# How to determine relative 2theta peak in XRD analysis?

1. Aug 31, 2009

### a_jop_rika

Hope somebody can help me.
Im currently doing analysis of determining single crystal of Zr2Cu on a surface of bulk metal.
Through XRD, i determine the peak and compared it with data from JCPDS card.
My problem is I dont know the correct way to compare the theoritical 2theta peak(from JCPDS) with the experiment 2theta peak. I mean how to prove that, like for example the 50degrees peak from experiment is comparable with 53degrees peak of theorotical 2theta, based on relative ratio calculation or sth like that? Sorry if I sound confusing pls tell me.

2. Sep 1, 2009

### PhaseShifter

Do you mean the reference uses a different wavelength than the experiment and you need to correlate reference peaks to experimental peaks?

$$2d~sin\theta=n\lambda$$

$${{sin\theta}\over{\lambda}}={n\over{2d}}$$

$${{sin\theta_{ref}}\over{\lambda_{ref}}}}={{sin\theta_{exp}}\over{\lambda_{exp}}}}$$

$${{\lambda_{exp}}\over{\lambda_{ref}}}{sin\theta_{ref}={sin\theta_{exp}$$

$$sin^{-1}({{\lambda_{exp}}\over{\lambda_{ref}}}{sin\theta_{ref})=\theta_{exp}$$

3. Sep 3, 2009

### a_jop_rika

Thank you so much for the quick reply.
Hm, im quite familiar with the equation you gave but the hint is intensity.
I think in order to say that "this 2theta from experiment is comparable with this 2theta from JCPDS it must have something to do with the intensity.

If I get to prove that for example, `the 50degrees from experiment is comparable with 53degrees of JCPDS(for Zr2Cu)', then I can use the hkl lattice data in JCPDS to build crystal model.

The intensity can be figure out by chi integration. The JCPDS data also have intensity(i) data, so the calculation must be around these two intensities?

Have any idea?

4. Jun 9, 2010

### Hidayat

sir i dont know means by 2 theta

5. Jun 9, 2010

### Rajini

Yes you can say...your exp. value matches jcpds data..If there are not peaks around your experimental 50 degree peak. But you should be careful in saying..How about the other peaks? does other peaks vary a lot ?
or try to find the allowed error deviation !
Intensity is just the area under the peak..So in practice you should fit the exp. peak with the theoretical peak (i actually dont know what peak (Lorentzian, gaussian, Voigt, etc) they use). Also try to assign all the peaks.

6. May 9, 2011

### PaulM2011

I was given a similar request. The objective of this is to make the intensities of each observed reflection (peak) in a given diffractogram comparable to the dominant reflection. Set the highest peak as 100% and all other peaks will be compared to that. So the next largest peak might be 80%, 45% and 5% etc... You can then make a judgement on any differences in orientation by comparing the relative intensities to the JCPDS relative intensities.