# Quantum Chemistry/Example 28

## Question[edit | edit source]

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 , , and ?

## Solution[edit | edit source]

(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 . Planck's constant is known to be . The length of the box is since one angstrom is known to be . The energy of the electron in the cube can be calculated as follows,

(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,

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,

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,

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

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

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.