Photons of energy 4.20 eV bombard a surface,which emits photoelectrons with kinetic energies for 0 to 1.80 eV. What is the threshold frequency for this surface?

Question

Quizplus · Accepted Answer

The work function (ϕ) is the difference between the photon energy and the maximum kinetic energy of the emitted photoelectrons: ϕ = 4.20 eV - 1.80 eV = 2.40 eV. The threshold frequency (f₀) is given by ϕ = h*f₀, where h is Planck's constant (4.135667696 × 10⁻¹⁵ eV·s). Solving for f₀ gives approximately 5.81 * 10¹⁴ Hz.

Photons of Energy 4 ×\times× 1014 Hz B) 5 ×\times× 1014 Hz C) 1

Photons of Energy 4 $\times$ 10¹⁴ Hz
B) 5 $\times$ 10¹⁴ Hz
C) 1