Re: de Broglie wave paradox?
From: Andr? Michaud (srp_at_microtec.net)
Date: 08/19/04
- Next message: Y.Porat: "Re: The electron shell model is in challenge"
- Previous message: Igor: "Re: de Broglie wave paradox?"
- In reply to: Pat Dolan: "Re: de Broglie wave paradox?"
- Next in thread: Pat Dolan: "Re: de Broglie wave paradox?"
- Reply: Pat Dolan: "Re: de Broglie wave paradox?"
- Messages sorted by: [ date ] [ thread ]
Date: 19 Aug 2004 06:11:02 -0700
patdolan834@cs.com (Pat Dolan) wrote in message news:<4f5d4135.0408181614.758eecd3@posting.google.com>...
> >
> > I have also seen that type of example in intro textbooks, but I
> > think that it is only meant to make the concept more tangible.
> >
> > In reality, it has real physical meaning only when applied to
> > orbiting elementary particles, in other words, to the orbiting
> > electron for example.
> >
> > The origin of the concept was the Bohr model that de Broglie was
> > analyzing, de Broglie's lambda is nothing else but the length of
> > the electron's orbit in ground state in the hydrogen atom.
> >
> > He first derived lamda=h/p in the following manner
> >
> > For photons, lambda is inversely proportional to nu (frequency)
> > so, lambda = 1 / nu
> >
> > so, lambda * nu = 1 (constant)
> >
> > lamba * nu = h / h (h = Planck's constant)
> >
> > he derived h * nu = h / lambda
> >
> > But for photons, E (energy) is equal to p (momentum)
> >
> > so we can write , E = h * nu and p = h / lambda
> >
> > and finally, lambda = h / p
> >
> > The real finding though was, when considering Bohr's model,
> > his intuition that classical p = mv could possibly apply,
> >
> > whence lambda_broglie = h / (mv)
> >
> > m being the mass of the electron, and v its velocity.
> >
> > Electron mass is well known through direct measurements
> > (9.10938188E-31 kg)
> >
> > and its velocity on the ground orbit of the Bohr model can
> > be obtained from the Coulomb equation
> >
> > v = q^2 / (2 eps_0 h) = 2187691.252 m/s
> >
> > (I could show you how to derive this velocity equation from the
> > standard coulomb equation if you wish)
> >
> > applying these values to de Broglie's lambda equation,
> >
> > lambda_broglie = h / mv = 3.324918462E-10 m
> >
> > which, when divided by 2 pi, gives you the well known
> > Bohr radius 5.291772085E-11 m
> >
> > You can also obtain the wavelengths of the complete Lyman series
> > from this equation
> >
> > Lambda_n = nh / mv (n=1,2,3...)
> >
> > One more note of interest is that the velocity of the electron
> > on the ground orbit of Bohr's model divided by the speed of light
> > gives the fine structure constant (alpha)
> >
> > alpha = v / c = 7.29735253E-3 = 1 / 137.0359998
> >
> > To my knowledge, this is how alpha was first derived.
> >
> > André Michaud
>
> Now I am more confused. I thought the reason for the baseball
> examples in QM textbooks are to indicate that matter has associated
> with it a wave property that is ultimately expressed in the psi
> function of quantum mechanics.
You are right about a link with QM of course, since it is this
idea of de Broglie that directly inspired Schrödinger to come
up with his equation. But this doesn't mean that it is formally
right to associate a wave property to just about any lump of matter.
The psi function applies only to elementary particles in motion.
> You seem to be using lambda_broglie to determine electromagnetic
> wavelengths, such as the Lyman series.
Yes in non-relativistic classical mechanics, Lambda_broglie determines
the wavelength of the energy that carries a particle, and can also be
used to calculate the classical orbits of the Bohr hydrogen atom (all
velocities on electronic orbits would be in the non-relativistic
range if the case applied).
For example, a photon (not associated to a particle) moves at
the speed of light and will have a wavelength lambda = c / nu,
but a particle with the same energy will have a de Broglie
wavelength lambda_broglie = v / nu
The energy frequency (nu) is directly related to the energy
level and not to the wavelength, contrary to assumption, and
will remain the same for both.
For non-relativistic velocities, the wavelength presently is
the distance covered per cycle of the frequency concerned.
V being 2187691.252 m/s as already calculated, an electron having
energy 27.21138345 eV (4.359743805E-18 Joules) which is the
energy induced at the Bohr ground orbit (Coulomb equation),
will cover a distance of v / 6.5796839717E15 Hz = 3.324918433E-10
meter as already shown, but a free photon of same energy
27.21138345 eV (4.359743805E-18 Joules) will cover a distance of
c / 6.5796839717E15 Hz = 4.556335256E-8 m
You will note here that lamda / lamda_broglie = alpha just like
the ratio of V on c.
> What quantity are we measuring, peak-to-peak, when we measure
> lambda_broglie?
Simply the distance that the particle will have have moved from
peak-to-peak of one cycle of the frequency of the energy that
the particle possesses on top of its rest energy.
> Is it volts/meter? or is it something like sqrt(probability)?
Way simpler than that. It simply measures the distance in meters
moved by the particle depending on its energy level.
> If it is the latter, then how does lambda_broglie corrolate to
> electron diffraction data?
> And what about a hydrogen proton near absolute zero, aren't
> we back to a relatively low mv, and hence, a pretty big
> lambda_broglie?
Metaphorically speaking yes. But since a proton is not elementary
(made up of three scatterable elementary particles, 2 quarks up
plus one quark down) a de Broglie wavelength for the proton seems
to me no more significant than one for a baseball, although
mathematically speaking, we can of course define a "de Broglie
wavelength" for just about anything.
Can you imagine what happens when the wave function of a baseball
collapses for example ? If such cases were real, according to
Copenhagen school interpretation of QM, the baseball would have
been a wave packet as it moved, and would suddenly appears as
physical object when its wave collapses. I guess you see the picture.
In the Bohr model as well as in physical reality, even at absolute
zero, the ground orbit of the electron remains the same.
However, if the quark velocities were non-relativistic (which is
doubtful), their de broglie wavelengths would definitely have to
continue to also be significant, because their motion can be the
only thing that maintains the integrity of the proton, which we
know is also maintained at absolute zero.
So, logically, these quarks definitely have to keep moving even at
absolute zero. Presumably, their wavelengths (straight de Broglie
or adjusted for relativistic contraction) at that temperature may
well be the same as at room temperature.
Extrapolating here.
André Michaud
- Next message: Y.Porat: "Re: The electron shell model is in challenge"
- Previous message: Igor: "Re: de Broglie wave paradox?"
- In reply to: Pat Dolan: "Re: de Broglie wave paradox?"
- Next in thread: Pat Dolan: "Re: de Broglie wave paradox?"
- Reply: Pat Dolan: "Re: de Broglie wave paradox?"
- Messages sorted by: [ date ] [ thread ]
Relevant Pages
|
