# 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?