What are your thoughts? Although this may not be homework, for homework-type questions we need to see your attempt at figuring things out yourself before giving comments.
#3
Rajini
619
4
Before posting your question please supply more information..
for e.g., what material, i mean polycrystalline or single crystal, etc.
So people here would clarify your doubts.
To me, there is no direct relation between particle size, gran size and spacing.
#4
kimmylsm
19
0
is polycrystalline that i am interested. from debye-scherrer formula where D = k(lamda)/(FWHM)(costheta). i wan to connect spacing and grain size. is it possible? i thought spacing is directly proportional to grain size.
#5
Rajini
619
4
Hi i am sure about the result but it will be helpful.
we know the following three formula:
<br />
{\rm 1.~}2d\sin\theta=n\lambda,<br />
\hspace{3mm}<br />
{\rm 2.~}d=\frac{a}{\sqrt{h^2+k^2+l^2}},<br />
\hspace{3mm}<br />
{\rm 3.~}D=\frac{K\lambda}{\beta\cos\theta}.<br />
From these three formulae you arrive at the following two formulae:
<br />
D=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda}{2d}\right)\right]}<br />
=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}\right)\right]}<br />
\theta and \beta are in degrees and radians, respectively.
This is how i related particle size and 'd' spacing.
good luck
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