Optical Phenomena & the Photoelectric Effect

Limitations of the Wave Model of Light

The classical wave model of light predicted that:

Any frequency of light, given enough intensity, should eject electrons from a metal surface
Higher intensity → more energetic electrons (higher KE)
There should be a time delay before electrons are emitted (wave energy builds up)

None of these predictions matched experimental observations. The photoelectric effect could not be explained by the wave model, prompting Einstein's revolutionary photon model in 1905.

Einstein's Photon Model

Einstein proposed that light consists of discrete packets of energy called photons. Each photon carries a fixed energy determined by its frequency:

E = hf = hc/λ

Symbol	Meaning	Value / Unit
E	Energy of one photon	J
h	Planck's constant	6.63 × 10⁻³⁴ J·s
f	Frequency of light	Hz
c	Speed of light	3.0 × 10⁸ m·s⁻¹
λ	Wavelength of light	m

Higher frequency = higher photon energy. Blue/UV photons carry more energy per photon than red/infrared photons.

The Photoelectric Effect

When light shines on a metal surface, photons can transfer their energy to electrons in the metal. If a photon's energy is sufficient, it can eject an electron from the surface.

Work function (W₀): the minimum energy required to eject an electron from the metal surface. Each metal has a unique W₀.
Threshold frequency (f₀): the minimum frequency of light that can eject an electron. W₀ = hf₀
If f < f₀: no electrons are ejected, regardless of intensity
If f ≥ f₀: electrons are ejected — maximum KE of ejected electrons is:

E_k(max) = hf − W₀ = hf − hf₀ = h(f − f₀)

Key observations that confirm the photon model:
• No time delay: ejection is instantaneous — one photon interacts with one electron
• Threshold frequency exists: below f₀, no electrons are ejected regardless of intensity
• Intensity increases number of electrons (not their energy) — more photons → more ejections
• Higher frequency → greater E_k(max) of ejected electrons
• Increasing intensity at fixed frequency → more electrons but same maximum KE

Worked Example — Photoelectric Effect

Problem: Light of frequency 8.0 × 10¹⁴ Hz shines on sodium (W₀ = 3.6 × 10⁻¹⁹ J). (a) Find the photon energy. (b) Find E_k(max) of ejected electrons. (c) Find the threshold frequency.

(a) E = hf = 6.63×10⁻³⁴ × 8.0×10¹⁴ = 5.30 × 10⁻¹⁹ J
(b) E_k(max) = hf − W₀ = 5.30×10⁻¹⁹ − 3.6×10⁻¹⁹ = 1.70 × 10⁻¹⁹ J
(c) f₀ = W₀/h = 3.6×10⁻¹⁹ / 6.63×10⁻³⁴ = 5.43 × 10¹⁴ Hz

Emission Spectra

When a gas is heated or subjected to an electrical discharge, its atoms absorb energy and electrons jump to higher energy levels. When they fall back down, they emit photons of specific frequencies — producing a line emission spectrum.

Each element has a unique set of spectral lines — like a fingerprint
Only specific frequencies (colours) are emitted — a series of bright lines on a dark background
Used to identify elements in stars and distant galaxies

ΔE = E_higher − E_lower = hf = hc/λ

Absorption Spectra

When white light (all frequencies) passes through a cool gas, atoms absorb exactly the same specific frequencies they would emit when excited. The result is a dark-line (absorption) spectrum — a continuous spectrum with dark lines at exactly the same positions as the emission lines for that element.

Emission = bright coloured lines on dark background. Absorption = dark lines at same positions on a continuous rainbow background.

Bohr Model of the Atom

Niels Bohr proposed that electrons occupy discrete (quantised) energy levels around the nucleus. Electrons can only exist at certain allowed energies — not in between.

Excitation: electron absorbs a photon and jumps to a higher level (E_photon = ΔE)
De-excitation: electron falls to a lower level, emitting a photon of exactly ΔE
Ground state: lowest energy level (n = 1)
Excited states: n = 2, 3, 4, … (higher energy)

E_photon = hf = hc/λ = E_higher − E_lower

For hydrogen: E_n = −13.6/n² eV (where 1 eV = 1.6 × 10⁻¹⁹ J)

IEB Extension — Wave-Particle Duality and de Broglie Wavelength

Einstein's photon model showed that light (previously thought to be a wave) has particle-like properties. de Broglie proposed the converse: matter particles also have wave properties. The de Broglie wavelength of a particle is:

λ = h/p = h/(mv)

This was confirmed by electron diffraction experiments. Wave-particle duality is a cornerstone of quantum mechanics. IEB may also reference Compton scattering — X-ray photons scatter off electrons and lose energy (increase in wavelength), confirming photons carry momentum p = h/λ = E/c.

Compton: Δλ = (h/m_ec)(1 − cos θ) (not required for calculation but context given)

Predict first: Can very bright light below the threshold frequency eject electrons? Then test it: Try a frequency below the metal's threshold, then above it, and compare.

Photoelectric Effect Simulator

Light Source

Frequency (×10¹⁴ Hz) 7.0 ×10¹⁴

Metal W₀ (×10⁻¹⁹ J) 3.6 ×10⁻¹⁹

Energy Analysis

Photon E

—×10⁻¹⁹J

Work fn W₀

—×10⁻¹⁹J

Ek max

—×10⁻¹⁹J

Adjust sliders to simulate

PhET Simulation — Photoelectric Effect

Shine light of adjustable frequency and intensity onto a real choice of target metals and watch electrons actually get ejected (or not) — a free, open simulation from the University of Colorado Boulder's PhET Interactive Simulations project. Complements the numeric energy analysis above with the visual behaviour it's modelling.

