Quantum Chemistry/Example 28

Question

For an electron in a 3D cube with a length of 6.00 Å, what is:

(a) the energy of its state?
(b) the degeneracy of this energy state?
(c) the total number of nodal planes for a particle in the $n_{\text{x}}=4$, $n_{\text{y}}=5$, and $n_{\text{z}}=3$?

Solution

(a) The mass of an electron must be known to determine the energy of the particle. To determine the energy state of an electron in a 3D box, the equation that relates the energy levels of a particle in a cube must be used with every variable in its SI units. The mass of an electron is known to be $9.1093837\text{×}10^{-31}\text{kg}$. Planck's constant is known to be $6.62607015\text{×}10^{-34}\text{m²·kg·s⁻¹}$. The length of the box is $6.00\text{x}10^{-10}\text{m}$ since one angstrom is known to be $10^{-10}\text{m}$. The energy of the electron in the cube can be calculated as follows,

$E=\left(\frac{h^2}{8m{L}^2}\right)\left(n_{\text{x}}^2+n_{\text{y}}^2+n_{\text{z}}^2\right)$

$E=\left(\frac{{(6.62607015\text{×}10^{-34}\text{m²·kg·s⁻¹})}^2}{8\text{×}(9.1093837\text{×}10^{-31}\text{kg})\text{×}{(6.00\text{x}10^{-10}\text{m})}^2}\right)\left(4^2+5^2+3^2\right)$

$E=(1.68\text{×}10^{-19}\text{J})\left(50\right)$

$E=8.38\text{×}10^{-18}\text{J}$

(b) Degeneracy is defined as an energy state with an identical energy to that of a particle with a different set of quantum numbers. In this question, the energy of degenerate states must equal the following,

$E=\left(\frac{h^2}{8m{L}^2}\right)\left(4^2+5^2+3^2\right)=\left(\frac{50h^2}{8m{L}^2}\right)$

To determine the degeneracy of a system, the simplest method is to use a matrix to ensure a systematic process is followed. In this case, the sum of the square of all quantum numbers must equal 50. Therefore, the matrix is as follows,

$n_{\text{x}}$	$n_{\text{y}}$	$n_{\text{z}}$	$n_{\text{x}}^2+n_{\text{y}}^2+n_{\text{z}}^2$
4	5	3	4² + 5² + 3² = 50
4	3	5	4² + 3² + 5² = 50
5	3	4	5² + 3² + 4² = 50
5	4	3	5² + 4² + 3² = 50
3	4	5	3² + 4² + 5² = 50
3	5	4	3² + 5² + 4² = 50

This results in the identification of 6 different sets of quantum numbers, and thus, this energy state is said to be 6-fold degenerate.

(c) To determine the number of nodal planes, each axis must be considered individually. Mathematically, to calculate the number of nodal planes associated with the x-axis, the following equation is used,

$N_{\text{nodal planes x-axis}}=n_{\text{x}}-1$

Therefore, in this example, the number of nodal planes in x-axis is,

$N_{\text{nodal planes x-axis}}=4-1=3$

Similarly, for the y-axis and the z-axis, the number of nodal planes is,

$N_{\text{nodal planes y-axis}}=n_{\text{y}}-1=5-1=4$

$N_{\text{nodal planes z-axis}}=n_{\text{z}}-1=3-1=2$

Therefore, the total number of nodal planes in this system is 9. This was determined by summing the number of nodal planes associated with each individual axis.