Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Analyse XRD Data

  1. Oct 19, 2009 #1
    Hello,
    I have X-Ray Diffraction Data: Intensity versus angle [tex] 2 \Theta[/tex] and shall find out the lattice constant and even better the crystal structure. The Data is from a [tex]\Theta-\Theta[/tex]Diffractometer. [tex]\lambda = 1,54 \cdot 10^{-10}m[/tex]

    I know that I have to find the peaks and can calculate d from the Bragg equation:
    [tex] d = n \lambda/2 \sin\theta[/tex]
    Is it correct to take half of the measured angle for the equation and to set n=1 in this case?

    Moreover my problem is, I don't know how to find out the lattice constant and the structure by calculating d. I don't have any data to compare d with (where do I get this data?).
    I know that the material ist Tungsten Carbide or Tungsten-Cobalt with hexagonal crystal structure, but I shouldn't use this information beforehand. What shall I do?

    Regards,
    Mr. Fogg
     
  2. jcsd
  3. Oct 19, 2009 #2

    drizzle

    User Avatar
    Gold Member

    Last edited by a moderator: Apr 24, 2017
  4. Oct 19, 2009 #3
    Hello,
    thanks!

    Do I also have to convert [tex]2 \Theta[/tex] into Radians when I use the Bragg's Law? In my calculation, I sometimes get a negative d , is that possible?

    How do I Index the peaks with Miller Indices?

    I don't know, how this equation helps me

    [tex] \frac{1}{d_{hkl}^2} = \frac{4}{3} \frac{h^2+hk+k^2}{a^2} + \frac{l^2}{c^2} [/tex]
     
  5. Oct 19, 2009 #4

    drizzle

    User Avatar
    Gold Member

    no, theta in Bragg’s law is in degrees. it's not possible at all to have d as a negative value



    look for the JCPDS file of this compound [it is a reference data that holds both the fixed values of d for all possible peaks of the material, and the corresponding (hkl) planes], use this equation along with Bragg’s law to solve, you know that two equations are needed to solve for two variables [which are the lattice constants a and c. of course you use the value of the measured d at a certain peak, that is a certain plane where it can be identified using the JCPDS card, do this for two peaks then solve], good luck!
     
    Last edited: Oct 20, 2009
  6. Oct 20, 2009 #5
    Thank You,
    where can I get the JCPDS file?

    When I convert the measured angle [tex] 2 \Theta[/tex] into the z-component of the wave vector with

    [tex] q_{z,i} = \frac{4 \pi}{\lambda} \sin(\alpha_i) [/tex]

    do I have to halve [tex] 2 \Theta [/tex] ?

    Mr. Fogg
     
  7. Oct 20, 2009 #6

    drizzle

    User Avatar
    Gold Member


    search the net! look for tungsten carbide (WC) JCPDS powder diffraction file

    ps. JCPDS = International Center for Powder Diffraction Data



    I don’t quite follow you, what is this for?
     
  8. Oct 20, 2009 #7
    One measured peak is for example at [tex] 2 \Theta = 157523,511 [/tex]°. In my calculation

    1) Division by 2 gives [tex] \Theta = 78761,756 [/tex] °
    2) Converting into radiants (for OpenOffice Calc) gives [tex] \Theta = 1374,652 [/tex]
    3) now calculating (with OpenOffice Calc) [tex] \sin(\Theta) = -0,979 [/tex]

    So I get a negative d now! What's wrong? Maybe I am too stupid to handle OpenOffice Calc in this case :tongue2:

    If I don't convert into radiants, the problem is still present concerning other peaks.

    Mr. Fogg
     
  9. Oct 20, 2009 #8
    The new wave vector in our experiment is final minus initial wave-vector:

    [tex] \vec{q} = \vec{k}_f - \vec{k}_i[/tex]

    and it's z-component is

    [tex] q_z = 2 k \sin(\alpha_i)[/tex]
     
  10. Oct 20, 2009 #9

    drizzle

    User Avatar
    Gold Member

    ehim :biggrin:, I think this is the value of the maximum intensity, right? in any XRD pattern the range of 2theta values is from 0 to 90 or so [and the horizontal axis is the one you look at to get the angle at which the peak occurs ;)]


    ps. actually, if you do convert the angle into radians instead of degrees it would give the same result. but B should be converted into radians before calculating the grain size g [as shown in the linked thread]
     
  11. Oct 20, 2009 #10
    :biggrin: :biggrin: :biggrin:

    Now I found my mistake.

    I didn't replace the dot with a comma in the file, I got. That's why there occured these incredible angles :smile:

    Now I will revise my analysis and keep your help in mind. When a question occurs, I will ask you again.

    Thank You!

    Mr. Fogg

    PS: Do You know what to do with the wave vector?
     
  12. Oct 20, 2009 #11

    drizzle

    User Avatar
    Gold Member

    I’m not really familiar with this, sorry. but I'm sure you'll get the help you need from other PF members...welcome anyways :smile:
     
  13. Sep 30, 2010 #12
    what equation is this called? what are the parameters 'l' and 'c'? how do u find lattice parameter if the cell is not cubic but tetragonal or orthorhombic?
     
  14. Oct 1, 2010 #13
    Hello,
    I already finished my work

    @ nyxynyx : Do you want to know that, or do you try to help me? Because everything is finished already. I don't need any replies to this thread, but thanks.

    Mr. Fogg
     
  15. Oct 1, 2010 #14
    i would like to know that! i believe it has got something to do with the extinction rules but I cant find any tables for extinction rules of tetragonal and orthorhombic
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Analyse XRD Data
  1. XRD database (Replies: 7)

  2. How to analyse XRD data? (Replies: 73)

  3. Help with XRD data (Replies: 2)

Loading...