Year 12 Physics Module 7: The Nature of Light Checkpoint 4 IQ4: Light — Wave or Particle?

Checkpoint 4

Test your understanding of the photoelectric effect and wave-particle duality. Covers Lessons 16 and 20.

L16: The Photoelectric Effect

Photon model; $E = hf$; work function; threshold frequency; Einstein's equation

L20: Wave-Particle Duality

Double-slit revisited; complementarity; synthesis; modern quantum technologies

Key Formulae

$E = hf = hc/\\lambda$Photon energy
$E_{k,max} = hf - \\phi$Photoelectric equation
$p = h/\\lambda = E/c$Photon momentum
Key Terms
PhotonA quantum of electromagnetic energy.
Work functionMinimum energy to eject an electron from a metal.
ComplementarityWave and particle aspects are mutually exclusive but complementary.
WavefunctionMathematical description of a quantum state.
1

Multiple Choice

15 questions — instant feedback

Understand Band 4+ Easy

Q1. The energy of a photon is proportional to:

Correct: B. $E = hf$. Photon energy is directly proportional to frequency and inversely proportional to wavelength.

Understand Band 5/6 Medium

Q2. In the photoelectric effect, the threshold frequency depends on:

Correct: C. $hf_0 = \\phi$. The threshold frequency is determined solely by the metal's work function, a material property.

Apply Band 5/6 Medium

Q3. Light of wavelength 400 nm shines on a metal with $\\phi = 2.0$ eV. The maximum kinetic energy of emitted electrons is approximately:

Correct: A. $E = hc/\\lambda = 1240/400 = 3.1$ eV. $E_{k,max} = 3.1 - 2.0 = 1.1$ eV.

Understand Band 5/6 Medium

Q4. (Enrichment — not examinable) The Compton effect demonstrates that photons:

Correct: D. Compton scattering shows photons transferring momentum to electrons in elastic collisions, behaving like billiard balls.

Apply Band 5/6 Medium

Q5. An X-ray of wavelength 0.100 nm backscatters ($\\theta = 180°$) from an electron. The wavelength shift is:

Correct: B. At $\\theta = 180°$, $\\cos\\theta = -1$, so $\\Delta\\lambda = 2\\lambda_C = 4.86$ pm. This is the maximum possible shift.

Understand Band 4+ Easy

Q6. (Enrichment — not examinable) de Broglie's hypothesis states that:

Correct: C. de Broglie proposed that all matter has wave properties with wavelength $\\lambda = h/p$, extending wave-particle duality beyond light.

Apply Band 5/6 Medium

Q7. (Enrichment — not examinable) An electron accelerated through 100 V has de Broglie wavelength approximately:

Correct: A. Using the shortcut $\\lambda$ (nm) $= 1.23/\\sqrt{V}$ (V): $\\lambda = 1.23/10 = 0.123$ nm.

Understand Band 5/6 Medium

Q8. Electron diffraction experiments (Davisson-Germer, Thomson) confirmed:

Correct: D. The observation of interference patterns from electrons confirmed de Broglie's hypothesis that matter has wave properties.

Understand Band 5/6 Medium

Q9. (Enrichment — not examinable) The Heisenberg uncertainty principle $\\Delta x \\Delta p \\geq h/(4\\pi)$ implies that:

Correct: B. If $\\Delta x$ is very small, $\\Delta p$ must be large to maintain the inequality. This is a fundamental property of nature, not a technological limitation.

Apply Band 5/6 Medium

Q10. (Enrichment — not examinable) A particle is confined to $\\Delta x = 0.1$ nm. The minimum uncertainty in its momentum is approximately:

Correct: A. $\\Delta p \\geq h/(4\\pi \\Delta x) = 6.63\\times10^{-34}/(4\\pi \\times 10^{-10}) = 5.3\\times10^{-25}$ kg·m/s.

