understand that the photoelectric effect provides evidence for a particulate nature of electromagnetic radiation while phenomena such as interference and diffraction provide evidence for a wave nature

Wave‑Particle Duality (Cambridge International AS & A Level Physics 9702 – 22.1 – 22.4)

Learning Objective

Explain how the photoelectric effect demonstrates the particulate nature of electromagnetic radiation, while interference, diffraction and (optionally) Compton scattering demonstrate its wave nature. Extend the discussion to atomic energy levels, line spectra and the de Broglie relation.

1. Photons – Energy, Momentum and de Broglie Wavelength (Syllabus 22.1)

• Energy of a photon: \(E = h\nu = \dfrac{hc}{\lambda}\)
• Momentum of a photon: \(p = \dfrac{h}{\lambda}\)
• de Broglie relation (general particle‑wave link): \(\lambda = \dfrac{h}{p}\)
• Constants: \(h = 6.626\times10^{-34}\,\text{J·s}\), \(c = 3.00\times10^{8}\,\text{m s}^{-1}\)

Key Equations

\$E = h\nu \qquad p = \frac{h}{\lambda} \qquad \lambda = \frac{h}{p}\$

Worked example – photon energy & momentum

Find the energy and momentum of a photon of wavelength 500 nm (green light).

1. Convert wavelength to metres: \(\lambda = 500\,\text{nm}=5.00\times10^{-7}\,\text{m}\).
2. Energy: \(E = \dfrac{hc}{\lambda}= \dfrac{(6.626\times10^{-34})(3.00\times10^{8})}{5.00\times10^{-7}} = 3.97\times10^{-19}\,\text{J}\)

   (≈ 2.48 eV).

3. Momentum: \(p = \dfrac{h}{\lambda}= \dfrac{6.626\times10^{-34}}{5.00\times10^{-7}} = 1.33\times10^{-27}\,\text{kg m s}^{-1}\).

2. Photoelectric Effect – Evidence for the Particle Nature (Syllabus 22.2)

Experimental observations

• The maximum kinetic energy of emitted electrons depends on the frequency of the incident light, not on its intensity.
• No electrons are emitted below a material‑specific threshold frequency \(\nu_{0}\), irrespective of intensity.
• The emission is essentially instantaneous – electrons are ejected as soon as a photon of sufficient energy reaches the surface.

Einstein’s photon model (1905)

Light consists of discrete packets (photons) each carrying energy \(E = h\nu\). A single photon can transfer its whole energy to one electron.

Photoelectric equation (including the electron charge)

\$eV_{s}=h\nu-\phi\$

• \(e\) – elementary charge (\(1.60\times10^{-19}\,\text{C}\)).
• \(V_{s}\) – stopping (retarding) potential measured experimentally.
• \(\phi\) – work function of the metal (minimum energy required to liberate an electron).
• \(K{\max}=eV{s}\) – maximum kinetic energy of the emitted electron.

Consequences for exam‑style questions

1. Intensity effect: Raising the light intensity increases the number of photons, therefore the photocurrent rises, but \(V{s}\) (and \(K{\max}\)) remain unchanged.
2. Frequency effect: Raising the frequency increases \(V{s}\) linearly; the slope of a \(V{s}\) vs \(\nu\) graph gives \(h/e\).

Typical classroom demonstration

1. Use a monochromatic LED or laser of known wavelength (hence known \(\nu\)).
2. Illuminate a clean metal plate (e.g., potassium) connected to a variable retarding voltage.
3. Measure the stopping potential \(V_{s}\) for at least three different frequencies.
4. Plot \(V_{s}\) (y‑axis) against \(\nu\) (x‑axis). The straight‑line fit yields:

   • Slope = \(h/e\) → allows calculation of Planck’s constant.
   • Intercept on the \(\nu\)‑axis = \(\phi/e\) → work function.

Sample data‑table layout

3. Interference, Diffraction and (Optional) Compton Scattering – Evidence for the Wave Nature (Syllabus 22.3)

3.1 Double‑slit interference

• Coherent light passing through two narrow, parallel slits produces alternating bright and dark fringes on a distant screen.
• Constructive (bright) condition: \(d\sin\theta = m\lambda\) \(m = 0,1,2,\dots\)
• Destructive (dark) condition: \(d\sin\theta = \left(m+\tfrac12\right)\lambda\).
• Fringe spacing on a screen at distance \(L\): \(y = \dfrac{m\lambda L}{d}\) (small‑angle approximation).

3.2 Single‑slit diffraction

• A single narrow aperture spreads the incident wavefront, giving a central maximum flanked by minima.
• First‑order minima: \(a\sin\theta = m\lambda\) \(m = 1,2,3,\dots\)
• For small angles, \(\sin\theta \approx \tan\theta = y/L\), so the position of the first minimum is \(y = \dfrac{\lambda L}{a}\).

