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Planck's constant

  1. Sep 12, 2006 #1
    According to quantum mechanics the energy of radiation is equal to multiple Planck's constant : E = h * v, where "v" is the radiation frequency.

    Since there is no limit to frequency, what is the energy of radiation of frequency 1/2 Hz?

  2. jcsd
  3. Sep 12, 2006 #2
    h/2 J.101010
  4. Sep 12, 2006 #3


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    Where h is expressed in SI units; Planck's constant has dimensions [tex]ML^2T^{-1}[/tex], so that would be [tex]Kg m^2 sec^{-1}[/tex].
  5. Sep 14, 2006 #4
    Thank you. I thought before that the energy must be procuct of integer number of 'h'.
  6. Sep 14, 2006 #5
    Energy is emitted or absorbed in integer multiples of [tex]h\nu[/tex].
  7. Sep 26, 2006 #6
    If this is true, than the oscillator working on the frequency ½ Hz can’t emit the radiation. Or am I wrong?
  8. Sep 26, 2006 #7
    n*h*(nu)...it's the 'n' that's restricted to integers, not nu.
  9. Nov 11, 2006 #8


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    If you look at those units for h, [tex]ML^2T^{-1}[/tex], you see that it does not represent a quantum of energy, but of action. This is why energy can increase or decrease by continuous amounts, it is only the action, e dt, that is quantized.
  10. Nov 12, 2006 #9
    if we assume photon can have the largest wavelength of the size of the universe and assume universe is 10^26m

    by forumla c=lamb*nu

    you can compute smallest nu and energy E=H*nu

    : Photon is an energy emitted from internal atom/molecule transitions. There is no other way i know of photon can be created.
  11. Nov 13, 2006 #10

    What about electron/positron pair annihilation, this creates gamma rays.
  12. Nov 13, 2006 #11
    "What about electron/positron pair annihilation, this creates gamma rays"

    Very good. I forgot about this.
  13. Jan 25, 2007 #12
    I thought any change of charged particle trajectory should generate photon.
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