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Planck's constant equaled 1 instead of 6.63x10^-34

  1. Apr 6, 2009 #1
    I've been faced with this question: how would the nature of atoms and quantum mechanics differ if h (planck's constant) equaled 1 instead of 6.63x10^-34. Basically, I'm looking for changes that would be observable, nothing too complicated or technical.
     
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  3. Apr 7, 2009 #2

    malawi_glenn

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    Re: h=1

    h = 1 Js you must mean, this constant also have units (in some unit systems)
     
  4. Apr 7, 2009 #3

    Redbelly98

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    Re: h=1

    Welcome to PF flip92.

    You could calculate the ground state energy of a hydrogen atom.

    Or, how often would a 100W visible light source emit a photon? You could use 550 nm as an average wavelength.

    p.s. malawi is correct, it's important to use the right units. This avoids confusion with alternative unit systems, such as eV-s or even no units at all! (Both of these get used by physicists.)
     
  5. Apr 7, 2009 #4

    malawi_glenn

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    Re: h=1

    yes, for example particle physicsts often use h-bar = 1 [unitless] which is a totally different statement that h-bar = 1 Js.
     
  6. Apr 7, 2009 #5
    Re: h=1

    No changes in the behavior and properties of particles will be observed.

    Planck's constant is just a value assigned such that the proportionality between energy of a particle and its frequency sufficiently equals..i.e.,

    Only when the constant of proportionality is 6.023 x 10^23, the energy equals to its frequency.
    When you assign the constant of proportionality (h) as 1, then Energy of a particle equals to 6.023 x 10^23 times its frequency.i.e.,
    E= 6.023 x 10^23 h ง

    So, no characteristic change will be observed on particles. But the analytical way of interpretation of its momentum, position, energy and other things just differ in their formula..:smile:
     
  7. Apr 7, 2009 #6

    malawi_glenn

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    Re: h=1

    Nope. look on how one determine QM from path integrals
     
  8. Apr 7, 2009 #7

    Avodyne

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    Re: h=1

    If you're going to change the value of [itex]h[/itex], you have to specify what you're holding fixed. For example, are you going to hold the electron mass [itex]m[/itex] fixed, or its Compton wavelength [itex]\lambda=h/mc[/itex]? (The latter is actually more natural from the point of view of quantum field theory.)

    If we hold all particle masses fixed, and all the constants of E&M, then the Bohr radius goes up by a factor of 10^68. But this just means atoms are much bigger, and so things made out of atoms (like us) are much bigger too; so would we notice any difference?

    To compare apples to apples, so to speak, we should look at dimensionless ratios. A very important one is the fine structure constant [itex]\alpha = e^2/(4\pi\epsilon_0\hbar c)=1/137.036[/itex]. This would become smaller by a factor of 10^34. The fine-structure constant controls lots of quantum phenomena in atoms and molecules; for example, the ionization energy of hydrogen, 13.6 eV, can be expressed as [itex]{1\over2}\alpha^2 mc^2[/itex]. All sorts of chemistry would be different in complicated ways.
     
  9. Apr 7, 2009 #8

    malawi_glenn

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    Re: h=1

    And what will then happen with gravitational effects ;-)
     
  10. Apr 7, 2009 #9

    Avodyne

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    Re: h=1

    Nothing. All gravitational effects that we observe are purely classical.
     
  11. Apr 8, 2009 #10

    malawi_glenn

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    Re: h=1

    if the em-coupling becomes smaller due to greater h, gravity effects will increase in atoms.
     
  12. Apr 8, 2009 #11

    Avodyne

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    Re: h=1

    Not under this scenario. The classical EM force and the classical gravitational force between two particles are both unchanged. So gravity is still much less important. Gravitational effects in atoms would be no easier to observe.
     
  13. Apr 8, 2009 #12
    Re: h=1

    See here:

    http://arxiv.org/abs/hep-th/0208093

    So, as others in this thread have pointed out, h is an irrelevant rescaling constant that only appears in equations because we are using quantities measured in different units in the same equation.

    It's like adding kilometers to miles. Then there must be a conversion factor of 1.60931 somewhere. And if one were stupid enough to assume that a length in England is physicaly different from a length in mainland Europe, that factor of 1.60931 would be dimensional too.
     
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