# Electron Velocity Diffraction Grating Problem

1. May 9, 2016

1. The problem statement, all variables and given/known data
Suppose that visible light incident on a diffraction grating with slit distance (space) of $0.01*10^{-3}$ has the first max at the angle of $3.6^{o}$ from the central peak. Suppose electrons can be diffracted with this same grating, which velocity of the electron would create the same diffraction pattern as this visible light?

2. Relevant equations

d*sin(theta) = m(lamba)

3. The attempt at a solution

I am thinking of using equations like $d\sin(\theta) = m\lambda$, but I am not sure where the angle comes into play here.

Obviously,

$(0.01 mm)(\sin(3.6)) = m\lambda$, but this doesn't help much?

2. May 9, 2016

### drvrm

use the grating equation to find out the wavelength of the diffraction pattern -given out with light.
suppose you wish to have the same pattern with electron beam - then the wavelength associated with electron should be same.
can the the electron have wave property?
if it can then how wavelength depends on its velocity/energy/momentum?
the lambda should be related with velocity

3. May 9, 2016

A friendly tip: to show your equations inline with text, enclose your LaTeX equations in ##, not $. To show them as separate "paragraphs", use $$. 4. May 9, 2016 ### Amad27 Thanks. Okay:$$0.01\sin(3.6) = m\lambda$$so m=1, and thus$$\lambda = 6.279*10^{-7} m$$Then by De Broglie,$$\lambda = h/mv$$Thus,$$v = \frac{6.63*10^{-34}}{(9.1*10^{-31})(6.279*10^{-7})} = 1.16 m/s$\$

Is this right?

5. May 9, 2016

### drvrm

i have not checked your numbers but theoretically the idea is same