(a) Find the angle $\mathit{\theta }$between adjacent nearest-neighbor bonds in the silicon lattice. Recall that each silicon atom is bonded to four of its nearest neighbors. The four neighbors form a regular tetrahedron—a pyramid whose sides and base are equilateral triangles. (b) Find the bond length, given that the atoms at the corners of the tetrahedron are 388pm apart.

• The silicon atom is bonded to four of its nearest neighbors.
• The atoms at the corners of the tetrahedron are apart by distance, a = 388 pm

Silicon has a diamond cubic crystal structure that is an FCC lattice. Using the concept of equilateral triangle equations and the tetrahedron formation in the silicon lattice, we can get the desired angle between the adjacent-neighbor atoms of silicon. Now using the resultant value of the vector starting from b to c, we can get the bond length which has an equal value to that of the vector.

Formulae:

The altitude value of an equilateral triangle with side a, $h=a\sqrt{3}/2$ (i)

The equation of Pythagoras theorem, $x^2+y^2=h^2$ (ii)

The representation of the vector form, $\overrightarrow{r}=x\hat{i}+y\hat{j}$ (iii)

The description in the problem statement implies that an atom is at the center point C of the regular tetrahedron, since its four neighbors are at the four vertices. The side length for the tetrahedron is given as: a = 388pm

Since each face is an equilateral triangle, the“altitude” or the slant height of each of those triangles (which not the altitude of thetetrahedron itself) is given using equation (i) as:$h\text{'}=a\sqrt{3}/2$

At a certain location along the line segment representing the “slant height” of each face is the centerC'of the face.

Now, for the line segment starting at atomAand ending at the midpoint of one of the sides, we know that it bisects the angle of the equilateral face, it is easy to see thatC'is a distance: $AC\text{'}=a/\sqrt{3}$.

Now, if we draw a line from C' all the way to the farthest point on the tetrahedron, then this new line is the altitude of the tetrahedron. Now, using equation (ii) and the value of altitude, we can get the value of the altitude as follows:

$$\begin{gathered} h=\sqrt{a^2-{\left(AC\right)}^2} \\ =\sqrt{a^2-{\left(\frac{a}{\sqrt{3}}\right)}^2} \\ =a\sqrt{\frac{2}{3}} \end{gathered}$$

Now after including the coordinates: the position of atomBon the+y-axis is at

$$\begin{gathered} y_b=h \\ =a\sqrt{2/3} \end{gathered}$$,

And atomAon the+x-axis is at

$$\begin{gathered} x_a=AC\text{'} \\ =a/\sqrt{3} \end{gathered}$$

Then pointC'is the origin.

The tetrahedron center pointCis on theyaxis at some value, that is equidistant fromAandB, so it is given as:

$$\begin{gathered} y_b-y_c=\sqrt{x_a^2+y_c^2} \\ a\sqrt{\frac{2}{3}}-y_c=\sqrt{{\left(\frac{a}{\sqrt{3}}\right)}^2+y_c^2} \\ {y}_c=a/2\sqrt{6} \end{gathered}$$

In unit vector notation, using the above values, we can get the vector starting at C’ and going to A using equation (iii) as follows:

$$\begin{gathered} {\overrightarrow{r}}_{ac}=x_a\hat{i}+\left(-y_c\right)\hat{j} \\ =\frac{a}{\sqrt{3}}\hat{i}-\frac{a}{2\sqrt{6}}\hat{j} \end{gathered}$$

Similarly, the vector starting at C and going to B is given using equation (iii) as follows:

$$\begin{gathered} {\overrightarrow{r}}_{bc}=\left(y_b-y_c\right)\hat{j} \\ =\frac{a\sqrt{3}}{2\sqrt{2}}\hat{j}...........................................\left(a\right) \end{gathered}$$

Thus, the angle between adjacent nearest-neighbor bonds in the silicon lattice is given using the above values of the vector as follows:

$$\begin{gathered} \theta ={\cos }^{-1}\left[\frac{\overrightarrow{r_{ac}}.\overrightarrow{r_{bc}}}{\left|\overrightarrow{r_{ac}}\right|\left|\overrightarrow{r_{bc}}\right|}\right] \\ ={\cos }^{-1}\left(\frac{-1}{3}\right) \\ =109.5° \end{gathered}$$

Hence, the value of the angle is $109.5°$.

The bond length is given by the length value of the vector $\overrightarrow{r_{bc}}$. Thus using the given data in equation (i), the vector value is given as follows:

$$\begin{gathered} \left|\overrightarrow{{\mathrm{r}}_{\mathrm{bc}}}\right|=\frac{\left(388\mathrm{pm}\right)\sqrt{3}}{2\sqrt{2}} \\ =237.6\mathrm{pm} \\ =238\mathrm{pm} \end{gathered}$$

Hence, the value of the bond length is 238 pm.