Question

9. When electromagnetic radiation of frequency 1.5 x 1014 Hz is incident on a metal surface, the

maximum kinetic energy of the emitted photoelectrons is found to be 3.8 x 10-20 J. Calculate the
work function of the metal.
10. Photons of electromagnetic radiation having energies of 1.0 eV, 2.0 eV and 4.0 eV are incident on a
metal surface having a work function of 1.7 eV.
a) Which of these photons will cause photoemission from the metal surface?
b) Calculate the maximum kinetic energies (in eV and J) of the liberated electrons in each of
those cases where photoemission occurs.

Answer 1

9. Using the equation KEmax = hf - Φ, where KEmax is the maximum kinetic energy of the emitted photoelectrons, h is Planck's constant, f is the frequency of the radiation, and Φ is the work function of the metal, we can rearrange to find Φ:

Φ = hf - KEmax

Φ = (6.63 x 10^-34 J s)(1.5 x 10^14 Hz) - 3.8 x 10^-20 J

Φ = 9.94 x 10^-20 J

Therefore, the work function of the metal is 9.94 x 10^-20 J.

10. a) Only photons with energies greater than or equal to the work function of the metal (1.7 eV) will cause photoemission. Thus, the photon with an energy of 2.0 eV and the photon with an energy of 4.0 eV will cause photoemission, but the photon with an energy of 1.0 eV will not.

b) For the photon with an energy of 2.0 eV:

KEmax = hf - Φ

KEmax = (6.63 x 10^-34 J s)(3.2 x 10^14 Hz) - 1.7 eV

KEmax = 3.23 x 10^-19 J or 2.0 eV

For the photon with an energy of 4.0 eV:

KEmax = hf - Φ

KEmax = (6.63 x 10^-34 J s)(6.4 x 10^14 Hz) - 1.7 eV

KEmax = 5.13 x 10^-19 J or 4.0 eV