# 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)?

Thank you for your help.

## Answers and Replies

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

Thank you for your assistance.