Ultraviolet light with wavelength λ = 300 nm falls on a cesium surface. the kinetic energy of emitted photoelectrons is 2.1 ev. what is the work function of cesium? what is the maximum wavelength of light for the photoelectric effect to occur for cesium?

The equation of photoelectric effect is given as,

hf = Kmax + work function

Kmax is the kinectic energy of the emitted electron, hf is photon energy transferred to a single electron in the material of the target.

From the equation above,

Work function = hf - Kmax.

hf = hc/wavelength

h = 6.63 x 10^-34J.S, c = 2.98 x 10^8m/s, wavelength = 300nm = 300 x 10^-9m, Kmax = 2.2eV = 2.2 x 1.6 x10^-19.

Work function = [( 6.63 x 10^-34 J.S * 2.28 x 10^8 m/s)/(300 x 10^-9)] - [2.2 * 1.6 x 10^-19]

work function = 3.07 x 10^-19V = 1.92eV

Maximum wavelength of light for photoelectric effect to occur in cesium

Wavelength = c/f. with f known, that wavelength can be calculated from the equation