Re: de Broglie wave paradox?

From: JLeeCforRP (jleecforrp_at_aol.com)
Date: 08/14/04 

Date: 14 Aug 2004 17:08:56 GMT

>In chap 1, page 31, of "Quantum Chemistry" (McQuarrie, 1983,
>University Science Books)an expamle calculation is made for the de Broglie
wavelength of a baseball traveling at 90 mph. Using the >relation lamda=h/p,
the calculation gives the value
>(6.626*10E-34)/(5.6kg*(m/s))= 1.2*10E-34 m as the baseball's
>"ridiculously small" de Broglie wavelength. My question concerns what
>happens to the baseball's de Broglie wavelength as it hits the ground and
rolls to a stop. At the moment just before the baseball comes to a halt, when
its velocity is say, 10E-34 m/s, the de Broglie relation
>in this example predicts a wavelength on the order of 1 meter. What
iterpretation should I make of this wavelength?
>
>Pat
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Pat,
Quantum Physics today has the de Broglie wavelength COMPLETELY wrong.
Why? because Theoretical Physicists today look at particles as "billiard Ball"
type entities (not all that unsimilar from their heads), and this is why they
can't understand how use the de Broglie equation to describe the orbital
wavelenghts of the atom.

With "Reality Physics" it has already been proven that matter (sub-atomic
particles ) is composed of light ("PHOTON SPIN"). This is why, when you use the
"Mass Decrease" of Reality Physics (NOT the "Mass Increase" of Relativity) to
find the "mass action" taken from the electron to form the first orbit of the
Bohr Hydrogen atom, which we find by, first of all, using Einstein's famous
equation m = E/c^2, and simply divide the known energy of the first orbit:
13.6ev by "c^2 to get: [2.42x10^-35 kg.], and then, second of all, put the
electron's velocity for the first orbit into the Fitzgerald Formula:
(1-[2.1885x10^6m/s]^2/ c^2)^1/2 to get:
[.999,973,39], and then multiply this result by "c" to obtain a reduced "photon
spin velocity within the electron of: [299,992,017 m/s] and then substitute
these CORRECT values from Reality Physics into the de Beoglie equation we get
the CORRECT value for the wavelength (w) of the first Bohr Hydrogen atom
orbit", such that:

w[N1]=h / 2.42x10^-35kg/299,992,017m/s

lamda [N1] = 9.12x10^-8 meters !

Wow! what a coincidence! Even more of a coincidence is that it works for all of
the electron pilot wavelengths of the Bohr Hydrogen atom. Imagine that?

Be sure not to tell anybody though as this completely disproves both current
Quantum Mechanics and Relativity.

al the best,

Jeff Lee CENTER for REALITY PHYSICS
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

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