Question about De Broglie's laws

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SUMMARY

The discussion focuses on calculating the wavelength of an electron traveling at 4.35 x 106 m/s using De Broglie's equation. The correct application of momentum, represented as p = mv, leads to the conclusion that λ ≈ 0.167 nm. The user initially attempted to use the kinetic energy equation E = 1/2mv2 in conjunction with E = hf, but this approach was flawed due to the incorrect relationship between frequency and wavelength. The correct relationship is f = v/λ, where v is the velocity of the electron, not the speed of light.

PREREQUISITES
  • Understanding of De Broglie's hypothesis and wave-particle duality
  • Familiarity with the equations for momentum (p = mv) and energy (E = hf)
  • Basic knowledge of kinetic energy calculations (E = 1/2mv2)
  • Concept of wavelength and frequency relationships (f = v/λ)
NEXT STEPS
  • Study the derivation of De Broglie's wavelength formula (λ = h/p)
  • Explore the implications of wave-particle duality in quantum mechanics
  • Learn about the photoelectric effect and its relation to energy quantization
  • Investigate the historical context of De Broglie's thesis and its acceptance by Einstein
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Students of physics, particularly those studying quantum mechanics, as well as educators looking to clarify the concepts of wave-particle duality and De Broglie's equations.

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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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