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

AI Thread Summary
The discussion centers on determining the correct slit width for X-ray diffraction using a wavelength of 10A and an angular width of 0.1. The initial calculation using Bragg's equation is deemed inappropriate as it applies to crystal diffraction rather than single slit diffraction. Participants suggest using the single slit diffraction equation, which relates slit width to the wavelength and angle of diffraction. The correct approach involves applying the equation a sin(θ) = mλ to find the appropriate slit width. Clarification on the correct methodology is emphasized to resolve the problem effectively.
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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