Electron Velocity Diffraction Grating Problem

  • Thread starter Amad27
  • Start date
  • #1
146
1

Homework Statement


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?

Homework Equations



d*sin(theta) = m(lamba)

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?
 

Answers and Replies

  • #2
963
213
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?

Homework Equations



d*sin(theta) = m(lamba)

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?
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
jtbell
Mentor
15,735
3,892
equations like $d\sin(\theta) = m\lambda$,
A friendly tip: to show your equations inline with text, enclose your LaTeX equations in ##, not $. To show them as separate "paragraphs", use $$.
 
  • #4
146
1
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
963
213
Thus,

v=6.63∗10−34(9.1∗10−31)(6.279∗10−7)=1.16m/s​
v = \frac{6.63*10^{-34}}{(9.1*10^{-31})(6.279*10^{-7})} = 1.16 m/s

Is this right?
i have not checked your numbers but theoretically the idea is same
 

Related Threads on Electron Velocity Diffraction Grating Problem

  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
877
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
2
Views
2K
Replies
10
Views
10K
Replies
5
Views
5K
Top