Determining Planck's constant experimentally?

Question

I performed an experiment finding the stopping potentials of Yellow, green, blue and violet 1 and 2 spectrums. I plotted this against the frequency of these wavelengths of light. I am supposed to determine Planck's constant experimentally but cannot figure out how...any ideas?

I performed an experiment and obtained a value of 0.75 for V at a frequency of 5.19E14....I'm supposed to find Planck's constant using this data...I'm not sure if my experiment completely failed or I'm not doing it correctly because my answer is way off of the known Planck's constant...any help?

Answer 1

So you have light of a certain color incident on a piece of metal. The light kicks off an electron. There is a certain energy cost to unsticking the electron from the metal--this is called the "work function," (W) and then the energy the electron has left depends on the frequency of the light. If the energy of a photon is hf (h=Planck's constant, f=frequency) then the energy of the electron=the stopping potential*the charge on an electron=hf-W. If you have plotted the stopping potential versus the frequency, then the slope of that line should be h/e, and the intercept should be W/e (where e is the charge on the electron).

Answer 2

It is not clear what is meant by "stopping potential". You can use E = h nu, or E = h/lambda if you know the energy of the photons and either the frequency or the wavelength of the radiation (the wavelength is easier to measure). If the "stopping potential" is measured in volts, then the energy involved can be in electron volts, and with appropriate conversion you can get it from that.

Another dodge would be to measure the radiation and voltage response of red and green light-emitting diodes. The green diodes will require more voltage, and if you measure the wavelengths, you can get Planck's constant as described above. The method isn't accurate, for two reasons: LED's don't have a narrow spectrum, and there are energy losses in the photon production process that can't easily be figured. But you should get a number that is within 50% of the correct value. The reason for using two different colors is to let you cancel out the lost energy, on the assumption (not necessarily true) that the energy loss is the same in the two diodes.

Answer 3

hf=eV, where h is Planck's constant, f is the frequency of the light and e is the charge on the electron and V, the potential (stopping). The slope of your graph will be V/f and when multiplied by the value of the constant e will give the Planck constant.

Answer 4

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