# Linear Algebra/Topic: Crystals/Solutions
## Solutions
### Problem 1

How many fundamental regions are there in one face of a speck of salt? (With a ruler, we can estimate that face is a square that is $0.1$ cm on a side.)

### Answer

Each fundamental unit is $3.34\times 10^{-10}$ cm, so there are about $0.1/(3.34\times 10^{-10})$ such units. That gives $2.99\times 10^{8}$, so there are something like $300,000,000$ (three hundred million) units.

### Problem 2

In the graphite picture, imagine that we are interested in a point $5.67$ Ångstroms up and $3.14$ Ångstroms over from the origin.
* Express that point in terms of the basis given for graphite.
* How many hexagonal shapes away is this point from the origin?
* Express that point in terms of a second basis, where the first basis vector is the same, but the second is perpendicular to the first (going up the plane) and of the same length.

### Answer

* We solve to get $c_{2}=\approx 4.42$ and $c_{1}\approx 0.16$.
    $c_{1}{\begin{pmatrix}1.42\\0\end{pmatrix}}+c_{2}{\begin{pmatrix}1.23\\0.71\end{pmatrix}}={\begin{pmatrix}5.67\\3.14\end{pmatrix}}\quad \implies \quad {\begin{array}{*{2}{rc}r}1.42c_{1}&+&1.23c_{2}&=&5.67\\&&0.71c_{2}&=&3.14\end{array}}$
* Here is the point located in the lattice. In the picture on the left, superimposed on the unit cell are the two basis vectors ${\vec {\beta }}_{1}$ and ${\vec {\beta }}_{2}$, and a box showing the offset of $0.16{\vec {\beta }}_{1}+4.42{\vec {\beta }}_{2}$. The picture on the right shows where that appears inside of the crystal lattice, taking as the origin the lower left corner of the hexagon in the lower left.   So this point is in the next column of hexagons over, and either one hexagon up or two hexagons up, depending on how you count them.
* This second basis makes the computation easier (we get $c_{2}\approx 2.21$ and $c_{1}\approx 3.99$), but it doesn't seem to have to do much with the physical structure that we are studying.
    $\langle {\begin{pmatrix}1.42\\0\end{pmatrix}},{\begin{pmatrix}0\\1.42\end{pmatrix}}\rangle$
    $c_{1}{\begin{pmatrix}1.42\\0\end{pmatrix}}+c_{2}{\begin{pmatrix}0\\1.42\end{pmatrix}}={\begin{pmatrix}5.67\\3.14\end{pmatrix}}\quad \implies \quad {\begin{array}{*{2}{rc}r}1.42c_{1}&&&=&5.67\\&&1.42c_{2}&=&3.14\end{array}}$

### Problem 3

Give the locations of the atoms in the diamond cube both in terms of the basis, and in Ångstroms.

### Answer

In terms of the basis the locations of the corner atoms are $(0,0,0)$, $(1,0,0)$, ..., $(1,1,1)$. The locations of the face atoms are $(0.5,0.5,1)$, $(1,0.5,0.5)$, $(0.5,1,0.5)$, $(0,0.5,0.5)$, $(0.5,0,0.5)$, and $(0.5,0.5,0)$. The locations of the atoms a quarter of the way down from the top are $(0.75,0.75,0.75)$ and $(0.25,0.25,0.25)$. The atoms a quarter of the way up from the bottom are at $(0.75,0.25,0.25)$ and $(0.25,0.75,0.25)$. Converting to Ångstroms is easy.

### Problem 4

This illustrates how the dimensions of a unit cell could be computed from the shape in which a substance crystalizes (see Ebbing 1993, p. 462).
* Recall that there are $6.022\times 10^{23}$ atoms in a mole (this is Avagadro's number). From that, and the fact that platinum has a mass of $195.08$ grams per mole, calculate the mass of each atom.
* Platinum crystalizes in a face-centered cubic lattice with atoms at each lattice point, that is, it looks like the middle picture given above for the diamond crystal. Find the number of platinums per unit cell (hint: sum the fractions of platinums that are inside of a single cell).
* From that, find the mass of a unit cell.
* Platinum crystal has a density of $21.45$ grams per cubic centimeter. From this, and the mass of a unit cell, calculate the volume of a unit cell.
* Find the length of each edge.
* Describe a natural three-dimensional basis.

### Answer

* $195.08/6.02\times 10^{23}=3.239\times 10^{-22}$
* $4$
* $4\cdot 3.239\times 10^{-22}=1.296\times 10^{-21}$
* $1.296\times 10^{-21}/21.45=6.042\times 10^{-23}$ cubic centimeters
* $3.924\times 10^{-8}$ centimeters.
* $\langle {\begin{pmatrix}3.924\times 10^{-8}\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\3.924\times 10^{-8}\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\3.924\times 10^{-8}\end{pmatrix}}\rangle$

## References

* Ebbing, Darrell D. (1993), General Chemistry (Fourth ed.), Houghton Mifflin.