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

So3 that’s what i think