What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.6*1015Hz?

To answer this question, we will use the following formula:

hf = phi + Emax

where

h = plancks constant = 6.626x10^-34 J s

f = frenquency of incident light = 1.6*10^15 Hz

phi = work function of the cesium = 2.14 eV = 3.428x10^-19 J

Emax = maximum kinetic energy of the emitted electrons.

Substitute with the givens in the above equation to calculate Emax as follows:

6.626x10^-34 x 1.6*10^15 = 3.428x10^-19 + Emax

Emax = 7.1736 x 10^-19 J