Question

Exercises 37 and 38 concern the crystal lattice for titanium, which has the hexagonal structure shown on the left in the accompanying

figure. The vectors\(\left( {\begin{array}{*{20}{c}}{2.6}\\{ - 1.5}\\0\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}0\\3\\0\end{array}} \right)\),\(\left( {\begin{array}{*{20}{c}}0\\0\\{4.8}\end{array}} \right)\)in\({\mathbb{R}^{\bf{3}}}\)form a basis for the unit cell shown on the right. The numbers here are Angstrom units\(\left( {1\mathop { A}\limits^{{\rm{ o}}} = 1{0^{ - 8}}cm} \right)\). In alloys of titanium, some additional atoms may be in the unit cell at the octahedral and tetrahedralsites (so named because of the geometric objects

formed by atoms at these locations).

The hexagonal close-packed lattice and its unit cell.

37. One of the octahedral sites is\(\left( {\begin{array}{*{20}{c}}{1/2}\\{1/4}\\{1/6}\end{array}} \right)\), relative to the lattice basis. Determine the coordinates of this site relative to the standard basis of\({\mathbb{R}^{\bf{3}}}\).

Short answer

The coordinates of this site relative to the standard basis of \({\mathbb{R}^{\bf{3}}}\) is \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{1.3}\\0\\{0.8}\end{array}} \right)\).

Step-by-step solution

State the coordinates of x relative to the standard basis

It is given that for the unit cell, thebasis is\(B = \left\{ {\left( {\begin{array}{*{20}{c}}{2.6}\\{ - 1.5}\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}0\\3\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}0\\0\\{4.8}\end{array}} \right)} \right\}\).

Obtain thecoordinatesof the octahedral site \({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}{1/2}\\{1/4}\\{1/6}\end{array}} \right)\) relative to the standard basis of\({\mathbb{R}^{\bf{3}}}\), as shown below:

Write the basis B in the matrix form as shown below:

\({P_B} = \left( {\begin{array}{*{20}{c}}{2.6}&0&0\\{ - 1.5}&3&0\\0&0&{4.8}\end{array}} \right)\)

State the coordinates of x relative to the standard basis

Compute the product of matrices\({\left( {\bf{x}} \right)_B} = \left( {\begin{array}{*{20}{c}}{1/2}\\{1/4}\\{1/6}\end{array}} \right)\)and\({P_B} = \left( {\begin{array}{*{20}{c}}{2.6}&0&0\\{ - 1.5}&3&0\\0&0&{4.8}\end{array}} \right)\)using the MATLAB command shown below:

\(\begin{array}{l} > > {\rm{A}} = \left( {{\rm{2}}{\rm{.6 0 0; }} - {\rm{1}}{\rm{.5 3 0; 0 0 4}}{\rm{.8}}} \right);\\ > > {\rm{B}} = \left( {{\rm{1/2 1/4 1/6}}} \right);\\ > > {\rm{M}} = {\rm{A}}*{\rm{B}}\end{array}\)

So, the output is

\(\begin{array}{c}{\bf{x}} = {P_B}{\left( {\bf{x}} \right)_B}\\ = \left( {\begin{array}{*{20}{c}}{2.6}&0&0\\{ - 1.5}&3&0\\0&0&{4.8}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{1/2}\\{1/4}\\{1/6}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{1.3}\\0\\{0.8}\end{array}} \right).\end{array}\)

Thus, the coordinates of this site relative to the standard basis of \({\mathbb{R}^{\bf{3}}}\) is \({\bf{x}} = \left( {\begin{array}{*{20}{c}}{1.3}\\0\\{0.8}\end{array}} \right)\).