# Maximum kinetic energy / cutoff potential problem

## Homework Statement

Light with a wavelength of 630nm is directed at a metallic surface that has a work function of 1.5 eV. Find:

a) The maximum kinetic energy of the emitted electrons.

b) The cutoff potential required to stop the photoelectrons.

Ek = hc / λ - W

Ek = eVo

## The Attempt at a Solution

a)
λ = 630nm = 6.3 x10-7m

W = 1.5eV (1.60x10-19J/eV)

W = 2.4 x10-19J

Ek = hc / λ - W

Ek = (6.63 x10-34J)(3.0 x108m/s) / (6.3 x10-7m) - 2.4 x10-19J

Ek = 7.57 x10-20J

The maximum kinetic energy is 7.57 x10-20J.

b)

Ek = eVo

7.57 x10-20J = (1.60 x10-19) Vo

Vo = 7.57 x10-20J / (1.60 x10-19)

Vo = 0.47 V

The cutoff is 0.47 V.

I am uncertain if my answer to part a) is correct, I have reviewed all my steps and can't find an error, however, for all of the questions I have done so far, the answer has always been x10-19, whereas my answer here, is 7.57 x10-20J is x10-20. Is this an acceptable answer? Is it possible for it to not always be to the -19J? or did I make an error somewhere in my calculations?

It looks right. Let me suggest for future reference that, for problems of this sort, you do your calculations in eV. Then all you have to do is convert the photon energy to eV. In this particular problem, the answers to (a) and (b) are identical in number because the max. kinetic energy of the electrons in eV is the same as the cutoff potential in Volts. You can always convert to Joules in the end if needed. 