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What is Planck's Quantization of Energy?

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    I don't understand what E = nhv means. What does it apply to and what does the n (energy level) mean in the equation.

    It says in the textbook, "According to Planck, the atoms of the solid osscillate with a definite frequency depending on the solid. But in order to reproduce the results of the experiements on glowing solids, he found it necessary to accept that an atom could only have certain energies of vibration, E, those allowed by the formula
    This was when they were discussing his experiment on testing the intensity of the light emitted by hot objects

    Later, they talk about the photoelectric effect and how E=hv is incorporated into that.

    They said how if a vibrating atom changed energies, the change in energy would be released as light energy and they described photons as particles with energy E proportional to the observed frwquency of the light.

    The definitions are confusing me.
    I don't get the difference between the 2 equations or what they mean, and what each of them apply to?
    Last edited: Sep 14, 2011
  2. jcsd
  3. Sep 15, 2011 #2


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    Classically, the energy for a wave of given frequency could take on any value, so the energy of a wave emitted from a blackbody could have any value from 0 to infinity. The blackbody emits a ton of waves at a frequency. If you measured the energy of the individual waves, you'd find you'd get a nice smooth distribution of energies that follows the Boltzmann distribution, if the system behaved classically. Planck's idea was that for a given frequency ν, an individual wave could only have an energy equal to some integer multiple of hν. Using this assumption, he was able to reproduce the observed behavior a blackbody.

    Einstein came along and said it's not so much that the waves can only have certain energies, it's that light comes in chunks, called photons, with energy equal to hν. Using this idea, he explained the photoelectric effect.
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