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,


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


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


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


 * $${E}=8.38\text{×}10^{-18} \text{J} $$

(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,


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

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,


 * {| class="wikitable" style="text-align: center; width: 300px; height: 150px;"

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