1. Photoelectric Effect In a metal, electrons are free to move within the metal, but bound at its surfaces. If an electron approaches the surface, it experiences a retarding potential V 0 it tries to escape. The retarding potential is generated be the vacancy that is produced in the cloud of conduction electrons the instant that the electron leaves the metal.

2. Think of light as an electromagnetic wave. The electron can absorb energy from the wave, so that it has enough kinetic energy T to overcome the retarding potential energy -eV 0. Then it can escape: The greater the intensity, the more energy can be delivered to an escaping electron.

3. Now do the experiment. The wavelength determines whether the electron can escape, not the intensity! The electron leaves with a kinetic energy T = h - eV 0. Light is absorbed like a particle, not a wave!

4. Let’s work some problems 2.23 Particles of a certain system can have energies E 0, 2E 0, or 3E 0 where E 0 = 0.025 eV. a) What is the ratio of the number of particles in each of the upper states to that in the ground state when the system is in equilibrium at 290 o K?

5. b) What is the average energy of a particle in this equilibrium distribution?

6. 3.34. In the photoelectric effect, electrons are emitted instantaneously even if the irradiating light is made extremely weak. Assume that 10 -5 W/m 2 falls on sodium and assume the radius of the sodium atom to be 2 Å. Classically, how long would it take for the atom to absorb 2 eV of energy, which is approximately what would be needed to free an electron? Light power absorbed by one atom The atom must absorb E = 2 eV = 3.2x10 -19 J to free an electron. The time to absorb this energy is T = E/P = (3.2x10 -19 J)/(1.3x10 -25 W) = 2.4x10 6 s ~ 1 month!

7. 3.39 The threshold wavelength for photoemission from a certain material is 0 = 6525 Å. Find the stopping potential when the material is irradiated with: a)light having a wavelength 1 =4000 Å the energy threshold (work function) is the kinetic energy of the emitted electron will be That kinetic energy is what must be cancelled by the stopping potential: V s = T 1 /e = 0.122 V.

8. b) light having twice the energy and three times the intensity of that in part a: The intensity does not affect the stopping potential! c) If a material having double the work function were used, what would then be the answers to parts a and b?