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
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.