Understand Band 5/6 Medium

Q11. In the double-slit experiment with single electrons, an interference pattern builds up because:

Correct: C. Even single electrons produce interference patterns. The wavefunction describes a probability wave that passes through both slits, and the detection probability at each point reflects the interference of these two paths.

Analyse Band 5/6 Medium

Q12. Which experiment provides evidence that light has both wave and particle properties?

Correct: D. No single experiment reveals both aspects. Wave experiments (interference, diffraction) show wave properties; particle experiments (photoelectric, Compton) show particle properties. Both are needed for a complete picture.

Analyse Band 6 Hard

Q13. According to the principle of complementarity:

Correct: B. Bohr's complementarity principle states that wave and particle aspects cannot be observed simultaneously in a single experiment, but both are needed for a complete description.

Apply Band 6 Hard

Q14. (Enrichment — not examinable) A proton and electron have the same de Broglie wavelength. This means they have:

Correct: A. $\\lambda = h/p$, so equal $\\lambda$ means equal $p$. Since $m_p \\neq m_e$, their speeds, KEs and masses are all different.

Analyse Band 6 Hard

Q15. (Enrichment — not examinable) The energy-time uncertainty principle $\\Delta E \\Delta t \\geq h/(4\\pi)$ explains:

Correct: C. Short-lived excited states have large energy uncertainty, producing a spread of photon energies — the natural linewidth of spectral lines.

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Answer all multiple choice to see your score
2

Short Answer Questions

5 questions — model answers revealed

Understand Band 5/6 Medium 3 marks

SAQ 1. (a) State Einstein's photoelectric equation. (b) Explain why the existence of a threshold frequency contradicts the wave model of light. (c) A metal has work function 2.5 eV. Calculate the threshold wavelength. (3 marks)

Model answer (3 marks):

(a) $E_{k,max} = hf - \\phi$ (1 mark).

(b) The wave model predicts that any frequency of light, given sufficient intensity and time, should eventually eject electrons. The existence of a threshold frequency — below which no electrons are emitted regardless of intensity — directly contradicts this prediction (1 mark).

(c) $\\lambda_0 = hc/\\phi = 1240/2.5 = 496$ nm (1 mark).

Apply Band 5/6 Medium 4 marks

SAQ 2. (Enrichment — not examinable) (a) Calculate the momentum of a photon with wavelength 600 nm. (b) An X-ray of wavelength 0.080 nm scatters at $90°$ from an electron. Calculate the scattered wavelength and the kinetic energy transferred to the electron. (c) Explain why the Compton effect is significant evidence for the particle nature of light. (4 marks)

Model answer (4 marks):

(a) $p = h/\\lambda = 6.63\\times10^{-34}/(600\\times10^{-9}) = 1.11\\times10^{-27}$ kg·m/s (1 mark).

(b) $\\Delta\\lambda = \\lambda_C(1 - \\cos 90°) = 2.43$ pm. $\\lambda' = 80 + 2.43 = 82.43$ pm = 0.0824 nm (0.5 mark). $E = 1240/0.080 = 15.5$ keV; $E' = 1240/0.0824 = 15.0$ keV. $E_k = 15.5 - 15.0 = 0.5$ keV = 500 eV (1 mark).

(c) The Compton effect treats photons as particles with momentum that collide elastically with electrons, transferring momentum and energy. The wavelength shift depends on scattering angle exactly as predicted by particle collision mechanics, not wave theory (1.5 marks).

Apply Band 5/6 Medium 4 marks

SAQ 3. (Enrichment — not examinable) (a) State de Broglie's hypothesis. (b) An electron is accelerated through 500 V. Calculate its de Broglie wavelength. (c) Calculate the speed of this electron and comment on whether relativistic effects are significant. (d) Explain why electron microscopes have much higher resolution than optical microscopes. (4 marks)

Model answer (4 marks):

(a) All particles have an associated wave with wavelength $\\lambda = h/p$ (1 mark).

