Question about De Broglie's laws

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Homework Help Overview

The original poster presents a problem involving the calculation of the wavelength of an electron traveling at a specified speed, utilizing De Broglie's laws. The discussion revolves around the application of momentum and energy equations in the context of wave-particle duality.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to solve the problem using both momentum and energy equations, questioning why the kinetic energy approach yields incorrect results. Participants discuss the correct relationships between frequency, wavelength, and velocity for electrons versus photons.

Discussion Status

Some participants provide clarifications regarding the equations used, noting the distinction between the frequency of light and that of an electron. There is a recognition of the historical context of De Broglie's hypothesis and its implications for understanding wave-particle duality.

Contextual Notes

The original poster expresses confusion over the application of energy equations, indicating a potential misunderstanding of the differences between photon and electron energy calculations. The discussion reflects on the foundational concepts of quantum mechanics without reaching a definitive conclusion.

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Homework Statement


Find the wavelength of an electron which is traveling at 4.35*10^6 m/s.

Homework Equations


p = h/λ
p = mv
E = hf
E = 1/2mv^2

The Attempt at a Solution


I know this can be easily solved using the momentum equation and De Broglie's law like this:
mv = h/λ
(9.109*10^-31)*(4.35*10^6)=(6.626*10^-34)*λ
λ ≈ 0.167nm

But here comes the actual question...
Why can't I solve this with the second law E = hf and the classical 1/2mv^2?
1/2mv^2 = hf, where f = (c/λ)
This gives me an incorrect result. If I wanted to use kinetic energy, I would have to first convert it to momentum
p = √(2Em), which I would use with p = h/λ.

After all, for instance the photoelectric effect can be calculated using kinetic energy with hf-W.
I think I've missed something relevant, and I can't seem to find the answer. Sorry if this is too obvious.
 
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It is giving the correct result : 1.6722e-010 = 0.16722 nm.
What result are you getting?
 
If you are using the following equation:

"1/2mv^2 = hf, where f = (c/λ)"

then it is incorrect because f ≠ c/λ, it is f = v/λ i.e.the velocity of the electron, not that of light.
 
You are on the right tracks.
In the early days of this work it was realized that the energy of a light photon was given
by E = hf (from photoelectric effect).
DeBroglies hypothesis was to equate Einstein's E = mc^2 with hf
So his hypothesis was that mc^2 = hf
which becomes mc = h/λ
De Broglie (controversially) said that this was a general relationship and since m x c looks like
'momentum' he produced his equation momentum =h/λ
ie mv = h/λ

Simple algebra but remarkable that it is true.
 
It makes more sense now. I guess I was all the time applying the energy of a photon, not an electron. Thank you, now I can sleep restfully at night.
 
De Broglie was a strange character! When he came up with his hypothesis, as part of his thesis, his professor did not think much of it and showed it to Einstein.
Einstein accepted the thesis completely !
There but for fortune...
Sleep tight
 

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