Analyse XRD Data

[Phileas.Fogg]

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

I know that I have to find the peaks and can calculate d from the Bragg equation:
d = n \lambda/2 \sin\theta
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

[drizzle]

Is it correct to take half of the measured angle for the equation and to set n=1 in this case?

yes. also you can check this thread (http://www.physicsforums.com/showthread.php?t=341880)for your other questions :)

[Phileas.Fogg]

Hello,
thanks!

Do I also have to convert 2 \Theta 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

\frac{1}{d_{hkl}^2} = \frac{4}{3} \frac{h^2+hk+k^2}{a^2} + \frac{l^2}{c^2}

[drizzle]

Hello,
thanks!

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

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



How do I Index the peaks with Miller Indices?

I don't know, how this equation helps me

\frac{1}{d_{hkl}^2} = \frac{4}{3} \frac{h^2+hk+k^2}{a^2} + \frac{l^2}{c^2}

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!

[Phileas.Fogg]

Thank You,
where can I get the JCPDS file?

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

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

do I have to halve 2 \Theta ?

Mr. Fogg

[drizzle]

Thank You,
where can I get the JCPDS file?


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

ps. JCPDS = International Center for Powder Diffraction Data



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

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

do I have to halve 2 \Theta ?

Mr. Fogg

I don’t quite follow you, what is this for?

[Phileas.Fogg]

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


One measured peak is for example at 2 \Theta = 157523,511 °. In my calculation

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

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

[Phileas.Fogg]

I don’t quite follow you, what is this for?

The new wave vector in our experiment is final minus initial wave-vector:

\vec{q} = \vec{k}_f - \vec{k}_i

and it's z-component is

q_z = 2 k \sin(\alpha_i)

[drizzle]

One measured peak is for example at 2 \Theta = 157523,511 °. In my calculation

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]

[Phileas.Fogg]

: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?

[drizzle]

: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?

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: