# Analyse XRD Data

1. Oct 19, 2009

### 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 $$\Theta-\Theta$$Diffractometer. $$\lambda = 1,54 \cdot 10^{-10}m$$

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

2. Oct 19, 2009

### drizzle

Last edited by a moderator: Apr 24, 2017
3. Oct 19, 2009

### 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

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

4. Oct 19, 2009

### drizzle

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
5. Oct 20, 2009

### 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

6. Oct 20, 2009

### drizzle

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?

7. Oct 20, 2009

### Phileas.Fogg

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

8. Oct 20, 2009

### Phileas.Fogg

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

$$q_z = 2 k \sin(\alpha_i)$$

9. Oct 20, 2009

### drizzle

ehim , 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]

10. Oct 20, 2009

### Phileas.Fogg

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

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?

11. Oct 20, 2009

### drizzle

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

12. Sep 30, 2010

### nyxynyx

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?

13. Oct 1, 2010

Hello,