Electron Velocity Diffraction Grating Problem

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Homework Help Overview

The problem involves a diffraction grating with a specified slit distance and an angle for the first maximum of visible light. The original poster is exploring how to determine the velocity of electrons that would produce a similar diffraction pattern using the same grating.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the grating equation \(d \sin(\theta) = m \lambda\) to relate the angle and slit distance to the wavelength. There are questions about how the angle factors into the calculations and the relationship between the electron's wavelength and its velocity.

Discussion Status

Some participants have provided calculations for the wavelength based on the given angle and slit distance, and there is a mention of using De Broglie's equation to relate wavelength to velocity. However, there is no explicit consensus on the correctness of the calculations presented.

Contextual Notes

There is an underlying assumption regarding the wave properties of electrons and how their wavelength is influenced by velocity, which is being explored in the discussion.

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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