In a photoelectric effect experiment, light of wavelength 300 nm shines on a metal surface with work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Planck constant h = 4.14 × 10^-15 eV·s, speed of light c = 3.00 × 10^8 m/s, 1 nm = 10^-9 m)

Correct answer

A. 2.14 eV

Principle or equation

Einstein's photoelectric equation: K_max = hf - φ, where hf is photon energy and φ is work function.

Why this answer is correct

First find photon energy: E = hc/λ = (4.14 × 10^-15 eV·s × 3.00 × 10^8 m/s) / (300 × 10^-9 m) = (1.242 × 10^-6 eV·m) / (3.00 × 10^-7 m) = 4.14 eV. Then K_max = 4.14 eV - 2.0 eV = 2.14 eV.

Example

For light of 400 nm (E = 3.10 eV) and φ = 2.0 eV, K_max = 1.10 eV.