luclaporte2
contestada

The light intensity incident on a metallic surface with a work function of 1.88 eV produces photoelectrons with a maximum kinetic energy of 1.2 eV.
if the frequency of the light is doubled, what is the maximum kinetic energy of the photoelectrons?
Answer in units of eV.

​

Respuesta :

Answer:

KE = KE (incidental) - KE of emitted photons

or KE = h * f - Wf

So   h * f = KE + Wf = 1.2 + 1.88 = 3.08    incident energy

If you double the frequency then h * f = 6.16

KE = 6.16 - 1.2 = 4.96 eV