In a photoelectric effect experiment, light of wavelength 400 nm shines on a metal surface with work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons? (Use h = 4.14 × 10⁻¹⁵ eV·s and c = 3.00 × 10⁸ m/s.)

Correct answer

A. 1.1 eV

Principle or equation

Einstein's photoelectric equation: K_max = hf - φ, where f = c/λ.

Why this answer is correct

First compute the photon energy: E = hc/λ = (4.14 × 10⁻¹⁵ eV·s)(3.00 × 10⁸ m/s) / (400 × 10⁻⁹ m) = 3.105 eV. Then K_max = 3.105 eV - 2.0 eV = 1.105 eV, approximately 1.1 eV.

Example

For λ = 300 nm and φ = 2.0 eV, photon energy = 4.14 eV, so K_max = 2.14 eV.