# De Broglie wavelenght as a function of an accelerating potential!

1. Jan 17, 2012

### NWSurvive

Hello everybody at physicsforums.com!

This is my very first topic in here, although I visit the forum for some time. I'm from Brazil and as english is not my mother language, i ask you not to be so harsh with my probables grammar mistakes XD.

Well, so let's get to business.

I have myself encoutered with a problem that it's giving me a hard time.
The thing is:

I have a particle of charge "e", rest mass "$m_{0}$", which moves at relativistic speed in a "V" potential. So, I have to show that the de Broglie wavelenght as a function of the accelerating potential V is equal to:

$\lambda$=$\frac{h}{\sqrt{2m_{0}eV}}$*$\left(1+\frac{eV}{2m_{0}c^{2}}\right)$$^{\frac{-1}{2}}$

Well, I think that the equations that i have to look for to start thinking in a solution are the follows:

1) $p=mv$

2) $\lambda$=$\frac{h}{p}$

3) $\lambda$=$\frac{hc}{pc}$

4) $\lambda=\frac{hc}{\sqrt{K\left(K+2mc^{2}\right)}}$

where "K" is the kinetic energy of the aprticle, "c" is the speed of light and "m" is the mass

5) $\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

where "γ" is the Lorentz tranform, "v" is the particle speed and "c" is the speed of light.

6) $p=\sqrt{2mE}$

7) $E=\sqrt{p^{2}c^{2}+m_{0}^{2}c^{4}}$

8) $pc=\sqrt{K+2Km_{0}c^{2}}$

Well, even with all these equations i was unable to find a proper solution. I mean, "K" is the kinetic energy and I have a potential "V", so "K" is equal to "eV"?

As the particle is a relativistic particle, so I must to consider $m=m_{0}\gamma$ in equation 6 for example?

I really wish to provide some previous result or attempt of solution, but I was unable to do so. All my efforts ended up in equations that don't even look like with the one that I have to achieve.

Many thanks to all who were able to read this far and if someone could provide a hint or something that guide me to the right result, then I will be very grateful.

Best regards,

NWSurvive

2. Jan 17, 2012

### Simon Bridge

Welcome to PF - you're English is better than many native speakers, don't worry.

You have a particle that has been accelerated through a potential V
You want to express the deBroglie wavelength in terms of the potential.

Normally you'd express it in terms of the speed of the particle - so can you find how the speed, depends on the potential?

3. Jan 17, 2012

### NWSurvive

Hi Simon, thanks for the reply

Well, I try to write the speed of the particle in terms of the potential, but ended up in a equation that relates the speed with the $\lambda$ and the rest mass of the particle.

I've proceeded this way:

1) $\lambda=\frac{h}{p}$

2) $\lambda=\frac{h}{m_{0}v}*\sqrt{1+\frac{v^{2}}{c^{2}}}$

After some algebra:

3) $\frac{\lambda^{2}m_{0}^{2}}{h^{2}}=\frac{c^{2}-v^{2}}{c^{2}v^{2}}$

Then:

4) $v^{2}=\frac{h^{2}c^{2}}{\lambda^{2}m_{0}^{2}c^{2}+h^{2}}$

Dividing right side by $h^{2}$ i finally get:

$v=c/\sqrt{1+\left(\frac{m_{0}c\lambda}{h}\right)}^{2}$

In the equation above "c" is being divided by the "root term".

Is it correct? Does that help in something?

NWSurvive

4. Jan 17, 2012

### Simon Bridge

$$\lambda=\frac{h}{p}$$
... relates the wavelength to the speed.

Comparing with the relation you have to prove means, you are being asked to show that:$$p=\sqrt{2m_0eV}\bigg (1+ \frac{eV}{2m_0c^2} \bigg )^{1/2}$$... it will help to simplify this expression.

If you look at the relation you are supposed to prove, you'll notice it does not have any speed terms in it?
This suggests a strategy - if you can write speed as a function of the potential, you can substitute and get rid of the v. I'll bet the result falls out from there.
But this means you need another equation that you have not thought of yet: how is speed related to accelerating potential?

Come on - the particle is accelerated through a potential V - how much kinetic energy does it gain? (are you given an initial speed?)

5. Jan 18, 2012

### NWSurvive

Hi Simon,

You mean write the potential energy in terms of kinetic energy right?

So, do I have to match the initial energy with the final energy?

I thought this way:

$E_{final}=E_{initial}$

$\frac{1}{2}mv^{2}=eV$,

Then I tought that at relativistic state I have to add one more term so:

$\frac{1}{2}mv^{2}+m_{0}c^{2}=eV$

Is that right?

It does not seem to be difficult, I think I'm making it appear so XD.

Thanks for the help again,

NWSurvive

6. Jan 18, 2012

### ehild

At relativistic speeds, the kinetic energy is KE=E-m0c2, the total energy - rest energy of the particle. eV, the energy gained from the electric field, adds to the rest energy, E=m0c2+eV, so KE=eV. You wrote the relation between energy and momentum in the first post: E2=p2c2+m02c4. Find p from E and then lambda =h/p.

ehild

Last edited: Jan 19, 2012
7. Jan 18, 2012

### Simon Bridge

That'll work - yes.
It is usually wise to assume the law of conservation of energy applies, yes.
Yes and no - ehild said what I was trying to lead you to.
It is a lot easier than most people expect.
... relativity is conceptually tricky at first (and at second and third really) but the math gets really simple.

8. Jan 19, 2012

### NWSurvive

Hi Simon and ehild,

Your help were extremely useful, thank you.

So, I made another attempt at the solution and I think that finally i was able to solve the problem.
Here is the solution i get:

1) According to Einstein:

$E=mc^{2}$

2) The relativistic kinetic energy "K" is then given as:

$K=mc^{2}-m_{0}c^{2}$

3) Energy relates with momentus as:

$p^{2}c^{2}=E^{2}-m_{0}^{2}c^{4}$

4) Combining all three equations above i got:

$p=\sqrt{\frac{k^{2}+2Km_{0}c^{2}}{c^{2}}}$

5) Putting $2Km_{0}$ in evidence:

$p=\sqrt{2Km_{0}\left(1+\frac{K}{2m_{0}c^{2}}\right)}$

6) Replacing $K=eV$, the above equation can also be writen as:

$p=\sqrt{2eVm_{0}}*\left(1+\frac{eV}{m_{0}c²}\right)^{\frac{1}{2}}$

Which is exactly what I need.

If I made something wrong or which doesn't make sense, please let me know.

Thanks again for the help!

NWSurvive

***There is a "2" in the denominator of the fraction, but I was unable to make the editor recognize it >:(.***

Last edited: Jan 19, 2012