The work function for potassium is 2.24eV.

Question

If potassium metal is illuminated with light of wavelength 310 nm,
find the maximum kinetic energy of the photoelectrons.

I have tried everything and still keep getting the wrong answer.

The speed of light is 3 × 108 and Planck’s constant is 6.62607 × 10−34
and the answer suppose to be in units of eV.

i used the equation MaxKE = h*c/ (wavelength x10^-9)

and then subtract (2.24 * 1.602x10^-19)

its not correct apparently, please help

Answer 1

i figured it out after all - thanks

Answer 2

Please include your units for physical quantities. Numbers are not enough.

Answer 3

use:
kmax = hf - e
kmax = h ( f becomes : c over lambda)- e

e = ( Evev * 1.6 *10exp -19)