Open directly on PhET ↗

Question 1 of 8

The energy of a photon of light with frequency 6.0 × 10¹⁴ Hz is (h = 6.63 × 10⁻³⁴ J·s):

Question 2 of 8

Light of frequency 5 × 10¹⁴ Hz shines on a metal with work function W₀ = 4.0 × 10⁻¹⁹ J. Which of the following is correct?

Question 3 of 8

If the intensity of light (above the threshold frequency) is doubled while the frequency is kept constant, what happens to the photoelectric effect?

Question 4 of 8

A metal has a threshold frequency of 6.0 × 10¹⁴ Hz. Its work function is (h = 6.63 × 10⁻³⁴ J·s):

Question 5 of 8

The dark lines in the absorption spectrum of hydrogen appear at exactly the same wavelengths as the bright lines in hydrogen's emission spectrum. This is because:

Question 6 of 8

An electron in a hydrogen atom falls from n = 3 to n = 2. The energies are E₃ = −1.51 eV and E₂ = −3.40 eV. The wavelength of the emitted photon is (h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m·s⁻¹, 1 eV = 1.6×10⁻¹⁹ J):

Question 7 of 8

Light of wavelength 400 nm strikes a metal with work function 2.3 eV. What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m·s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J)

Question 8 of 8

Two experiments use the SAME metal. Experiment 1 shines very intense red light on it (frequency below the threshold frequency). Experiment 2 shines very dim violet light on it (frequency above the threshold frequency). According to the photon model, what happens?

IEB Extended Questions

IEB Question 1

An electron is accelerated through a potential difference of 100 V. Its mass is 9.11×10⁻³¹ kg. Using λ = h/p and KE = eV (e = 1.6×10⁻¹⁹ C), the de Broglie wavelength of the electron is approximately:

Use h = 6.63 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m·s⁻¹, 1 eV = 1.6 × 10⁻¹⁹ J. Show all substitutions with correct powers of 10 and units.

Question 1 — Photoelectric Effect Calculations (8 marks)

Ultraviolet light of wavelength 200 nm shines on a platinum surface (work function W₀ = 8.0 × 10⁻¹⁹ J). (a) Calculate the frequency of the UV light. (b) Calculate the energy of one photon. (c) Calculate the threshold frequency of platinum. (d) Determine whether electrons will be ejected. Justify your answer. (e) If electrons are ejected, calculate the maximum kinetic energy. (f) Calculate the maximum speed of the ejected electrons (m_e = 9.11 × 10⁻³¹ kg). (g) Explain what would happen if light of the same wavelength but twice the intensity were used. (h) Explain what would happen if light of wavelength 300 nm were used instead.

Question 2 — Emission Spectra and Energy Levels (6 marks)

The energy levels of hydrogen are given by E_n = −13.6/n² eV. (a) Calculate E₁, E₂, E₃, and E₄ in eV. (b) Convert E₁ and E₂ to joules. (c) Calculate the energy of the photon emitted when an electron drops from n=4 to n=2. (d) Calculate the wavelength of this photon. (e) In which part of the electromagnetic spectrum does this wavelength fall? (f) Explain why the absorption spectrum of hydrogen shows dark lines at exactly the same wavelengths as its emission spectrum.

Question 3 — Wave Model vs Photon Model (5 marks)

A physics learner says: "If I shine brighter red light on the metal, the electrons will eventually be ejected because the energy builds up." (a) Identify the model of light the learner is using. (b) State two predictions of the wave model that contradict the photoelectric effect. (c) Explain, using the photon model, why increasing the intensity of red light (below threshold) NEVER ejects electrons. (d) What change to the light source is needed to eject electrons, and why? (e) State which scientist proposed the photon model and in which year.

Question 4 — Spectroscopy Application (5 marks)

An astronomer observes the spectrum of a distant star. The hydrogen-alpha line (normally at 656 nm) appears at 689 nm. (a) Is this an emission or absorption line? Explain. (b) Calculate the energy of a photon at 656 nm. (c) Calculate the energy of a photon at 689 nm. (d) Explain whether the star is moving toward or away from Earth, and name this phenomenon. (e) Using Δλ/λ ≈ v/c, calculate the speed of the star relative to Earth.

Question 5 — Finding h from Stopping Potential Data (8 marks)

A student shines light of different frequencies onto a metal surface and measures the stopping potential V_s (the minimum reverse voltage needed to reduce the photocurrent to zero) for each:

Frequency f (×10¹⁴ Hz)	6.0	7.0	8.0	9.0
Stopping potential V_s (V)	0.41	0.83	1.24	1.66

(a) The photoelectric equation can be written eV_s = hf − W₀. Rearranged as V_s = (h/e)f − W₀/e, explain why a graph of V_s against f should be a straight line, and state what its gradient and its f-intercept represent physically.
(b) Using the first and last data points, calculate the gradient of the V_s vs f relationship.
(c) Since this gradient equals h/e, use it to estimate Planck's constant h (e = 1.6 × 10−19 C). Compare your value to the accepted value 6.63 × 10−34 J·s.
(d) Using the pattern in the table (not a new formula), estimate the threshold frequency f₀ of this metal — the frequency at which V_s would be zero.