1bPhotoelectric Effect
Explore how light ejects electrons from metals, why classical physics failed to explain it, and how Einstein's photon model earned the Nobel Prize.
Einstein's photon model of the photoelectric effect proved light has particle-like properties — a discovery that underlies solar panels, digital cameras, and every light sensor in modern technology.
Lesson Overview
The photoelectric effect occurs when light shining on a metal surface ejects electrons. Classical wave theory predicted that any frequency of light — given enough intensity — should eject electrons. Experiment proved otherwise: only light above a threshold frequency works, regardless of intensity. In 1905, Einstein explained this by treating light as particles called photons, each carrying energy E = hf. This explanation earned him the 1905 Nobel Prize in Physics and confirmed the quantum nature of light.
Key Concepts
Photon
A quantum (particle) of light with energy E = hf
Work Function (φ)
Minimum energy needed to eject an electron from a metal surface
Threshold Frequency
f₀ = φ/h — minimum frequency of light that can eject electrons
Kinetic Energy of Electron
KE_max = hf − φ — energy left after overcoming work function
Stopping Potential
Voltage V_s that stops the fastest electrons: eV_s = KE_max
Applications
Solar cells, photodetectors, digital cameras, night-vision devices
Light of frequency 8.0×10¹⁴ Hz strikes a metal with work function φ = 2.0 eV. Find the maximum kinetic energy of ejected electrons. (1 eV = 1.6×10⁻¹⁹ J)
A metal has a work function of 4.5 eV. What is the threshold frequency for the photoelectric effect?
Explain why increasing the intensity of red light cannot eject electrons from a metal with threshold frequency in the UV range.
Electrons ejected from a surface have maximum KE = 1.8 eV. What stopping potential is needed to halt them?
UV light of wavelength 200 nm hits a sodium surface (φ = 2.3 eV). Find (a) photon energy and (b) maximum KE of ejected electrons.
Light of wavelength 400 nm hits a metal with work function 1.9 eV. Does the photoelectric effect occur? If so, find KE_max.
Hint: First find photon energy: E = hc/λ. Then compare to φ. If E > φ, electrons are ejected with KE_max = E − φ.
A metal has threshold wavelength λ₀ = 350 nm. What is its work function in eV?
Hint: At threshold, all photon energy goes to overcoming the work function: φ = hf₀ = hc/λ₀.
If the stopping potential for a metal is 2.5 V, what is the maximum kinetic energy of the ejected electrons in joules?
Hint: KE_max = eV_s where e = 1.6×10⁻¹⁹ C.
Two metals A and B are illuminated with the same UV light. Metal A ejects electrons; metal B does not. What can you conclude about their work functions?
Hint: Compare the photon energy to each metal's work function. Which metal has a higher work function?
Doubling the frequency of light used in the photoelectric effect — what happens to (a) the number of electrons ejected and (b) their maximum kinetic energy?
Hint: Number of electrons depends on intensity (number of photons). KE_max = hf − φ — how does doubling f affect this?
Key Vocabulary
Photon
A discrete quantum (particle) of electromagnetic radiation with energy E = hf, where h is Planck's constant and f is frequency.
Example: A photon of blue light (f ≈ 7×10¹⁴ Hz) carries about 4.6×10⁻¹⁹ J of energy.
Work Function (φ)
The minimum energy required to remove an electron from the surface of a specific metal.
Example: Sodium has φ ≈ 2.3 eV; gold has φ ≈ 5.1 eV — gold requires higher-energy photons to eject electrons.
Threshold Frequency (f₀)
The minimum photon frequency needed to eject an electron: f₀ = φ/h. Below this frequency, no electrons are emitted regardless of intensity.
Example: For a metal with φ = 3.0 eV, f₀ = (3.0 × 1.6×10⁻¹⁹) / 6.626×10⁻³⁴ ≈ 7.25×10¹⁴ Hz.
Stopping Potential (V_s)
The reverse voltage required to stop the most energetic photoelectrons: eV_s = KE_max.
Example: If KE_max = 2.0 eV, the stopping potential is 2.0 V.
Interactive Practice — 5 Questions
In the photoelectric effect, what determines whether electrons are ejected?
The maximum kinetic energy of photoelectrons is given by:
Einstein's explanation of the photoelectric effect proposed that light:
Stopping potential V_s is related to maximum kinetic energy by:
Which application directly uses the photoelectric effect?
Independent Practice
Light of frequency 1.2×10¹⁵ Hz strikes a metal with work function 3.5 eV. Calculate the maximum kinetic energy of ejected electrons in both joules and eV.
A metal has a threshold wavelength of 520 nm. Calculate its work function in eV.
Explain why the photoelectric effect cannot be explained by classical wave theory, and how Einstein's photon model resolves each discrepancy.
If the stopping potential for a photoelectric experiment is 1.6 V, what is the maximum speed of the ejected electrons? (m_e = 9.11×10⁻³¹ kg)
★ A photovoltaic solar cell uses silicon with a band gap of 1.1 eV (similar to a work function). Calculate the maximum wavelength of light that can generate electricity in this cell, and explain why silicon solar cells are efficient for sunlight.
ChallengeCommon Mistakes
Thinking that brighter (more intense) light will eventually eject electrons even below threshold frequency.
Intensity only increases the number of photons, not their energy. If hf < φ, no individual photon can eject an electron — ever.
Confusing stopping potential (volts) with kinetic energy (joules or eV).
KE_max = eV_s. The stopping potential in volts numerically equals KE_max in eV (since e × V_s in eV = V_s).
Math Tips
When working in eV: photon energy E(eV) = hf / (1.6×10⁻¹⁹). Then KE_max(eV) = E(eV) − φ(eV). The stopping potential in volts equals KE_max in eV — a convenient shortcut.