Week 38 (May 13 - 17)

Reading: Atomic physics (Chap. 42)

Key Topics:
the Balmer series and atomic spectra, the photoelectric effect, early models of the atom and atomic transitions,


Homework Problems:

  1. Balmer series problem: Find the longest and the shortest wavelength in the Balmer series. You will need to use the Balmer formula, which allows one to find the wavelengths of light emitted by a hydrogen gas. Answer: the longest wavelength (lowest energy photon) corresponds to the transition from n=3 to n=2. Plugging these values into the Balmer formula gives lambda = 656 nm. The shortest wavelength (highest energy photon) corresponds to the transition from n=infinity to n=2. This gives lambda = 364 nm.
  2. Photon problem: What is the energy of a single photon of red light (wavelength = 632 nm)? Answer: since E = Planck's constant * frequency, we get E = hc/lambda = 3.14 e -19 joules or 1.962 electron-volts.
  3. Photoelectric effect problem: A 10 milliWatt blue laser (wavelength 400 nm—note that I changed this!) is shone on a gold surface. (i) what is the energy of each blue photon? (ii) how many photons per second strike the surface? (iii) what is the work function of gold (the minimum amount of energy required to knock an electron from the surface) (iii) how much kinetic energy does each emitted electron have? (Let's assume that the photon energy all goes into knocking off electrons, and not heating the gold) (iv) what is the speed (in meters per second) of each emitted electron? Answer: (i) energy = 4.97 e -19 joules or 3.1 eV. (ii) number per second = energy per second / energy per photon = 2 e 16 photons per second. (iii) the work function of gold is 5.31 eV. (iv). So the electrons cannot even be ejected from gold because the photons don't have enough energy to liberate them. If we used potassium, with a work function of 2.29 eV, then the electrons would have energy 0.81 eV and a speed of 534 km/second.
  4. Bohr model problem:
General College Physics