Electron Velocity Diffraction Grating Problem

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To solve the diffraction problem, the equation d*sin(theta) = m*lambda is used, where d is the slit distance, theta is the angle of the first maximum, and m is the order of the maximum. Given a slit distance of 0.01 mm and an angle of 3.6 degrees, the wavelength of the visible light is calculated to be approximately 6.279 x 10^-7 m. To achieve the same diffraction pattern with electrons, the De Broglie wavelength formula (lambda = h/mv) is applied. By rearranging this formula, the velocity of the electrons needed to match the diffraction pattern is found to be about 1.16 m/s. The theoretical approach is confirmed, though numerical accuracy should be verified.
Amad27
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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?
 
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Amad27 said:
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
 
Amad27 said:
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 $$.
 
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?
 
Amad27 said:
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
 

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