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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 "[itex]m_{0}[/itex]", 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:

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

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

1) [itex]p=mv[/itex]

2) [itex]\lambda[/itex]=[itex]\frac{h}{p}[/itex]

3) [itex]\lambda[/itex]=[itex]\frac{hc}{pc}[/itex]

4) [itex]\lambda=\frac{hc}{\sqrt{K\left(K+2mc^{2}\right)}}[/itex]

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

5) [itex]\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex]

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

6) [itex]p=\sqrt{2mE}[/itex]

7) [itex]E=\sqrt{p^{2}c^{2}+m_{0}^{2}c^{4}}[/itex]

8) [itex]pc=\sqrt{K+2Km_{0}c^{2}}[/itex]

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 [itex]m=m_{0}\gamma[/itex] 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

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

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