Example:Particle spacing from peak position

Consider the case of trying to measure the particle-particle spacing from the q-value of a particular peak. The interpretation of the q value of course depends upon the packing of the particles; i.e. the unit cell. Consider a cubic unit cell (SC, BCC, FCC). Note that in general:

${\begin{alignedat}{2}\left({\frac {1}{d_{hkl}}}\right)^{2}=\left({\frac {h}{a}}\right)^{2}+\left({\frac {k}{b}}\right)^{2}+\left({\frac {l}{c}}\right)^{2}\end{alignedat}}$

Since $q=2\pi /d$, and since $a=b=c$, the realspace spacing of planes is:

${\begin{alignedat}{2}d_{hkl}&={\frac {a}{\sqrt {h^{2}+k^{2}+l^{2}}}}\end{alignedat}}$

BCC 110

${\begin{alignedat}{2}d_{110}&={\frac {a}{\sqrt {1^{2}+1^{2}+0^{2}}}}\\&={\frac {a}{\sqrt {2}}}\end{alignedat}}$

Note that for BCC, the particle-particle distance is given by:

$d_{nn}={\sqrt {3}}a/2$

So we expect:

${\begin{alignedat}{2}d_{nn}&={\frac {{\sqrt {3}}a}{2}}\\&={\frac {{\sqrt {3}}d_{110}{\sqrt {2}}}{2}}\\&={\frac {{\sqrt {6}}d_{110}}{2}}\\&={\frac {{\sqrt {6}}(2\pi /q_{110})}{2}}\\&={\frac {\pi {\sqrt {6}}}{q_{110}}}\\\end{alignedat}}$

Of course, we could also have written:

${\begin{alignedat}{2}d_{110}&={\frac {a}{\sqrt {2}}}\\&={\frac {2d_{nn}/{\sqrt {3}}}{\sqrt {2}}}\\&={\frac {2d_{nn}}{\sqrt {6}}}\\\end{alignedat}}$

FCC 111

${\begin{alignedat}{2}d_{111}&={\frac {a}{\sqrt {1^{2}+1^{2}+1^{2}}}}\\&={\frac {a}{\sqrt {3}}}\end{alignedat}}$

And:

$d_{nn}={\sqrt {2}}a/2$

So:

${\begin{alignedat}{2}d_{nn}&={\frac {{\sqrt {2}}a}{2}}\\&={\frac {{\sqrt {2}}d_{111}{\sqrt {3}}}{2}}\\&={\frac {{\sqrt {6}}d_{111}}{2}}\\&={\frac {\pi {\sqrt {6}}}{q_{111}}}\\\end{alignedat}}$

Or: \begin{alignat}{2} d_{110}

 & = \frac{a}{\sqrt{ 2 }} \\
 & = \frac{ 2 d_{nn} / \sqrt{3} }{\sqrt{ 2 }} \\
 & = \frac{ 2 d_{nn} }{\sqrt{ 6 }} \\

\end{alignat} </math>