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XRay Diffraction

  1. Dec 1, 2016 #1
    1. The problem statement, all variables and given/known data
    X-rays of wavelength λ are incident on a cubic crystal with spacing d as shown in the figure. The incident angle is fixed at θi degrees. A detector is rotated from perpendicular to the 100 plane (90 degrees (grazing)) until the first maximum is detected at θs degrees due to interference from the 100 plane. What is the lattice spacing, d, of the cubic crystal?

    2. Relevant equations
    Bragg's Law : mλ=2dsinθ

    3. The attempt at a solution
    Known: m=1, λ
    The angle to be used for Bragg's Law is not the incident/reflected angle because θs does not equal θi. So I figured out the angle between the incoming and reflected ray (180-θsisplit) and that would not be the angle we are looking for either. So i divided that θsplit by 2, and subtracted it from 90. To find what I thought was the correct angle but apparently not.

    If this is not the angle, what is and how do i get to it?
    Is m≠1 despite the problem stating "first order maximum"?
    is my θs not the correct angle?

    Thank you in advance. also sorry about the images, they work better if you drag them into a new tab.
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2016 #2

    Charles Link

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    Homework Helper

    This is not (IMO) characteristic of what are typically called Bragg maxima, which need to meet two conditions : 1) The angle of incidence equals angle of reflection, and 2) The reflected rays(waves) from adjacent planes must constructively interfere. Here, the first condition is not met. (You will get a peak, and it could perhaps even be called a Bragg peak, but it is not the extremely intense type of Bragg peak that you get when conditions (1) and (2) are both met). ## \\ ## For this one, simple interference principles and equations will apply (and get you the answer) with a slight modification. The simple interference/diffraction equations usually are written with zero degree angle of incidence and you get ## m \lambda=d \sin(\theta_s) ## for the interference maxima. ## \\ ## Question for you (I am not allowed to give you the answer, at least not without some effort on your part) How would the interference equation be modified for a bunch of equally spaced slits if the incident angle were ## \theta_i ## instead of zero degrees? For a couple hints on this, at normal incidence, ( ## \theta_i=0 ##), why is the path distance difference between adjacent slits (or scatterers) equal to ## d \sin(\theta_s) ## ? What is the additional path distance difference (between rays at adjacent slits) that occurs if the incident angle is some non-zero ## \theta_i ##? How would you then get the total path distance difference between rays from adjacent slits(or scatterers) in the case of a non-zero ## \theta_i ## plus a ## \theta_s ##, and what is the condition for constructive interference between the rays from adjacent slits (or scatterers)? How does your modified equation for constructive interference read for this (more general non-zero ## \theta_i ##) case? From this equation you can compute the distance ## d ## between the crystal planes.
     
    Last edited: Dec 1, 2016
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