The work function of a metal is 2 eV. What is the maximum kinetic energy of the photoelectrons emitted when light of frequency 8 x 10¹⁴ Hz is incident on the metal? (h = 6.626 x 10⁻³⁴ Js)
Correct: A
The energy of the incident photon E = hf = (6.626 x 10⁻³⁴) * (8 x 10¹⁴) = 5.30 x 10⁻¹⁹ J. Converting the work function to Joules: 2 eV = 2 * 1.6 x 10⁻¹⁹ = 3.2 x 10⁻¹⁹ J. Maximum KE = E - work function = 5.30 x 10⁻¹⁹ - 3.2 x 10⁻¹⁹ = 2.10 x 10⁻¹⁹ J. Converting to eV: KE = (2.10 x 10⁻¹⁹) / (1.6 x 10⁻¹⁹) = 1.31 eV.