Structure factor for each reflection is the sum of the contributions of each atom F=Σ(fi*exp(2πi(hxi+kyi+lzi)) and intensity of the reflection is proportional to FF*. Contribution of the chosen atom to the chosen reflection can be both positive and negative, depending on the value of (hxi+kyi+lzi). When you partially substitute Ca by Mg, sign of the corresponding contribution remains the same, but the absolute value decreases (atomic factor of Mg is lower). If the contribution was positive, intensity will decrease. If the contribution was negative, intensity will decrease.

You can look at the corresponding oxides as a simple example. Reflections with even indices are quite strong in case of CaO, but in case of MgO contributions of Mg and O almost completely compensate each other and intensities become almost zero.

Structure factor for each reflection is the sum of the contributions of each atom F=Σ(fi*exp(2πi(hxi+kyi+lzi)) and intensity of the reflection is proportional to FF*. Contribution of the chosen atom to the chosen reflection can be both positive and negative, depending on the value of (hxi+kyi+lzi). When you partially substitute Ca by Mg, sign of the corresponding contribution remains the same, but the absolute value decreases (atomic factor of Mg is lower). If the contribution was positive, intensity will decrease. If the contribution was negative, intensity will decrease.

You can look at the corresponding oxides as a simple example. Reflections with even indices are quite strong in case of CaO, but in case of MgO contributions of Mg and O almost completely compensate each other and intensities become almost zero.

Dear Vijay Mishra,

As the doping level of a dopant (Mg) in HAp  increase,starins  and defects increase and  we will have XRD broading.

Regards,

Abbas

Dear  Alexander B. Missyul,

If the contribution was positive, intensity will decrease. If the contribution was negative, intensity will decrease???

Alexander's description was spot on other than the typo that was rudely (in my opinion) pointed out...positive contribution yields increased intensity, whereas a negative contribution yields decreased intensity.  And the CaO/MgO example is excellent, I have used it myself numerous times to explain the effect of atom substitution on a site (both for position of peaks and peak intensity).

This is probably the predominant reason for your changes in intensity rather than strain/defects.

Probably, I missed some details in my explanation, so it became somewhat unclear. Let us take the same example of MgO/CaO and (111) peak. The structure factor for will be

FMO=fCa*exp(2πi(1*0+1*0+1*0)) + fCa*exp(2πi(1*0.5+1*0.5+1*0)) + fCa*exp(2πi(1*0.5+1*0+1*0.5)) + fCa*exp(2πi(1*0+1*0.5+1*0.5)) + fO*exp(2πi(1*0.5+1*0+1*0)) + fO*exp(2πi(1*0+1*0.5+1*0)) + fO*exp(2πi(1*0+1*0+1*0.5)) + fO*exp(2πi(1*0.5+1*0.5+1*0.5)) = 4fM-4fO (here M is either Ca or Mg).

As long as fCa>fMg, structure factor for CaO will also be larger than for MgO. Thus,   intensity of (111) will be less for MgO than for CaO. This is what I was talking about when I said that intensity will decrease after substitution if contribution of the metal to intensity is positive.

Alexander,

position of affected peaks should also be shifting, and peak asymmetry developed, in a manner depending on substitution ratio, isn't it?

Feri

Ferenc,

Yes, peaks will be shifted (but if you dope with a small amount of some element that has similar size, such shift will not be easy to notice). As for asymmetry, it is usually determined by the diffractometer optics, not the material. Asymmetrical peak broadening is not impossible, but at least it is the good reason to ask yourself 'Do I really want to deal with such sample?'

Thanx to all for fruitfull and healthy discussion on my problem...

Occupying the lattice site (ionic radii,valency) or grain boundary,increased occupancy-concentration quenching-solubility may throw light depends upon the system