Photoelectric Effect and Intensity
A metal surface has a work function of 2.3 eV. It is illuminated first with light of wavelength 450 nm at low intensity, then with light of wavelength 300 nm at the same intensity, then with 450 nm light at ten times the intensity. A student predicts that increasing intensity at 450 nm will eventually eject electrons because 'more energy per second is delivered', and that the 300 nm light will eject electrons with the same maximum kinetic energy as any other light that ejects electrons. Planck's constant is 6.63 × 10⁻³⁴ J s and the speed of light is 3.00 × 10⁸ m/s.
1. Calculate the photon energy for 450 nm and 300 nm and determine whether each ejects electrons.
Read the worked answer
E = hc/λ.
450 nm: E = (6.63×10⁻³⁴ × 3.00×10⁸)/(450×10⁻⁹) = 1.989×10⁻²⁵/4.50×10⁻⁷ ≈ 4.42×10⁻¹⁹ J. In eV: 4.42×10⁻¹⁹/1.60×10⁻¹⁹ ≈ 2.76 eV.
300 nm: E ≈ 1.989×10⁻²⁵/3.00×10⁻⁷ ≈ 6.63×10⁻¹⁹ J ≈ 4.14 eV.
Work function 2.3 eV. At 450 nm, 2.76 eV > 2.3 eV, so electrons are ejected (max KE ≈ 0.46 eV). At 300 nm, 4.14 eV > 2.3 eV, so electrons are ejected (max KE ≈ 1.84 eV). Both wavelengths exceed the threshold, so both eject electrons.
2. Critique the student's two claims using photon theory.
Read the worked answer
Claim 1 (intensity at 450 nm): wrong. Whether an electron is ejected depends on whether a single photon has energy above the work function, not on total energy per second. At 450 nm each photon already has 2.76 eV > 2.3 eV, so electrons are ejected even at low intensity. Increasing intensity increases the number of photons and hence the number of photoelectrons per second (current), not the maximum kinetic energy or the threshold. If the wavelength were below threshold, no intensity would help.
Claim 2 (same max KE): wrong. Maximum kinetic energy is KE_max = hf − φ, so it depends on photon frequency. The 300 nm light (4.14 eV) gives KE_max ≈ 1.84 eV, much larger than the 0.46 eV from 450 nm. The student confused 'ejects electrons' with 'ejects them with the same energy'.