Determine the maximum kinetic energy of a potassium photoelectron when it is illuminated by beams

Determine the maximum kinetic energy of a potassium photoelectron when it is illuminated by beams with a wavelength of 400 nm, if the work function of electrons for potassium is 2.26 eV

Initial data: λ (wavelength of rays illuminated by the potassium photoelectron) = 400 nm (400 * 10 ^ -9 m); A (work function of electrons for potassium) = 2.26 eV (2.26 * 1.6 * 10 ^ -19 J).

Reference values: h (Planck’s constant) = 6.626 * 10 ^ -34 J * s; C (speed of light) = 3 * 10 ^ 8 m / s.

The maximum energy of the potassium photoelectron is determined from the photoelectric effect equation: h * ν = A + Ek, whence Ek = h * ν – A = h * C / λ – A.

Calculation: Eк = 6.626 * 10 ^ -34 * 3 * 10 ^ 8 / (400 * 10 ^ -9) – 2.26 * 1.6 * 10 ^ -19 = 1.3535 * 10 ^ -19 J.