how to determine relative 2theta peak in XRD analysis?

[a_jop_rika]

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 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.

[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\the ta_{exp}}\over{\lambda_{exp}}}}

{{\lambda_{exp}}\over{\lambda_{ref}}}{sin\theta_{r ef}={sin\theta_{exp}

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

[a_jop_rika]

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?