Frequency of a de Brogli Wave

In summary, de Broglie waves are a concept in quantum mechanics that describe the wave-like nature of particles. The frequency of a de Broglie wave can be calculated using the formula f = v/λ, where v is the speed of the particle and λ is the wavelength. However, there is some confusion as to whether the phase velocity or group velocity should be used in this calculation. The phase velocity gives the usual meaning of frequency, while the group velocity may not have a clear interpretation. Additionally, in relativistic quantum mechanics, the total energy of the particle can be calculated using the equation E = γmc^2, where γ is the relativistic factor and m is the relativistic mass. This equation can also be used
  • #1
yogifromindia
5
0
I am confused about de Broglie waves.

we know λ = h/mv

What is the frequency of the dB wave?

Generally, speed of any wave v = fλ.

Is the frequency of dB wave f = v / λ, where v is the speed of the particle (ignoring Special relativity effects)?
 
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  • #2
Here is a tricky question: How do you define or measure the frequency of a de-Broglie wave?

With v = fλ, you get f=v / λ = mv2/h.
 
  • #3
So if the wavelength of an electron is 1.2 x 10e(-10), what is its frequency?

Knowing λ, I can calculate the electron's velocity as v = h/mλ. I get v = 6.07 x 10e(6).

And now f = v/λ, which comes to 5.05 x 10e(16).

Is this right?
 
  • #4
yogifromindia said:
So if the wavelength of an electron is 1.2 x 10e(-10), what is its frequency?

Knowing λ, I can calculate the electron's velocity as v = h/mλ. I get v = 6.07 x 10e(6).

And now f = v/λ, which comes to 5.05 x 10e(16).

Is this right?

Yeep!
 
  • #5
mfb said:
Here is a tricky question: How do you define or measure the frequency of a de-Broglie wave?

With v = fλ, you get f=v / λ = mv2/h.

Using this relation, energy of wave = hf = mv2.

What does mv2 mean? We know mv2/2 is K.E. of the particle. Why is the energy of dB wave twice the K.E.?
 
  • #6
yogifromindia said:
Why is the energy of dB wave twice the K.E.?
It is not the energy, it is just an expression which is used in calculations.

In relativistic quantum mechanics, this is easier (!): you get the total energy of the particle.
 
  • #7
mfb said:
It is not the energy, it is just an expression which is used in calculations.

In relativistic quantum mechanics, this is easier (!): you get the total energy of the particle.

Do you mean to say if i plugged in the relativistic formula for Mass, hf will be equal to the rest energy (m0c2) plus kinetic energy of particle mv2/2 (where m0 is rest mass and m is the relativistic mass)?

I also understand dB waves have phase and group velocities (and phase velocity is generally greater than c). While calculating frequency of dB wave using f = v/λ, is v the phase velocity or the group velocity? I would think phase velocity would be right as that is the actual wave velocity (but I am also confused because we generally disregard phase velocity as it is greater than c).
 
  • #8
There is the problem.
v in the momentum is group velocity, but if you use the group velocity in your frequency equation it is unclear (at least to me) what that "frequency" means. Phase velocity gives the usual meaning of frequency.

yogifromindia said:
Do you mean to say if i plugged in the relativistic formula for Mass, hf will be equal to the rest energy (m0c2) plus kinetic energy of particle mv2/2 (where m0 is rest mass and m is the relativistic mass)?
mv^2/2 is not the relativistic kinetic energy. The total energy is ##\gamma m c^2## and you'll see this expression in the time-dependence of the particle.
 
  • #9
Ok i did some math, but am no close to solution. I am able to reconcile Special Relativity energies, and I am also able to relate it to the dB wave's energy. But not sure if i still understand it. Here is what I did.

Take an electron traveling at 2.5 x 108 m/s (relativistic effects will be significant).

First the Special Releativity:

Rest Mass m0 = 9.1 x 10-31 kg

Relativistic Mass mr = m0/√1-(v2/c2) = 1.64x10-30 kg

Momentum P = mr x v = 1.64 x 10-30 x 2.5 x 108 = 4.12 x 10-22 kg-m/s

Rest Energy = m0c2 = 8.19 x 10-14 J
Total Energy = mrc2 = 1.48 x 10-13 J

As per S/R:

(mrc2)2 = (m0c2)2 + (cp)2

This relation reconciles.

Now the dB part:

dB Wavelength λ = h/p = 6.63 x 10-34 / 4.12 x 10-22 = 1.6 x 10-12 m

Since in this discussion we are not sure whether vphase or v should be used in frequency calculation, I have tried doing with both.

Phase Velocity vphase = c2/v = 3.6 x 108 m/s

Frequency calculated using Phase Velocity
fphase = vphase / λ = 2.25 x 1020 Hz

Frequency calculated using Group Velocity
fgroup = v / λ = 1.56 x 1020 Hz

(Note v is same as vgroup)

Energy calculated using Frequency based on Phase Velocity
Ephase = hfphase = 6.63 x 10-34 x 2.25 x 1020 J = 1.49 x 10-13 J

Energy calculated using Frequency based on Group Velocity
Egroup = hfgroup = 6.63 x 10-34 x 1.56 x 1020 J = 1.03 x 10-13 J

Now if you see here Ephase is same as TotalEnergy

i.e. hvphase/λ = mrc2

But I am still confused...what sense can be made of this? What exactly is Vphase?
 

1. What is the concept of frequency of a de Broglie wave?

The frequency of a de Broglie wave refers to the number of oscillations or cycles that a matter wave undergoes in a given time period. This concept is based on the theory of wave-particle duality, which states that all particles, including matter, have both wave-like and particle-like properties.

2. How is the frequency of a de Broglie wave related to the energy of a particle?

The frequency of a de Broglie wave is directly proportional to the energy of a particle. This means that as the energy of a particle increases, its frequency also increases. This relationship is described by the de Broglie equation, which states that the wavelength of a particle is equal to Planck's constant divided by the particle's momentum.

3. What factors affect the frequency of a de Broglie wave?

The frequency of a de Broglie wave is affected by the momentum and mass of a particle. As the momentum or mass of a particle increases, its frequency also increases. Additionally, the type of particle and the medium through which it is traveling can also affect its frequency.

4. How is the frequency of a de Broglie wave measured?

The frequency of a de Broglie wave can be measured using various experimental techniques, such as diffraction or interference experiments. These experiments involve measuring the wavelength and momentum of a particle and using the de Broglie equation to calculate its frequency.

5. What is the significance of the frequency of a de Broglie wave in modern physics?

The frequency of a de Broglie wave is significant in modern physics as it provides a deep understanding of the wave-particle duality of matter. It has also been used to explain various phenomena, such as electron diffraction and the behavior of particles in quantum systems. Additionally, the concept of de Broglie waves has been instrumental in the development of quantum mechanics and its applications in various fields of science and technology.

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