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

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
Quelsita
Messages
41
Reaction score
0
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?
 
Physics news on Phys.org
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

[tex] a\ \sin\theta =m\lambda[/tex]

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?