3.3 Compton scattering (optional extension – Syllabus 22.3)

When X‑ray photons collide with loosely bound electrons, the scattered photons emerge with a longer wavelength. The shift depends only on the scattering angle \(\theta\).

\$\Delta\lambda = \lambda' - \lambda = \frac{h}{m_{e}c}\,(1-\cos\theta)\$

• \(m_{e}=9.11\times10^{-31}\,\text{kg}\) – electron rest mass.
• \(c = 3.00\times10^{8}\,\text{m s}^{-1}\).
• \(\dfrac{h}{m_{e}c}=2.43\times10^{-12}\,\text{m}\) – the Compton wavelength of the electron.

Observation of the wavelength shift confirms that photons carry momentum \(p = h/\lambda\), a particle property.

Interpretation

All three phenomena require the principle of superposition and give quantitative relationships that involve the wavelength \(\lambda\). They therefore provide strong experimental evidence that light behaves as a wave under the appropriate conditions.

4. Atomic Energy Levels, Line Spectra and the Wave Nature of Electrons (Syllabus 22.4)

Quantised atomic energy levels

• Electrons in atoms occupy discrete energy states \(E_{n}\) (n = 1, 2, 3 …).
• A transition between an initial level \(E{i}\) and a final level \(E{f}\) involves absorption or emission of a photon.

Photon energy for an atomic transition

\$h\nu = \Delta E = |E{i}-E{f}|\$

Thus the frequency (or wavelength) of the emitted/absorbed radiation is a direct fingerprint of the energy difference.

Emission and absorption spectra

• Emission spectrum: Electrons drop from higher to lower levels, producing bright lines at characteristic wavelengths.
• Absorption spectrum: Electrons absorb photons and jump to higher levels, leaving dark lines in an otherwise continuous spectrum.

Example – Hydrogen Balmer series

The visible lines obey the Rydberg formula:

\$\frac{1}{\lambda}=R_{H}\!\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right),\qquad n=3,4,5,\dots\$

• \(R_{H}=1.097\times10^{7}\,\text{m}^{-1}\).
• Resulting wavelengths: H\(\alpha\) (656 nm), H\(\beta\) (486 nm), H\(\gamma\) (434 nm), …

Link to wave‑particle duality

• The quantised energy levels arise from the wave nature of electrons (de Broglie condition: an integer number of electron wavelengths fits around the nucleus).
• The photons emitted or absorbed behave as particles, each carrying the precise energy \(h\nu\) dictated by the level separation.

5. Synthesis – Complementarity

Bohr’s principle of complementarity states that wave and particle descriptions are mutually exclusive in a single experiment but together give a complete picture of quantum phenomena. The photoelectric effect, interference/diffraction, Compton scattering and atomic spectra collectively illustrate this dual character.

6. Summary Table

7. Suggested Classroom Activities

• Photoelectric experiment: Measure stopping potentials for three monochromatic LEDs, plot \(V_{s}\) vs. \(\nu\) and determine \(h/e\) and \(\phi/e\).
• Double‑slit with a laser: Record the fringe pattern, measure fringe spacing \(y\) on a screen at distance \(L\), and calculate \(\lambda = \dfrac{dy}{mL}\).
• Single‑slit diffraction: Use a slit of known width, locate the first minima, and verify \(a\sin\theta = \lambda\).
• Compton scattering demonstration (optional): Use a monochromatic X‑ray source and a movable detector to measure the wavelength shift at different angles.
• Spectrum analysis: Observe the hydrogen emission spectrum with a diffraction grating, identify Balmer lines and calculate the Rydberg constant.

8. Key Equations for Revision

• \(E = h\nu = \dfrac{hc}{\lambda}\) – photon energy
• \(p = \dfrac{h}{\lambda}\) – photon momentum
• \(\lambda = \dfrac{h}{p}\) – de Broglie wavelength (general particle‑wave relation)
• \(eV_{s}=h\nu-\phi\) – photoelectric (stopping‑potential) equation
• \(d\sin\theta = m\lambda\) – double‑slit constructive interference
• \(a\sin\theta = m\lambda\) – single‑slit diffraction minima
• \(\Delta\lambda = \dfrac{h}{m_{e}c}(1-\cos\theta)\) – Compton wavelength shift
• \(h\nu = |E{i}-E{f}|\) – photon energy for atomic transitions
• \(\displaystyle\frac{1}{\lambda}=R{H}\!\left(\frac{1}{n{1}^{2}}-\frac{1}{n_{2}^{2}}\right)\) – Rydberg formula for hydrogen

9. Concluding Remarks

The photoelectric effect, interference, diffraction, Compton scattering and atomic spectra together provide compelling, experimentally verified evidence that electromagnetic radiation (and, by extension, all quantum particles) possesses a dual wave‑particle character. Mastery of the observations, equations and the way they interlink is essential for success in the Cambridge AS & A Level Physics examinations and forms the foundation for further study of quantum mechanics.