130_notes.dvi

Lets compare the above SI units numbers to my suggested method of using the fine
structure constant...Putting in the alpha charge of 2e.

U=

6 Ze^2

8 πǫ 0 R

=

3 Zα ̄hc

R

=

(3)(79)(1973)

(137)(5)

eV = 683 eV

This is easier.

2.7 Sample Test Problems

1. What is the maximum wavelength of electromagnetic radiation whichcan eject electrons from
   a metal having a work function of 3 eV? (3 points)
   Answer
   The minimum photon energy needed to knock out an electron is 3 eV. We just need to convert
   that to wavelength.

E = hν= 3eV

ν =

c

λ

2 π ̄hc

λ

= 3eV

λ=

2 π ̄hc

3 eV

=

2 π(1973eV ̊A)

3 eV

≈ 4000 ̊A

2.*Based on classical electromagnetism and statistical mechanics, Rayleigh computed the energy

density inside a cavity. He found that, at a temperature T, the energy density as a function

of frequency was

u(ν,T) =

8 πν^2

c^3

kBT.

Why is this related to black body radiation? Why was this in obvious disagreement with

observation?

3. What is the maximum wavelength of electromagnetic radiation whichcan eject electrons from
   a metal having a work function of 2 eV?

4.*State simply what problem with black-body radiation caused Plank to propose the relation

E=hνfor light.

5. The work function of a metal is 2 eV. Assume a beam of light of wavelengthλis incident upon
   a polished surface of the metal. Plot the maximum electron energy (ineV) of electrons ejected
   from the metal versusλin Angstroms. Be sure to label both axes with some numerical values.

6.