Skip to main content
Physics 4% exam weight

Modern Physics: Photoelectric Effect

Part of the WAEC WASSCE study roadmap. Physics topic phy-16 of Physics.

→ See in Roadmap WAEC WASSCE Study Plans Build 1-Month Plan
By Last updated 4% exam weight

Modern Physics: Photoelectric Effect

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

The photoelectric effect is the ejection of electrons from a metal surface when light of frequency above a critical value strikes it. Each photon delivers a quantum of energy E = hf (h = 6.63 × 10⁻³⁴ J·s). Emission happens only when f > f₀, where f₀ is the threshold frequency tied to the metal’s work function (φ) by φ = hf₀.

  • Einstein’s photoelectric equation: KE_max = hf − φ = eV₀, where V₀ is the stopping potential.
  • Increasing intensity raises the number of photoelectrons per second, not their kinetic energy.
  • Increasing frequency raises the kinetic energy of emitted electrons.

For WAEC Paper 2, expect a 4–6 mark structured question asking you to compute φ, f₀, λ₀, KE_max or V₀ from given numbers.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Einstein’s Quantum Explanation

Classical wave theory predicted that higher light intensity should eject electrons of greater energy from any metal. Experiments by Lenard and Millikan contradicted this: electron energy depended only on frequency, and emission stopped below a sharp threshold frequency f₀. Einstein (1905) resolved this by treating light as photons — discrete packets each carrying energy E = hf.

The photon is absorbed by a single bound electron. Part of its energy overcomes the metal’s work function φ (the minimum energy needed to liberate an electron); the remainder becomes the electron’s kinetic energy.

Core Formulas

FormulaMeaning
E = hfPhoton energy (J), with h = 6.63 × 10⁻³⁴ J·s
KE_max = hf − φEinstein’s photoelectric equation
φ = hf₀ = hc/λ₀Work function linked to threshold frequency f₀ and threshold wavelength λ₀
KE_max = eV₀Maximum KE equals charge × stopping potential

Standard Problem Types in WAEC

  • Find work function when f₀ is given: φ = hf₀.
  • Convert between threshold wavelength and frequency: λ₀ = c/f₀ (c = 3 × 10⁸ m/s).
  • Compute maximum KE of ejected electrons from incident wavelength.
  • Calculate stopping potential V₀ = KE_max / e, often expressed in volts (1 eV = 1.6 × 10⁻¹⁹ J).

Common student error: plugging λ (in nm) directly into E = hc/λ without converting to metres. Always convert nanometres to metres before substituting.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Adjacent Connections

The threshold wavelength λ₀ marks the boundary where photon energy exactly equals the work function; electrons are liberated with zero kinetic energy, so the photoelectric current just begins. Just above f₀, KE_max grows linearly with frequency, producing a straight-line graph of KE_max against f whose slope equals h and whose x-intercept equals f₀. This graph is a favourite WAEC drawing question.

The effect links directly to wave-particle duality: photoelectric emission demonstrates the particle nature, while interference demonstrates the wave nature. Threshold considerations also underpin solar cell design, since semiconductors are engineered with low work functions to capture visible photons efficiently.

Common Examiner Traps

TrapCorrect Idea
”Brighter light → faster electrons”Intensity changes count, not speed
”Below f₀, intense light still ejects electrons”Below f₀, no electrons are emitted at all
Confusing V₀ with anode voltageV₀ is the retarding potential that stops electrons
Mixing units of eV and J in E = hfConvert: 1 eV = 1.6 × 10⁻¹⁹ J

Worked Micro-Example

Light of wavelength 400 nm falls on a metal whose work function is 1.9 eV. Photon energy E = hc/λ = (6.63 × 10⁻³⁴)(3 × 10⁸)/(400 × 10⁻⁹) = 4.97 × 10⁻¹⁹ J ≈ 3.11 eV. Then KE_max = 3.11 − 1.9 = 1.21 eV, and V₀ = 1.21 V.

Practice Prompts

  1. A metal has φ = 2.5 eV. Find the threshold wavelength and the stopping potential when λ = 300 nm is used.
  2. Sketch the KE_max versus frequency graph for two metals and explain how h and f₀ are obtained from it.

Continue your study

Content adapted based on your selected roadmap duration. Switch tiers using the selector above.

Sources & verification