Question

Rubidium has a minimum energy of 4.72x10^-21 J/electron for the photoelectric effect to be observed. Compare each situation in low intensity light with the photons having an energy of 4.72x10^-21 J/photon (referred to as Lamp A).

a. Rubidium is illuminated with a medium intensity light; the photons that have an energy of 6.28x10^-22 J/photon. I. How many electrons are ejected? II. What will be the kinetic energy of the ejected electrons? b. Rubidium is illuminated with a high intensity light; the photons that have an energy of 3.42x10^-22 J/photon. I. How many electrons will be ejected? II. What will be the kinetic energy of the ejected electrons?

Answer 1

1.a. Since the photons have energy less than yhe minimum ionisations energy needed to knock out the electron frm the atom, the electrons wont be ejected out. So 0 electrons will be ejected out. In case of the 1st case mentioned the energy is just sufficient but the ejected electrons wld collide with the inner atoms of the metal nd hence wld come to rest. so the KE wld anyway be 0.

1.b. KE = 0.

2.a/b Same as in 1. Increasing intensity hs no effect unless the energy per photon is higher than the minimum required energy to observe photoelectric effect

so in both the cASES PHOTOELECTRIC EFFECT WONT BE OBSERVED.