(b) $\\lambda = 1.23/\\sqrt{500} = 1.23/22.4 = 0.055$ nm = 55 pm (0.5 mark).

(c) $v = \\sqrt{2eV/m_e} = \\sqrt{2(1.6\\times10^{-19})(500)/(9.11\\times10^{-31})} = 1.33\\times10^7$ m/s. $v/c = 0.044$, so relativistic effects are small ($\\gamma \\approx 1.001$) (1 mark).

(d) Resolution is limited by diffraction: $\\theta \\sim \\lambda/d$. Electron wavelengths ($\\sim 0.05$ nm) are thousands of times shorter than visible light ($\\sim 500$ nm), so electron microscopes can resolve much finer detail, even atomic structures (1.5 marks).

Analyse Band 6 Hard 5 marks

SAQ 4. (Enrichment — not examinable) (a) State the Heisenberg uncertainty principle for position and momentum. (b) An electron is confined to a region of size 0.5 nm. Estimate the minimum uncertainty in its momentum and the corresponding minimum kinetic energy. (c) Explain why this means an electron cannot simply sit at rest at the centre of an atom. (d) An excited state lives for $5.0\\times10^{-9}$ s. Calculate the minimum uncertainty in the energy of emitted photons. (5 marks)

Model answer (5 marks):

(a) $\\Delta x \\Delta p \\geq h/(4\\pi)$ (1 mark).

(b) $\\Delta p \\geq h/(4\\pi \\Delta x) = 6.63\\times10^{-34}/(4\\pi \\times 0.5\\times10^{-9}) = 1.06\\times10^{-25}$ kg·m/s (0.5 mark). $E_k \\approx (\\Delta p)^2/(2m_e) = (1.06\\times10^{-25})^2/(2 \\times 9.11\\times10^{-31}) = 6.2\\times10^{-21}$ J = 0.039 eV (1 mark).

(c) If the electron were at rest at a point, $\\Delta x = 0$ and $\\Delta p = 0$, violating the uncertainty principle. Confinement requires momentum and hence kinetic energy — this is called confinement energy (1 mark).

(d) $\\Delta E \\geq h/(4\\pi \\Delta t) = 6.63\\times10^{-34}/(4\\pi \\times 5.0\\times10^{-9}) = 1.06\\times10^{-26}$ J = $6.6\\times10^{-8}$ eV (1.5 marks).

Analyse Band 6 Hard 5 marks

SAQ 5. (a) Define wave-particle duality. (b) For each of the following, state which model (wave, particle, or both) explains it and give a brief reason: (i) photoelectric effect, (ii) electron diffraction (Enrichment — Module 8), (iii) Compton scattering (Enrichment — Module 8), (iv) Young's double-slit with light. (c) Explain Bohr's principle of complementarity. (d) Describe how the observation of which path a particle takes in a double-slit experiment affects the outcome. (5 marks)

Model answer (5 marks):

(a) Wave-particle duality is the principle that all quantum entities exhibit both wave and particle properties, with which aspect is observed depending on the experiment (0.5 mark).

(b)(i) Photoelectric effect: particle model — requires photons with discrete energy $hf$ to overcome work function (0.5 mark).

(b)(ii) Electron diffraction: wave model — electrons produce interference patterns, requiring wave behaviour (0.5 mark).

(b)(iii) Compton scattering: particle model — photons collide with electrons like billiard balls, transferring momentum (0.5 mark).

(b)(iv) Double-slit: wave model — interference pattern arises from superposition of waves from both slits (0.5 mark).

(c) Complementarity states that wave and particle descriptions are mutually exclusive in any single experiment but both are necessary for a complete description of quantum systems (1 mark).

(d) When which-path information is obtained, the wavefunction collapses from a superposition of both paths to a definite single path. The interference pattern disappears and the screen shows particle-like distributions behind each slit (1.5 marks).

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