De Broglie wavelength of electron and proton

In summary, the de Broglie wavelength of a proton will be smaller than that of an electron due to its larger mass and momentum. The relation E2 = P2c2 + m02c4 is valid for calculating the total energy of a particle, but it is important to use the correct notation for the mass and momentum. The rest energy of a particle can also be calculated using the formula E=mc2.
  • #1
Jahnavi
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Homework Statement


De broglie.jpg


Homework Equations

The Attempt at a Solution



de Broglie wavelength λ of a particle = h /P

a) since mass of proton is more than electron and speed is same , momentum of proton is more . De Broglie wavelength of proton will be less .

b) wavelengths will be same .

c) Using P = √(2Km) . Since energy is same and mass of proton is more , it's momentum will be more . Hence wavelength of proton will be less .

Is that correct ?

I would like to know whether I should have used the relation E2 = P2c2 + m02c4 ?

When is the above relation valid ? Is it only when the particles are moving at the speed of light ?

I am not sure when are we supposed to use the above relation . Should we use the above relation or using P = √(2Km) was okay ?
 

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  • #2
Jahnavi said:

Homework Statement


View attachment 227152I would like to know whether I should have used the relation E2 = P2c2 + m02c4 ?

When is the above relation valid ? Is it only when the particles are moving at the speed of light ?

I am not sure when are we supposed to use the above relation . Should we use the above relation or using P = √(2Km) was okay ?
It depends what did the problem mean on "energy". The total energy, or the kinetic energy.
In the formula E2 = P2c2 + m02c4 energy means the total energy of the particle, E=mc2=m0γc2.
And you know quite well, that particles with non-zero mass can not move with the speed of light.
When the speed of a particle is much less than that of the speed of light the kinetic energy KE=P2/2m. Otherwise, you should use the relativistic quantities and formulas.
 
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  • #3
ehild said:
It depends what did the problem mean on "energy". The total energy, or the kinetic energy.
In the formula E2 = P2c2 + m02c4 energy means the total energy of the particle, E=mc2=m0γc2.

How did you get E=mc2 ?

What is γ ?
 
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  • #5
Jahnavi said:
How did you get E=mc2 ?

Just to make it clear, ##m_{o}## is the actual mass of the particle, not ##m## which is just ##\gamma m_{o}##. Outdated books says ##m=\gamma m_{o}##, is the relativistic mass. "Don't use this term its wrong, and don't think of it as mass"

To get ##m=\gamma m_{o}##, write E as sum kinetic energy and rest energy and then substitute rest mass and kinetic energy (relativistic form). You will get the same answer as ehlid.
 
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  • #6
ehild said:
You never saw Einstein's famous equation E=mc2?

I have .

But I thought this was obtained from the formula E2 = P2c2 + m02c4 by putting P = 0 .

I thought E = mc2 was valid for a particle at rest i.e for a particle having zero momentum .
 
  • #7
Phylosopher said:
To get ##m=\gamma m_{o}##, write E as sum kinetic energy and rest energy and then substitute rest mass and kinetic energy (relativistic form). You will get the same answer as ehlid.

Could you show me how you get this result .

What is rest energy ?
 
  • #8
Jahnavi said:
I have .

But I thought this was obtained from the formula E2 = P2c2 + m02c4 by putting P = 0 .

I thought E = mc2 was valid for a particle at rest i.e for a particle having zero momentum .
Well, recently m means the rest mass (invariant mass) of the particle. When I studied relativity the rest mass was denoted by m0 and m was the relativistic mass. The formula E2 = P2c2 + m02c4 is not an axiom, it can be derived from the expression of energy and momentum.
 
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  • #9
Jahnavi said:
Could you show me how you get this result .

What is rest energy ?
m0c2.
 
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  • #10
ehild said:
In the formula E2 = P2c2 + m02c4 energy means the total energy of the particle, E=mc2=

Are the two expressions equivalent ?

What should we put for P (momentum ) ?

Should it be mv ?

I am studying basics of photons , so I am quite hesitant to use mv for momentum for other particles as well .
 
  • #11
Jahnavi said:
Are the two expressions equivalent ?

What should we put for P (momentum ) ?

Should it be mv ?

I am studying basics of photons , so I am quite hesitant to use mv for momentum for other particles as well .
Other people say that I must not use the relativistic mass. So the momentum is P=mγv, and E=mγc2.
See https://en.wikipedia.org/wiki/Energy–momentum_relation, "heuristic approach for massive particles"
 
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  • #12
Jahnavi said:
Could you show me how you get this result .

$$E=E_{kinetic}+E_{rest}$$

let ##m_{o}## be the rest mass. Then rest energy and kinetic energy are as follow:

$$E_{kinetic}=(\gamma -1) m_{o} c^{2}$$
$$E_{rest}= m_{o}c^{2}$$

Thus ##E=\gamma m_{o} c^{2}##

Jahnavi said:
Are the two expressions equivalent ?

What should we put for P (momentum ) ?

Should it be mv ?

I am studying basics of photons , so I am quite hesitant to use mv for momentum for other particles as well .

From your questions, it seems that you didn't yet study relativity fully. Maybe you are starting your study on the subject? Look at the wiki links ehlid kindly provided.
 
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  • #13
The question is not very specific in part (c). When it says the two particles have the same energy, they do not specify relativistic energy (with ## E^2=p^2c^2+m_o^2c^4 ##) or simply kinetic energy. I think it was a good observation on your part @Jahnavi that you recognized that there are two possibilities here. ## \\ ## The answer is different depending on which type of energy is implied.
 
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  • #14
Charles Link said:
The answer is different depending on which type of energy is implied.

In the OP in part c) I have assumed energy to be kinetic energy in which case the momentum of proton was more and wavelength smaller .

But if I consider energy to be relativistic energy in part c) , then momentum of photon comes out to be less and wavelength of proton is longer .

I hope you agree with this :smile:

Completely different result .

Interesting !
 
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  • #15
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Related to De Broglie wavelength of electron and proton

What is the De Broglie wavelength of an electron?

The De Broglie wavelength of an electron is a fundamental concept in quantum mechanics that describes the wave-like behavior of matter. It is given by the equation λ = h/mv, where h is Planck's constant, m is the mass of the electron, and v is the velocity of the electron.

What does the De Broglie wavelength tell us about the electron?

The De Broglie wavelength tells us that even particles with mass, like electrons, have wave-like properties. This means that they can exhibit interference and diffraction, similar to waves. It also relates the momentum of the electron to its wavelength, showing the duality of particles and waves in the quantum world.

How is the De Broglie wavelength of an electron different from that of a proton?

The De Broglie wavelength of an electron is typically much smaller than that of a proton because electrons have a much lower mass and higher velocity. This means that the wave-like behavior of electrons is more noticeable in their interactions, while protons behave more like classical particles.

Can the De Broglie wavelength of an electron be measured?

Yes, the De Broglie wavelength of an electron can be measured using diffraction experiments, where the electron is passed through a small slit and its diffraction pattern is observed. This pattern can then be used to calculate the wavelength of the electron.

What are the practical applications of the De Broglie wavelength of an electron?

The De Broglie wavelength of an electron is crucial in understanding the behavior of matter at the atomic and subatomic level. It has practical applications in fields such as quantum computing, electron microscopy, and particle accelerators. It also helps to explain phenomena such as electron diffraction and the stability of atoms.

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