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For Deriving De Broglie' Wavelength

by puneeth
Tags: broglie, deriving, wavelength
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puneeth
#1
Mar15-08, 01:07 AM
P: 9
1. The problem statement, all variables and given/known data
The De Broglie Wavelength of any object in motion is given by
[tex]\lambda[/tex]=[tex]\frac{h}{P}[/tex] where h is Planck's constant and P is the body's momentum. for heavy masses this wavelength is too small to be observed, nevertheless it is still there.... I have seen a derivation for this which is not convincing and any clarification which is convincing will be received with thanks...


2. Relevant equations
E=mc^2
E=h[tex]\nu[/tex]

3. The attempt at a solution
The derivation is as follows --
from the above 2 equations
mc^2 =h[tex]\nu[/tex]
but [tex]\nu[/tex]=c/[tex]\lambda[/tex]
hence mc=h/[tex]\lambda[/tex]
rearranging terms, [tex]\lambda[/tex]=h/mc = h/p
where p is the momentum of the body like electron travelling with high speed.

the above relation can be used for finding the wavelength of a moving electron or even a slow moving proton - (as mass of proton is 1837 times electron for having the same momentum it travells at a speed 1837 times lesser than that of electron).

how can we take momentum of any body in place of "p" in the formula when it is actually kept in the place of "mc"? during the derivation "c" is cancelled on both sides. if we actually begin the derivation taking a body travelling with speed "v" then we cannot cancel "c" and "v". the formula is applied even to electrons moving at "c/2" or "c/3" speeds. thus cancelling the speeds on both sides would be a really bad approximation.
even the energy of a body travelling with c/3 etc would not turn out to be exactly mc^2. then how is the derivation valid? please clarify... i am unable to think it out
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Jivesh
#2
Dec24-09, 01:59 AM
P: 18
I would like to give a somewhat valid derivation of the de Broglie equation. I had given a similar derivation on another thread discussing the same problem.
Here goes:
In order to come up with the wavelength of the matter wave of a particle, we need to consider the wave properties of that particle, like phase velocity, group velocity, crests, etc. Now, the phase velocity is given by V=w/k, where w is the angular frequency and k is the wave number and the group velocity is v=dw/dk, a differential. The phase velocity is associated with a single wavelength and the group velocity is associated with a wave having a number of waves, each of different frequency.
Now, first considering the virtual photon of the particle. A virtual photon is actually a quantum of energy exchanged between the electron and the proton in and atom or between electron and electron. It is emitted by all particles, like pions, neutrons, protons, etc. As it is energy and energy travels at the speed of light, we have the relation of energy in the virtual photon as E = mc^2 and also as E = hc/lambda, which gives the first result as λ = h/mc.
De Broglie hypothesized that the particle itself was not a wave, but always had with it a pilot wave, or the virtual photon or a wave that helps guide the particle through space and time. He postulated that the group velocity of the wave was equal to the actual velocity of the particle.
However, the phase velocity would be very much different. He saw that the phase velocity was equal to the angular frequency divided by the wavenumber. Since he was trying to find a velocity that fit for all particles (not just photons) he associated the phase velocity with that velocity. He equated these two equations:
V = ω/k(which the eqn for the wave) = E/p(which is the eqn for the particle)
From this new equation from the phase velocity we can derive:
V = mc^2/m v = c^2/v
Applied to Einsteinís energy equation, we have:
E = pV = mv(c^2/v)
Also, the wavelenght of the matter wave (or any wave), is given as the phase velocity divided by the frequency of the wave. Hence, λ = V/f.
Now we can get to an actual derivation of the De Broglie equation:
p = E/V
p = (hf)/V
p = h/λ
With a little algebra, we can switch this to get λ = h/mv.
I hope this helps in understanding the concept.


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