# De Broglie Momentum doesn't seem in Agreement with p=mv

• I
• Benoit
In summary, an electron has a wavelenght depending on its velocity, and the wavelength can be calculated by equating the two expressions for momentum. There is a factor of 2 error in the calculation, which suggests that there may be an error in the formula.
Benoit
Hi there.

So I had this lab last week about De Broglie hypothesis. In a simulation, we plugged in the electron velocity and the computer gave back a beautiful wavefunction, from which I can measure the wavelength. So here I have an electron going at 0.6 m/s with a wavelenght of 0.00060606 meters. When we calculate the momentum, we have two choices, either p=mv or p=h/λ. I respectively found 5.466E^(-31) and 1.093291E^(-30). The link ? A factor of 1/2 between the two values. I did this with many other wavelengths and velocities. What is the point I am missing ? Shouldn't the two equations give the same answers ?

Benoit said:
the computer gave back a beautiful wavefunction

What was the wavefunction?

A classical sine wave, we were only considering the problem in one dimension with only one velocity at a time. We then went further with wavepackets, but I had problems way before going into that. I measured myself the wavelenghts. It's a Matlab simulation that our Prof gave us, it seemed legit, that's why I ask myself these questions. Here is a link to see the distribution. Don't be scared because there are some french words, the math is the same https://docs.google.com/document/d/1k0BKPiMQSJ7jxplDCH7-yf0z7-08c3mhDWBgW8cWzVU/edit?usp=sharing

Last edited:
Benoit said:
I measured myself the wavelenghts.

Did you measure a full wavelength (sine function going from 0, to 1, to 0, to -1, to 0) or just a half wavelength?

A full wavelength. Actually, I counted 33 λ over 20 mm, so there is the value I gave you.

If you are sure those 2 equations are coherent together, I'll just assume I'm the cause of the error and maybe I'll try to talk to my teacher, I just didn't want to look silly in front of him.

Benoit said:
Hi there.

[So] I had this lab last week about De Broglie hypothesis. In a simulation, we plugged in the electron velocity and the computer gave back a beautiful wavefunction, from which I can measure the wavelength. So here I have an electron going at 0.6 m/s with a wavelenght of 0.00060606 meters. When we calculate the momentum, we have two choices, either p=mv or p=h/λ. I respectively found 5.466E^(-31) and 1.093291E^(-30). The link ? A factor of 1/2 between the two values. I did this with many other wavelengths and velocities. What is the point I am missing ? Shouldn't the two equations give the same answers ?
If you used software to do the calculations, then you should send a bug report to the makers.

Benoit said:
I counted 33 λ over 20 mm

I get the same thing from the image at the link. But the image doesn't give the formula that was used to calculate the wave given the electron's velocity. What formula was used?

I don't know, it is not given. The more velocities we were plugging, the more the distribution was narrowing but no equation was given. It's a lab, we had to "measure" ourselves I suggest.

Mentz114 said:
If you used software to do the calculations, then you should send a bug report to the makers.
Yes, I tend to believe I did the error before doing so, but I'll surely do it if I'm sure there is an error.

Benoit said:
I don't know, it is not given.

That makes me wonder, since the obvious way to calculate the wavelength from the velocity is by equating the two expressions for momentum, i.e., ##m v = h / \lambda##. So the fact that there is a factor of 2 error makes me suspect that there is an error in the formula somewhere, and the fact that it isn't given makes me suspect that it has not been carefully checked.

Benoit
Alrigth, thanks guys \

Hey, finally it turned out the distribution was squarred, because it's a probability distribution. When you square a sine function, the frequency double, hence the wavelenght becomes the half.

## 1. How does De Broglie momentum differ from classical momentum?

De Broglie momentum is a concept in quantum mechanics that describes the momentum of a particle in terms of its wavelength, while classical momentum is defined as the product of an object's mass and velocity. In other words, De Broglie momentum takes into account the wave-like nature of particles, while classical momentum only considers their particle-like behavior.

## 2. Can De Broglie momentum be derived from the classical formula p=mv?

No, De Broglie momentum cannot be derived from the classical formula p=mv. This is because the classical formula does not take into account the wave-like nature of particles, while De Broglie momentum does. The two formulas are based on different principles and cannot be equated.

## 3. How is De Broglie momentum experimentally verified?

De Broglie momentum has been experimentally verified through a phenomenon known as electron diffraction. This occurs when a beam of electrons is passed through a diffraction grating, resulting in a diffraction pattern that can only be explained by the wave-like properties of electrons. This confirms the relationship between momentum and wavelength proposed by De Broglie.

## 4. Does De Broglie momentum apply to all particles?

Yes, De Broglie momentum applies to all particles, including macroscopic objects. However, the effects of De Broglie momentum are negligible for larger objects due to their relatively large mass and low velocities. It is most noticeable for particles with very small mass, such as electrons.

## 5. How does De Broglie momentum impact our understanding of particle behavior?

De Broglie momentum plays a significant role in our understanding of particle behavior, particularly in quantum mechanics. It helps to explain phenomena such as electron diffraction, the uncertainty principle, and the wave-particle duality of matter. It also has practical applications in fields such as nanotechnology, where the behavior of particles on a microscopic scale is crucial.

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