# De Broglie wavelength of electron and proton

## 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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ehild
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## Homework Statement

View attachment 227152

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 ?
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.

Jahnavi
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 ?

ehild
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Jahnavi
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 Outdated books says ##m=\gamma m_{o}##, is the relativistic mass.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.

Jahnavi
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 .

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 ?

ehild
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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.

Jahnavi
ehild
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Could you show me how you get this result .

What is rest energy ?
m0c2.

Jahnavi
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 .

ehild
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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"

Jahnavi
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}##

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.

Jahnavi
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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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Jahnavi and Phylosopher
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

Completely different result .

Interesting !