Example:Particle spacing from peak position

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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:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q=2 \pi / d} , and since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=b=c} , the realspace spacing of planes is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} d_{hkl} & = \frac{a}{\sqrt{ h^2 + k^2 + l^2 }} \end{alignat} }

BCC 110

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} d_{110} & = \frac{a}{\sqrt{ 1^2 + 1^2 + 0^2 }} \\ & = \frac{a}{\sqrt{ 2 }} \end{alignat} }

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{nn} = \sqrt{3}a /2}

So we expect:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

Of course, we could also have written:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} }

FCC 111

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} d_{111} & = \frac{a}{\sqrt{ 1^2 + 1^2 + 1^2 }} \\ & = \frac{a}{\sqrt{ 3 }} \end{alignat} }

And:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d_{nn}=\sqrt{2}a/2}

So:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{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{alignat} }

Or:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{2} d_{111} & = \frac{a}{\sqrt{ 3 }} \\ & = \frac{ 2 d_{nn} / \sqrt{2} }{\sqrt{ 3 }} \\ & = \frac{ 2 d_{nn} }{\sqrt{ 6 }} \\ \end{alignat} }

Comparison

Notice that the BCC 110 and FCC 111 have the same 'conversion factor' from q-position into particle-particle distance. Thus, even without knowing which unit cell the particles are packing into, one can perform the conversion, provided one is confident about the peak being either the BCC 110 or the FCC 111.

See Also

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