Determine the maximum kinetic energy of a photoelectron of potassium when it is illuminated with light with

Determine the maximum kinetic energy of a photoelectron of potassium when it is illuminated with light with a wavelength of 400nm if the work function of electrons in potassium is 3.6 * 10 ^ -19

Data: λ (length of the light wave illuminating the photoelectron) = 400 nm = 400 * 10 ^ -9 m; A (work function) = 3.6 * 10 ^ -19 J.

Constants: Vc (light wave speed) = 3 * 10 ^ 8 m / s; h (Planck’s constant) = 6.626 * 10 ^ -34 J * s.

To determine the maximum kinetic energy of a photoelectron, we use the equation of the photoelectric effect: h * ν = A + Eк and Eк = h * ν – A = h * Vc / λ – A.

Calculation: Eк = 6.626 * 10 ^ -34 * 3 * 10 ^ 8 / (400 * 10 ^ -9) – 3.6 * 10 ^ -19 = 4.9695 * 10 ^ -19 – 3.6 * 10 ^ -19 = 1.3695 * 10 ^ -19 J.