What Is the Correct Slit Width for X-Ray Diffraction with a 10A Wavelength?

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Homework Help Overview

The discussion revolves around the diffraction of X-rays when passing through a fine slit, specifically focusing on determining the appropriate slit width for X-rays with a wavelength of 10Å to achieve a specific diffraction pattern. The problem involves concepts from wave optics and diffraction theory.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Bragg's equation and question its relevance to single slit diffraction. There are suggestions to consider the single slit diffraction equation and the grating equation, with inquiries about the maximum angle involved.

Discussion Status

The discussion is active, with participants exploring different equations relevant to the problem. Some guidance has been offered regarding the appropriate equations to use, but there is no clear consensus on the correct approach yet.

Contextual Notes

Participants note the distinction between Bragg's law, which pertains to crystal diffraction, and the equations relevant for single slit diffraction. There is an emphasis on ensuring the correct context and assumptions are applied to the problem.

Quelsita
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Problem:
Diffraction of X rays has been observed when a beam strikes an extremely fine slit. What must be the width if the slit if X rays of wavelength 10A are to display a diffraction pattern with a central beam of angular width 0.1?

I used Bragg's equation to obtain:
d=n(lambda)/2sin(theta)
d=1*10A/2sin(0.1)=2864.8A= 246.48nm?

Is this correct?
 
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Hi Quelsita,

Quelsita said:
Problem:
Diffraction of X rays has been observed when a beam strikes an extremely fine slit. What must be the width if the slit if X rays of wavelength 10A are to display a diffraction pattern with a central beam of angular width 0.1?

I used Bragg's equation to obtain:
d=n(lambda)/2sin(theta)
d=1*10A/2sin(0.1)=2864.8A= 246.48nm?

Is this correct?

I don't believe you are using the correct equation here. Bragg's law describes diffraction by a crystal; here the process is single slit diffraction. What equation would describe that?
 
Could you use the grating equation d[sin(thetam)+sin(thetai)]=m(lambda)?
But what is thetamax?
 
Quelsita said:
Could you use the grating equation d[sin(thetam)+sin(thetai)]=m(lambda)?
But what is thetamax?

I think you should use the single slit diffraction equation. The equation for the minima is

<br /> a\ \sin\theta =m\lambda<br />

I would think that your book has at least a section on single slit diffraction, so it can tell you how to apply this to your problem. What do you get?
 

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