# Quantum Chemistry/Example 28

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## 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 ${\displaystyle {n_{\text{x}}}=4}$, ${\displaystyle {n_{\text{y}}}=5}$, and ${\displaystyle {n_{\text{z}}}=3}$?


## 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 ${\displaystyle {9.1093837{\text{×}}10^{-31}{\text{kg}}}}$. Planck's constant is known to be ${\displaystyle {6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}}}$. The length of the box is ${\displaystyle {6.00{\text{x}}10^{-10}{\text{m}}}}$ since one angstrom is known to be ${\displaystyle 10^{-10}{\text{m}}}$. The energy of the electron in the cube can be calculated as follows,

${\displaystyle {E}=\left({\frac {h^{2}}{8mL^{2}}}\right)\left({n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}\right)}$
${\displaystyle {E}=\left({\frac {\left(6.62607015{\text{×}}10^{-34}{\text{m²·kg·s⁻¹}}\right)^{2}}{8{\text{×}}\left(9.1093837{\text{×}}10^{-31}{\text{kg}}\right){\text{×}}\left(6.00{\text{x}}10^{-10}{\text{m}}\right)^{2}}}\right)\left({{4^{2}}+{5^{2}}+{3^{2}}}\right)}$
${\displaystyle {E}=\left(1.68{\text{×}}10^{-19}{\text{J}}\right)\left(50\right)}$
${\displaystyle {E}=8.38{\text{×}}10^{-18}{\text{J}}}$

(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,

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

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,

${\displaystyle {n_{\text{x}}}}$ ${\displaystyle {n_{\text{y}}}}$ ${\displaystyle {n_{\text{z}}}}$ ${\displaystyle {n_{\text{x}}^{2}+n_{\text{y}}^{2}+n_{\text{z}}^{2}}}$
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,

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

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

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

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

${\displaystyle {N_{\text{nodal planes y-axis}}=n_{\text{y}}-1=5-1=4}}$
${\displaystyle {N_{\text{nodal planes z-axis}}=n_{\text{z}}-1=3-1=2}}$

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.