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Physics · 8. Quantum physics and photons

Light quanta, photons and the photoelectric effect

Investigate photon energy, metal work function, stopping voltage and experimental determination of Planck’s constant h.

40 minutesContent version: 2.1
Start the interactive pathwayOpen the virtual experiment

Driving question

Why does intense red light fail to liberate electrons, whereas faint violet light instantly triggers photoemission?

Curriculum coverage

Vietnam 2018 Physics 12 (Quantum physics); Cambridge A-Level Physics 9702 Topic 22; AP Physics 2.

Model scope and limitations

Considers clean metallic photoemission in vacuum; neglects multiple inelastic subsurface scattering.

Learning objectives

  • Apply Planck-Einstein quantum hypothesis: photon energy E = hf = hc/λ.
  • Apply Einstein’s photoelectric equation hf = A + Ek_max to find threshold wavelength λ0 and stopping potential Uh.
  • Determine Planck’s constant h from the gradient of stopping potential Uh versus light frequency f.

Prerequisite knowledge

  • Electromagnetic wave speed, wavelength and frequency (c = λf).
  • Work by electric field and electron-volt unit conversion (1 eV = 1.6×10⁻¹⁹ J).

Core knowledge

Light photon

Light consists of discrete quanta (photons) of energy E = hf. Intensity scales with photon flux, leaving individual photon energy unchanged.

Work function and threshold

An electron is ejected only if absorbed photon energy equals or exceeds the work function A (hf ≥ A, λ ≤ λ0 = hc/A).

Stopping potential

A reverse voltage Uh that halts the fastest photoelectrons satisfies eUh = Ek_max = hf − A. The Uh–f slope is precisely h/e.

Worked example

Shining λ = 300 nm (f = 1.0×10¹⁵ Hz) onto metal with work function A = 2.2 eV: photon energy E = 4.13 eV. Max KE = 4.13 − 2.2 = 1.93 eV, yielding stopping voltage Uh = 1.93 V.

Misconceptions to avoid

  • Increasing light intensity boosts the number of ejected electrons per second (saturation current), but does not alter maximum kinetic energy.
  • Photoemission occurs near-instantaneously (<10⁻⁹ s), contradicting wave theory which predicts a prolonged energy accumulation delay.

Virtual experiment procedure: Photoelectric effect and stopping potential

  1. Select cathode material (e.g. Cs, Na, or Zn) inside the photocell.
  2. Vary monochromatic frequency f and adjust reverse voltage until galvanometer current reads zero (Uh).
  3. Tabulate Uh versus f, plot the linear trend line and compute h = e × (gradient).

Safety and cautions

  • Mercury excitation lamps emit UV radiation; avoid direct eye exposure and wear UV-rated protective goggles.

Evidence to collect

  • Tabulated data of wavelength λ, frequency f, and measured stopping voltage Uh.
  • Linear plot of Uh vs f demonstrating slope near 4.14×10⁻¹⁵ V·s and y-intercept −A/e.
Open the virtual experiment
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Diagnostic check

1. If the intensity of light above the threshold frequency is doubled, which quantity doubles?

  1. Saturation photocurrent
  2. Maximum electron kinetic energy
  3. Stopping potential
Show answer and explanation

Correct answer: Saturation photocurrent

Doubling intensity doubles incident photon flux and thus doubles the number of ejected electrons.

2. What relation links stopping potential Uh to maximum kinetic energy of photoelectrons?

  1. eUh = Ek_max
  2. Uh = e × Ek_max
  3. Uh = Ek_max / c
Show answer and explanation

Correct answer: eUh = Ek_max

The electric field work eUh completely halts the fastest electrons reaching the anode.

Application practice

1. A metal has work function A = 2.5 eV. What is its threshold wavelength λ0 (hc ≈ 1240 eV·nm)?

  1. 496 nm
  2. 310 nm
  3. 620 nm
Show answer and explanation

Correct answer: 496 nm

λ0 = hc / A = 1240 / 2.5 = 496 nm (visible spectrum).

2. A Uh vs f graph has gradient 4.14×10⁻¹⁵ V·s. With e = 1.6×10⁻¹⁹ C, what is Planck’s constant h?

  1. 6.62×10⁻³⁴ J·s
  2. 4.14×10⁻³⁴ J·s
  3. 1.60×10⁻³⁴ J·s
Show answer and explanation

Correct answer: 6.62×10⁻³⁴ J·s

h = e × gradient = 1.6×10⁻¹⁹ × 4.14×10⁻¹⁵ ≈ 6.62×10⁻³⁴ J·s.

3. Light at 400 nm hits a cathode of work function 2.0 eV (photon energy 3.1 eV). What is the stopping voltage?

  1. 1.1 V
  2. 3.1 V
  3. 5.1 V
Show answer and explanation

Correct answer: 1.1 V

eUh = hf − A = 3.1 − 2.0 = 1.1 eV, so Uh = 1.1 V.

Academic references

  1. Cambridge IGCSE Physics 0625 — official syllabus and resources
  2. Cambridge AS & A Level Physics 9702 — official syllabus
  3. OpenStax Physics — peer-reviewed high-school physics
  4. BIPM — SI Brochure, 9th edition

This lesson is maintained against the cited sources below. Simulations are learning models and do not replace supervised physical-laboratory safety procedures.

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WebLab STEM
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