I have further observed that lattice constant has decreased at 3%

Dear Ramesh K Sharma

Size increase up to 3 at.% dopant concentration, beyond this concentration the crystallite size decreases may be due to lattice disorder produced in the samples at higher dopant concentration. This is mainly due to difference in the ionic radii of

Zn2+, Ni2+

Thanks for your valuable answer

This is due to the difference in ionic sizes. At low doing level, dopants do not induce any lattice strain. But higher doing level introduce strain.

Dear R K Gupta

Thanks for your answer

The ionic size of Zn2+ is 0.74 Å

and  Ni 2+ is 0.70Å. 

If higher doping induces strain it must increase with impurity . but in present case it is decreasing at 5% Ni.  

at 3% it is 5nm and at 5% it is 2nm , undoped  sample is also 2nm. I am not able to correlate the observations.

I will appreciate if you elaborate your explanation.

Thanks and regards

Let us write down the global free energy of the polycrystalline sample, which is the sum over the  surface plus the bulk region assuming that we have equi-axed grains. Some arithmetic manipulations yields.

Global Gibbs =  Vol. { 3/ R  Gs   +   Gb}   where  Vol = Np (4 Pi R^3 /3) .  Where  Gs  is specific surface free energy  of grain boundaries,  Gb  is the volumetric bulk free energy. Gb includes chemical plus the stored strain energies. Np is the number of grains.

Hypothesis.  i) Initial increase in Ni  content  causes increase  the surface energy  Gs by segregating at GB regions without  to much affecting  bulk region. This increase can be compensated by the increase in the grain size R  to  keep the global Gibbs free energy constant.

ii)  The further increase in Ni content  by replacing  Zn  in the bulk start to show decrease in volumetric bulk free energy due to the reduction in stored strain energy due to the size differences. If one presumes that for high segregation at GB  results precipitation NiS without affecting the surface specific energy any more  then it is clear that  bulk reduction can be compensated by the increasing  in the surface contribution just reducing  R  particle size. 

Don' t forget,  I  have normalized the global Gibbs equation with respect to the total volume of the  bulk, which is assumed to be an invariant quantity. Del Vol= 0

Thank you Dr Tarik, Your detailed explanation has given me a very solid base to work upon.

 Dear Professor  Tarik,

 I am working on the hypothesis you have suggested. I will appreciate if you kindly suggest me some related research papers.

Further the  document containing the global Gibbs energy formula which you have quoted will certainly help me.  pl suggest me such document.

Thanks and regards

Dear Ramesh,  There is a   standard formula  for the nucleation theory, where the global Gibbs free energy is written for a single spherical particle in terms of  the bulk and surface energies. All I have done extended this formula to cover the ensemble of Np equi-axed grains enclosed by a  fixed volume (conservation of mass).  Then I got the above formula by normalizing with respect to the total volume assuming that we have spherical grains.  Actually we can further normalized with respect to the specific Gibbs free energy of the bulk  assuming that it doesn't change very much with doping compared to the surface since the volumetric ratio of the surface to bulk is extremely small (thickness of grain boundary layer is fraction of a nm).

global G=  Vol. Gb  { 1 +  3/R    (Gs/Gb)}  and   let:   f S =Gs/Gb

Let us the variance of above expression which should be zero for the stability:

Var global G=  3/R Var (fs )  -  3/R2 fs Var R  =0  Stability Criterium

Var fs  /  fs   =  Var R/ R  That means increase in  f s  results  increase in R  visa versa:

Above expression may be also written as  in terms of  the logarithmic variance:

  Var  Lnfs  = Var  LnR

Actually since we kept  the volume constant one should use the Helmholtz free energy rather the Gibbs.  This mistake is constantly made in the literature. See our article on the variational  formulation of irreversible thermodynamics  of deformable solids.

Thanks Professor Tarik for your valuable and detailed reply. It will really help me.

Thanks and regards