Question

The work function of cesium is 1.96 eV. If radiation of wavelength 4.00 × 10^2 nm is incident on the surface, find the kinetic energy of the ejected photoelectrons in eV and the speed of the ejected electrons.

Answer 1

hc/lamda=work funtion+0.5*mv^2

Answer 2

Here
h=planks's constant
Lamda=4*10^2nm=4*10^2*10^-6m
Work function=1.96ev=1.96*1.6*10^-19J

Answer 3

c=light velocity
m=mass of electron=9.1*10^-31kg
v=?

Answer 4

Response plz!!!!

Answer 5

I don't know the velocity

Answer 6

Frm my given equation can u try to find out v=?

Answer 7

Plz feel free to ask for clearing ur confusion!!!!