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Prof's Bonus Question for XRay Diffraction

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi guys, so here's a bonus question on my assignment for materials course.

    " A metallic crystal is irradiated with Xrays of a fixed wavelength. Bragg reflection is observed at angles of 23 and 73.5 degrees, as well as another angle phi which is intermediate between the above two values. Assuming the same set of crystal planes (separation d) are responsible for scattering in all three cases, calculate a possible value of phi in degrees. Are there other possible values for phi apart from the value that you found?


    2. Relevant equations

    [tex]
    2dsin(\theta)= m \lambda
    [/tex]
    i think this is the only equation we need for Bragg peaks because we didn't learn anything else
    3. The attempt at a solution
    k so i figured if the wavelenth, distance are constant then only m and [tex] \theta [/tex] are changing so i got:

    [tex]
    m1 = 2dsin23/ \lambda = 0.781 d/ \lamda
    [/tex] and
    [tex] m2= 2dsin73.5/ \lambda = 1.918 d/ \lambda
    [/tex]

    I have absolutely no clue how to go from here? any help is much appreciated
     
  2. jcsd
  3. Jan 23, 2010 #2
    The complete solution was deleted (it's not allowed). So if you didn't see it:

    You can get [tex] m_1 / m_2 [/tex] from the two last equations.
    Then you should select a pair of integers which fit this ratio.
    Any integer number between them will give you [tex] \varphi [/tex].
     
  4. Jan 23, 2010 #3
    Ok thanks i'll give that a try, and no I didn't see the complete solution. I'll report back when I get it
     
  5. Jan 23, 2010 #4
    so heres what i got:

    [tex]
    \frac{m_{1}}{m_{2}} = 0.407
    [/tex]
    from here I concluded that that ratio is approximately [tex] m_{1} = 2 [/tex] and [tex] m_{2} = 5 [/tex]
    so the possible values for [tex] m_{\phi}= [/tex] 3 and 4
    Then using [tex] m_{1}=2 [/tex] I plugged that back into first equation above and got [tex] \frac{d}{\lambda} = 2.56 [/tex] so [tex] \frac{\lambda}{d} = 0.391 [/tex]
    So using [tex] m_{\phi}= 3[/tex] I got [tex] \phi = 35.9 [/tex] and using [tex] m_{\phi}= 4[/tex] I got [tex] \phi = 51.4 [/tex]
    And these values work great! Thanks alot
     
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