Homework Help: De broglie wavelength

ariana13 · May 8, 2008

How do you calcuate at what energy do a photon and an electron have the same de broglie wavelength?

Hootenanny · May 8, 2008

ariana13 said: ↑

How do you calculate at what energy do a photon and an electron have the same de broglie wavelength?

Hi ariana13 and welcome to PF,

I'm assuming that this is a homework question, in which case for future reference we have Homework & Coursework problems for such questions. Don't worry about it now, your thread will get moved there in due course.

Now for your question. What do you know about the de Broglie wavelength? How is it claculated?

ariana13 · May 8, 2008

Sorry if I put this in the wrong thread! For a photon, wavelength is just lambda=h*c/energy. I think for an electron you use the relativistic equation lambda=h/mv*sqt(1-v^2/c^2). I've tried equating these, but i ended up with a horrible equation to solve because i don't have velocity of the electron. I think i must have gone wrong somewhere.

Hootenanny · May 8, 2008

Since we are taking about energies here, it may be more useful to use an alternative form for the de Broglie wavelength for the electron,

[tex]\lambda = \frac{hc}{pc} = \frac{hc}{\sqrt{T^2+2Tm_0c^2}}[/tex]

ariana13 · May 8, 2008

Thanks for your help. Sorry if this is a stupid question, but i've never seen that equation before, where does it come from? What does T stand for?

Hootenanny · May 8, 2008

ariana13 said: ↑

Thanks for your help. Sorry if this is a stupid question, but i've never seen that equation before, where does it come from? What does T stand for?

There are no stupid questions

Anyway, it comes from the relativistic energy eqaution,

[tex]E^2 = \left(pc\right)^2 + \left(m_0c^2\right)^2[/tex]

Where E it the total energy. In the expression in my previous post T represents the kinetic energy of the electron. Of course if one would prever to calculate the total energy of the electron (including rest energy) one may rewrite the previous equation,

[tex]\lambda = \frac{hc}{pc} = \frac{hc}{\sqrt{E^2 - m_0^2c^4}}[/tex]

ariana13 · May 8, 2008

Thanks for clarifying that, i think i do it now. :)