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## Main Question or Discussion Point

is interplanar spacing or interatomic spacing directly proportional to grain size?

- Thread starter kimmylsm
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- #1

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is interplanar spacing or interatomic spacing directly proportional to grain size?

- #2

Mapes

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- #3

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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.

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Hi i am sure about the result but it will be helpful.

we know the following three formula:

[tex]

{\rm 1.~}2d\sin\theta=n\lambda,

\hspace{3mm}

{\rm 2.~}d=\frac{a}{\sqrt{h^2+k^2+l^2}},

\hspace{3mm}

{\rm 3.~}D=\frac{K\lambda}{\beta\cos\theta}.

[/tex]

From these three formulae you arrive at the following two formulae:

[tex]

D=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda}{2d}\right)\right]}

=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}\right)\right]}

[/tex]

[tex]\theta[/tex] and [tex]\beta[/tex] are in degrees and radians, respectively.

This is how i related particle size and 'd' spacing.

good luck

we know the following three formula:

[tex]

{\rm 1.~}2d\sin\theta=n\lambda,

\hspace{3mm}

{\rm 2.~}d=\frac{a}{\sqrt{h^2+k^2+l^2}},

\hspace{3mm}

{\rm 3.~}D=\frac{K\lambda}{\beta\cos\theta}.

[/tex]

From these three formulae you arrive at the following two formulae:

[tex]

D=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda}{2d}\right)\right]}

=\frac{K\lambda}{\beta\cos\left[\sin^{-1}\left(\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}\right)\right]}

[/tex]

[tex]\theta[/tex] and [tex]\beta[/tex] are in degrees and radians, respectively.

This is how i related particle size and 'd' spacing.

good luck

Last edited:

- #6

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thanks a lot.

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