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## 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.