Homework on The Photoelectric Effect (solutions)

Feel free to work on the homework in groups. The work you hand in, however, should reflect your understanding of the material and be in your own words. Students who turn in identical (or close to identical) homework assignments will be asked to explain their answers orally to the TA or prof. A student who cannot explain how he or she arrived at a given answer will be charged with academic dishonesty.

You should justify all of your answers for full credit.

You pass light from a hydrogen lamp (as opposed to the mercury lamp from our class) through a diffraction grating to separate it into its component wavelengths. You then shine the light on a silicon surface to examine the photoelectric effect for aluminum in the hopes of making an inexpensive photocell. You find the following data:

Color	Wavelength (nm)	Stopping Potential (V)
red	653	0.210
blue	486	0.905
violet	433	1.12
violet	411	1.37

Color	Relative Intensity	Stopping Potential (V)
red	100%	0.205
	80%	0.204
	60%	0.205
	40%	0.202
	20%	0.198

1.	What is the value of Planck's constant (h) as measured by your experiment? You may find it useful to adapt the spreadsheet from the activity, but you should include all graphs and equations in your homework submission. See attached spreadsheet for work. Particle theory says that the energy absorbed by an electron is proportional to the frequency of light: E_abs = hf. Some of this energy goes into the work function to remove the electron, the rest is kinetic energy of the electrons. We thus have the expression hf = K + f K = hf - f. The graph plots K as the y-variable and f as the x-variable. Thus the slope is h, Planck's constant, and the intercept is -f, the work function. From our graph we find a slope of h= 6.63E-34 Js - it works!
2.	What is the workfunction of aluminum as measured by your experiment? To what cutoff frequency does this correspond? Remember to show your work. Looking at the intercept from our graphs, we find a work function of 2.70E-19 J = 1.8 eV. Plugging this back in and setting E = 0, we get f = f/h; (1.8 eV) / (4.662 x 10^-15 eV s) = 3.86 x 10¹⁴ Hz for the cutoff frequency.
3.	What part(s) of this experiment support(s) the wave theory of light? Explain. The use of a diffraction grating to generate an interference pattern; particles do not interfere.
4.	What part(s) of this experiment support(s) the particle theory of light? Explain. The dependece of energy on frequency and the independence of energy from intensity; the energy a wave carries is dependent on intensity and independent of frequency.
5.	Using your own words (and not those of a Power Point slide and/or your classmate), explain how the current quantum theory of light resolves this controversy. Quantum theory says that light is neither wave nor particle, but it has properties of both. In general, it acts like a particle when interacting with other particles or when being measured, but like a wave when traveling.