# Double-slit/Kinetic Energy

1. Apr 2, 2009

### dcramps

1. The problem statement, all variables and given/known data
A double-slit interference experiment is performed with photons of energy Eγ. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (that is, the spacing between maxima is the same)? Express your answer in terms of Eγ, me (electron mass), and c (speed of light).

2. Relevant equations

3. The attempt at a solution
I really hate to do this, but I have absolutely no idea what it is asking me here or how to do it.

How does using photons and electrons change the outcome? I am unsure how to calculate the kinetic energy...do I modify the dsinΘ=mλ equation somehow?

I am really lost here!

2. Apr 2, 2009

### rl.bhat

Energy of photon = hc/lambda. hence lambda(p) = hc/Ey
For electrons lambda(e) = h/mv = h/(2mEe)^1/2
Equate lambda(p) and lambda(e) and solve for Ee

3. Apr 2, 2009

### dcramps

For electrons, that's the de Broglie wavelength, correct? I had just happened across that in my notes, but didn't get so far as to change momentum to (2mEe)1/2

Thank you for responding at such a late hour >_< I was not expecting such a quick reply!