Question

When a metal was exposed to one photon of light at a frequency of 4.55× 1015 s–1, one electron was emitted with a kinetic energy of 4.10× 10–19 J. 1)Calculate the work function of this metal. 2) What is the maximum number of electrons that could be ejected from this metal by a burst of photons (at some other frequency) with a total energy of 5.11× 10–7 J?

Answer 1

1) work function = hv/1.602 x 10^-19

                          = 6.626 x 10^-34 x 4.55 x 10^15/1.602 x 10^-19

                          = 18.82 eV

2) Ephoton = hv - KE

                   = 6.626 x 10^-34 x 4.55 x 10^15 - 4.10 x 10^-19

                   = 2.605 x 10^-18 J

Maximum number of electrons ejected = 5.11 x 10^-7/2.605 x 10^-18 = 1.96 x 10^11 electrons