Question

Calculate the threshold frequency for gold having photoelectric work function equal to 4.8 eV. If light of wavelength 2220 Å falls on gold, what will be maximum kinetic energy of the photo electrons coming out?

Answer 1

Energy of the light photon = 12, 000 / 2220 = 5.58 eV. Out of this, 4.8 eV would be used for dislodging the electron and the balance would represent its kinetic energy.

E_max = 5.58 – 4.8 = 0.78 eV

Alternatively λ₀ =12, 400 / 4.8 = 2583 Å. Hence, we may use