How to determine relative 2theta peak in XRD analysis?

Click For Summary
SUMMARY

This discussion focuses on determining the relative 2theta peak in X-ray Diffraction (XRD) analysis for the single crystal Zr2Cu on a bulk metal surface. The user seeks guidance on comparing experimental 2theta peaks with theoretical values from the JCPDS card, specifically addressing the importance of intensity ratios in this comparison. Key equations discussed include the relationship between wavelength and sin(theta) for both experimental and reference peaks. The conversation emphasizes the necessity of fitting experimental peaks to theoretical models and comparing relative intensities to validate findings.

PREREQUISITES
  • Understanding of X-ray Diffraction (XRD) principles
  • Familiarity with JCPDS data and its application
  • Knowledge of peak fitting techniques (Lorentzian, Gaussian, Voigt)
  • Basic proficiency in calculating intensity ratios from diffraction data
NEXT STEPS
  • Learn how to perform peak fitting in XRD analysis using software tools
  • Study the significance of intensity ratios in XRD data interpretation
  • Explore the use of chi integration for intensity calculations
  • Investigate the JCPDS database for additional reference materials and data
USEFUL FOR

Researchers, materials scientists, and crystallographers involved in XRD analysis and those seeking to validate crystal structures against theoretical models.

a_jop_rika
Messages
4
Reaction score
0
Hope somebody can help me.
I`m 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 don`t know the correct way to compare the theoretical 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.
 
Engineering news on Phys.org
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}
 
Thank you so much for the quick reply:smile:.
Hm, i`m 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?
 
sir i don't know means by 2 theta
 
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 don't know what peak (Lorentzian, gaussian, Voigt, etc) they use). Also try to assign all the peaks.
 
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.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
2
Views
4K
Replies
7
Views
17K