Question

A cubical box of widths ${\mathbf{L}}_{\mathbf{x}}\mathbf{=}{\mathbf{L}}_{\mathbf{y}}\mathbf{=}{\mathbf{L}}_{\mathbf{z}}\mathbf{=}\mathbf{L}$contains eight electrons. What multiple of ${\mathbf{h}}^{\mathbf{2}}\mathbf{/}\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}$ gives the energy of the ground state of this system? Assume that the electrons do not interact with one another, and do not neglect spin.

Short answer

The multiple value of ${\mathrm{h}}^2/8{\mathrm{mL}}^2$that gives the energy of the ground state of the system is 42.

Step-by-step solution

The given data:

A cubical box of widths ${\mathrm{L}}_x={\mathrm{L}}_y={\mathrm{L}}_2=\mathrm{L}$ contains eight electrons.

Understanding the concept of ground state configuration of the system:

The low state of the quantum-mechanical system is its fixed state of extremely low power; the power of the lower state is known as the zero power of the system.

Pauli's exclusive principle states that no two electrons in the same atom can have the same values in all four of their quantum numbers.

Using the concept of ground-state energy of the system of a cube, define the individual energies of the particle in the lowest single-particle energy level of the cubical system that contributes to the ground state. This state occupies two electrons according to Pauli's exclusion principle. Now, for the next 6 electrons to be in the next lowest energy level, they accommodate a three-fold degenerate state. Thus, the total energy contributes to the ground state energy of the 8-particle system.

Formula:

The single particle energy levels of a cubical system in terms of $n_x,n_y,n_z$,

${\mathrm{E}}_{\mathrm{x},\mathrm{y},\mathrm{z}}=\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)\left({\mathrm{n}}_{\mathrm{x}}^2+{\mathrm{n}}_{\mathrm{y}}^2+{\mathrm{n}}_{\mathrm{z}}^2\right)$ ….. (1)

Calculation of the multiple value of h2/8mL2 :

According to the concept, with eight electrons, the ground-state energy of the system is the sum of the energies of the individual electrons in the system’s ground state configuration.

Now, from equation (1), the lowest single-particle energy level is possible for the state is,

${\mathrm{n}}_{\mathrm{x}}=1,{\mathrm{n}}_{\mathrm{y}}=1,{\mathrm{n}}_{\mathrm{z}}=1$

This energy is with two electrons at the lowest level and is given by,

${\mathrm{E}}_{1,1,1}=\left(\frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)$

The remaining 6 electrons occupy the next lowest energy state, and the energy of this three-fold degenerate state is given using equation (1) as follows:

$\begin{array}{l}{\mathrm{E}}_{1,1,1}={\mathrm{E}}_{1,2,1}\\ ={\mathrm{E}}_{1,2,1}\\ =\left(\frac{6{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)\end{array}$

Thus, the total energy contributed to the ground state by the eight-electron system is given by:

$$\begin{gathered} {\mathrm{E}}_{\mathrm{ground}}=2\left(\frac{3{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right)+6\left(\frac{6{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \\ =42\left(\frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\right) \end{gathered}$$

Hence, the value of the multiple is 42 .