Question

UV radiation having a wavelength of 125 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kinetic energy of the ejected photoelectrons?

Answer 1

1) Find the photon energy, E.
E=hc/(lambda) = 6.63x10^-34 x 3x10^8 / (125x10^-9)
=1.5912x10^-18 J

Turn this to eV using 1eV 1.6x10^-19J
E = (1.5912x10^-18) / (1.6x10^-19) = 9.945 eV

The energy from the photon gets completely 'used up'. Some is used knocking out the electron(the work function). Whatever is left is given to the electron as kinetic energy. (In fact the electron may lose some of its kinetic energy on the way out, colliding with atoms). In equation form:

(Photon energy) = (Work function) + (Max KE of electron) (This is Einstein's photoelectric equation)
E = phi + KE-max
9.945 = 4.82 + KE.max
KE_max = 9.945 - 4.820 = 5.125eV