# EV vs V, Cutoff Potential Question

## Homework Statement

Light with a wavelength of 630 nm 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

## Homework Equations

Ek = [(h * c) / λ] - W
Ek = e V0, where V0 is the cutoff potential and e = 1.60*10^-19

## The Attempt at a Solution

Ek = [(h * c) / λ] - W = [(1.989*10^-25 * 3.00*10^8) / 6.30*10^-7] - W
Ek = 3.16*10^-19 - (1.5 * 1.60*10^-19) = 3.16*10^-19 - 2.4*10^-19
Ek = 0.76*10^-19 Joules

a) maximum kinetic energy = 0.76*10^-19 Joules

Ek = e V0 therefore V0 = Ek / e
V0 = = 0.76*10^-19 / 1.60*10^-19 = 0.47 V

b) cutoff potential = 0.47 Volts

(b) is where I think I am confused, are volts and electron volts the same? And should the value be negative (electrons have negative charge and are repulsed by negative charges)?

trollcast
Gold Member
are volts and electron volts the same?

The electron volt is a unit of energy equal to 1.6 × 10-19 joules.

So the e in the equation above is not for the conversion of joules to electron volts, but the elementary charge? And are my calculations correct?

trollcast
Gold Member
a)
OK I ran through your calculation quickly and I'm getting a different answer to you.

What are you using as plancks constant, h? You seem to be using 1.989E-25 ?

b)

Imagine the electron leaving the metals surface at a point X and being detected at a point Y.

A potential is applied between point X and Y to deccelerate the electrons. Now since the electron is slowing down work is being done on the electron.

The formula for work done on an electron is, W=e * V , where W is work done, e is the elementary charge and V is the potential the electron is travelling.

The cutoff potential is the voltage between X and Y that just stops any electrons reaching Y. So at that point, W = KE , where KE is the kinetic energy of your electrons.

So re arranging you get Vo = KE / e .

Sorry - I copied it over wrong. The 1.989E-25 is (h * c), where h = 6.63E-34, so it should read

Ek = [(h * c) / λ] - W = [1.989*10^-25 / 6.30*10^-7] - W
Ek = 3.16*10^-19 - (1.5 * 1.60*10^-19) = 3.16*10^-19 - 2.4*10^-19
Ek = 0.76*10^-19 Joules