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mf92

Speed of a particle is 200 000m/s and has a wavelength of 150nm how much does it weight?

Elroch

Well, it is slow enough not to have to worry if relativity matters. The formula for Compton wavelength (if that is what you mean) gives:

m = h_bar/lambda x c

 =6.62e-34 /(1.5e -7 x 3e8) = 1.47 e-35 kg

I don't think this particle exists, as its mass seems to be between that of neutrinos and the electron.

mf92

yep, however instead c you use v then m=2.2 e-32

This was fairly simple, thought if anyone has interesting calculations then post it here

Elroch

Yes, it's h_bar (which was the value I used), but it is c, not v, I think. The Compton wavelength does not become infinite if a particle stops.

KairavJoshi

I like the lamda symbol lol

Elroch

@RainbowRising, sorry, another typo by me. Smile What I intended to say was that you are right, it is meant to be h in the formula, but although I accidentally called it h_bar I used the value for h and believe the answer was correct. Do we agree that the speed is irrelevant? The Compton wavelength is independent of the speed of  a particle. (Eg see here or here)

Elroch

The Compton wavelength involving c is a more fundamental property, essentially relating the mass of any particle to a wavelength which appears in equations in quantum mechanics. The de Broglie wavelength obviously is not a fundamental property of a particle but of a particle with a particular speed and is relevant to how it diffracts and such things.

Elroch

Well, as well as it appearing in any of the quantum mechanical equations for a particle (such as Schroedinger's), I would refer you to the last section of the Wikipedia article, which says that the Compton wavelength appears in the formula for the interaction of fermions with photons. Interaction cross sections are given in units of area, so I presume (without wrapping my brain around the details this late at night) that it is the square of the Compton wavelength that actually appears.

Elroch

Yes, it is not immediately apparent to the eye. Also, it is the reduced Compton wavelength that appears (Compton wavelength / 2 pi). If you rearrange the Schroedinger equation and write it in terms of the electromagnetic coupling constant alpha, you get the form:

The second term constant is -1/(4*pi) x Compton wavelength. Of course if you wanted to, you could move this dimensionless constant to the other two terms. My general point is that the Compton wavelength is a fundamental property of a particle that appears in physical laws, whereas the de Broglie wavelength is a property of a particle moving at a particular speed which determines other aspects of its behaviour under such circumstances (such as what happens when it hits a regular lattice).

Elroch

Sorry, it is a special situation - an electron in an atom (hence the r). The main other unusual choice is to put the fine structure constant in it to make it a little simpler. I am using an example added to the Wikipedia article by someone who is more of an expert on these things than I am. If it's not clear, Z is the atomic number and e the electron charge.

Elroch

Well it's all rather arbitrary. In any version of the Shroedinger equation, The d/dt term (curly d's) is traditionally written with a constant of (i h_bar), and the Del^2 term has a -((h_bar)^2)/2m constant. If you wish to make the second multiplier h_bar/mc (the reduced Compton wavelength), the multiplier for the d/dt term would be (-2 i / c), which is not really any more complicated than the usual one. The reason for the usual choice is to make the constant have units of energy, whereas making it the reduced Compton wavelength obviously gives it units of length. But the freedom to do this makes it not very significant really.