Question

An electron is trapped in an octahedral hole in a closest packed array of aluminum atoms (assume they behave as uniform hard spheres). In this situation, the energy of the electron is quantized and the lowest-energy transition corresponds to a wavelength of 9.50nm . Assuming that the hole can be approximated as a cube, what is the radius of a sphere that will just fit in the octahedral hole? Reference Exercise 164 in Chapter 12 for the energy equation for a particle in a cube.

Short answer

The radius of the sphere that fits in the octahedral hole is 46.5 pm.

Step-by-step solution

Determine the energy of the two states

$E_{xyz}=\frac{h^2\left(n_x^2+n_y^2+n_z^2\right)}{8m{L}^2}$

First state energy:

$$\begin{gathered} E_{111}=\frac{h^2\left(1^2+1^2+1^2\right)}{8m{L}^2} \\ =\frac{3h^2}{8m{L}^2} \end{gathered}$$

Second state energy:

$$\begin{gathered} E_{112}=\frac{h^2\left(1^2+1^2+2^2\right)}{8m{L}^2} \\ =\frac{6h^2}{8m{L}^2} \end{gathered}$$

Determine the value of L

The value of L is determined by equating first state energy and second state energy.

$$\begin{gathered} \frac{3h^2}{8m{L}^2}=2.09\times {10}^{-17} \\ L=\left(\frac{3{\left(6.626\times {10}^{-34}Js\right)}^2}{2.09\times {10}^{-17}\times 8\times 9.109\times {10}^{-31}kg}\right)\times \frac{{10}^{12}pm}{1m} \end{gathered}$$

Determine the radius of the sphere

$$\begin{gathered} r=\frac{L}{2} \\ r=\frac{93.0pm}{2} \\ r=46.5pm \end{gathered}$$