. Supplied data (Frequency / 10¹⁴ Hz; Photon energy / 10⁻¹⁹ J): (0, 0), (5, 3.313), (10, 6.626).

  1. (a)

    Calculate each wavelength, photon energy and photon count. Identify the approximate spectral region of each pulse.

    3 marks
  2. (b)

    Explain why interference supports a wave model while photon-counting detection needs a particle-like model. Evaluate the claim that all electromagnetic radiation has identical biological effects.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overview(a) 600 nm, 3.313×10⁻¹⁹ J, 2.0×10¹³ photons; 300 nm, 6.626×10⁻¹⁹ J, 1.0×10¹³ photons. (b) Different models describe different observations; common vacuum speed does not imply identical interaction or hazard.
01

(a)

3 marks

Calculate each wavelength, photon energy and photon count. Identify the approximate spectral region of each pulse.

How to approach it

Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.

  1. 1

    Use λ = c/f: the lower frequency has wavelength 6.0×10⁻⁷ m and the higher 3.0×10⁻⁷ m.

  2. 2

    Ephoton = hf gives 3.313×10⁻¹⁹ and 6.626×10⁻¹⁹ J. Dividing equal total pulse energy by these values gives 2.0×10¹³ and 1.0×10¹³ photons.

  3. 3

    600 nm is visible light and 300 nm is ultraviolet. Both propagate at c in vacuum; the higher-frequency pulse contains fewer photons for the same total energy.

02

(b)

3 marks

Explain why interference supports a wave model while photon-counting detection needs a particle-like model. Evaluate the claim that all electromagnetic radiation has identical biological effects.

How to approach it

Build a chain of claim, evidence and physics reasoning. Address more than one relevant factor, identify a limitation or assumption, and finish with a conclusion that is conditional on the evidence rather than absolute.

  1. 1

    Spatial interference patterns follow phase and superposition, which a simple stream of classical particles does not explain.

  2. 2

    Localized detector events with frequency-dependent energy are described by photons. These are complementary descriptions of quantum light rather than literal tiny balls riding classical waves.

  3. 3

    Absorption, penetration, dose and photon energy depend on wavelength and material. Ultraviolet can cause photochemical damage; exposure assessment cannot be inferred from vacuum speed alone.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

  1. Use λ = c/f: the lower frequency has wavelength 6.0×10⁻⁷ m and the higher 3.0×10⁻⁷ m.
  2. Ephoton = hf gives 3.313×10⁻¹⁹ and 6.626×10⁻¹⁹ J. Dividing equal total pulse energy by these values gives 2.0×10¹³ and 1.0×10¹³ photons.
  3. 600 nm is visible light and 300 nm is ultraviolet. Both propagate at c in vacuum; the higher-frequency pulse contains fewer photons for the same total energy.
  4. Spatial interference patterns follow phase and superposition, which a simple stream of classical particles does not explain.
  5. Localized detector events with frequency-dependent energy are described by photons. These are complementary descriptions of quantum light rather than literal tiny balls riding classical waves.
  6. Absorption, penetration, dose and photon energy depend on wavelength and material. Ultraviolet can cause photochemical damage; exposure assessment cannot be inferred from vacuum speed alone.

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Your self-assessment

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