Problem 28 - Fundamentals of Physics

Question

A cubical box of widths \(L_{x}=L_{y}=L_{z}=L\) contains an electron. What multiple of \(h^{2} / 8 m L^{2}\), where \(m\) is the electron mass, is (a) the energy of the electron's ground state, (b) the energy of its second excited state, and (c) the difference between the energies of its second and third excited states? How many degenerate states have the energy of (d) the first excited state and (e) the fifth excited state?

Step-by-Step Solution

Answer

(a) 3; (b) 6; (c) 3; (d) 3 states; (e) 6 states.

Step 1: Understanding the Problem

We are calculating the energy levels of an electron in a 3D box where the dimensions are all the same ( L_{x}=L_{y}=L_{z}=L ). We will express energies as multiples of h^{2} / 8 m L^{2} .

Step 2: Calculate Ground State Energy

For a 3D cubical box, the ground state energy corresponds to quantum numbers n_{x}=1, n_{y}=1, ext{ and } n_{z}=1. The energy is given by:\[ E_{111} = \frac{h^{2}}{8mL^{2}} \left(1^{2} + 1^{2} + 1^{2}\right) = \frac{3h^{2}}{8mL^{2}} \] Thus, the multiple of h^{2} / 8 m L^{2}is 3.

Step 3: Find Energy of Second Excited State

The second excited state is the third energy level, which corresponds to combinations of quantum numbers like (2,1,1) or (1,2,1)or (1,1,2).Each configuration results in the energy:\[ E_{211} = E_{121} = E_{112} = \frac{h^{2}}{8mL^{2}} (2^2 + 1^2 + 1^2) = \frac{6h^{2}}{8mL^{2}} \] Thus, the multiple is 6.

Step 4: Determine Energy Difference between Second and Third Excited States

The third excited state can be seen with the next possible set of quantum numbers like (2,1,2)or (1,2,2)or (2,2,1), with energy:\[ E_{212} = E_{122} = E_{221} = \frac{h^{2}}{8mL^{2}} (2^2 + 1^2 + 2^2) = \frac{9h^{2}}{8mL^{2}} \] The difference between the second and third excited states is 9 - 6 = 3.

Step 5: Count Degenerate States for First Excited State

The first excited state corresponds to (2,1,1), (1,2,1), or (1,1,2) which are just the second energy levels seen before. The degeneracy is the number of different permutations of these quantum numbers, which is 3.

Step 6: Count Degenerate States for Fifth Excited State

The fifth excited state involves configurations such as (3,2,1), (3,1,2), (2,3,1), etc., which correspond to the sum 3^2 + 2^2 + 1^2 = 14. The permutations of this combination give a degeneracy of 6, as each of these numbers can be arranged in different positions.

Key Concepts

Particle in a Box

The "particle in a box" model is a fundamental concept in quantum mechanics. It describes a particle that is confined within an impenetrable box with perfectly rigid walls, where the particle is free to move only within the defined boundaries. In the box, the potential energy of the particle is zero, but it becomes infinite at the walls.
The particle shows quantized behavior, which means it can only occupy specific energy levels. The dimensions of the box (indicated by lengths such as \(L_x\), \(L_y\), and \(L_z\)) influence the allowed energy states. In our provided problem, the box is cubical, i.e., \(L_x = L_y = L_z = L\), simplifying calculations due to symmetrical dimensions. This simple model helps visualize more complex scenarios in quantum mechanics.

Quantum Numbers

Quantum numbers are crucial in defining the states of a quantum system. For a particle in a 3D box, we need three quantum numbers: \(n_x\), \(n_y\), and \(n_z\). These numbers correspond to the energy levels along each spatial dimension of the box.

• The quantum numbers must be positive integers (1, 2, 3, ...).
• They express the quantization condition that the particle satisfies inside the box.

Each combination of \(n_x\), \(n_y\), and \(n_z\) determines the particle's energy level. Lower quantum numbers mean lower energy levels, denoting the ground or minimal energy state, whereas higher numbers signify excited states. In the given exercise, the ground state corresponds to (1,1,1), where each quantum number is at its minimum.

Energy Levels

Energy levels in a particle-in-a-box system are determined by the quantum numbers \(n_x\), \(n_y\), and \(n_z\). The energy formula is derived as follows:
\[ E_{n_x n_y n_z} = \frac{h^2}{8mL^2} (n_x^2 + n_y^2 + n_z^2) \] where \(h\) is Planck's constant, \(m\) is the mass of the particle (like an electron in this problem), and \(L\) is the width of the box.
This formula shows that the energy depends on the square of the quantum numbers, meaning that as the quantum numbers increase, the energy increases rapidly.
The numerical coefficients for excited states arise directly from sums of squares of these quantum numbers. For instance, in the second excited state (or third energy level), combinations of (2,1,1) yield an energy of \(\frac{6h^2}{8mL^2}\). This pattern follows for higher states as well.

Degenerate States

Degenerate states are different quantum states that have the same energy level. In the particle-in-a-box model, states are degenerate when different permutations of quantum numbers result in the same energy. This commonality occurs due to the symmetry of the box dimensions.

• For example, in a cubic box, the first excited state has combinations (2,1,1), (1,2,1), and (1,1,2), all leading to the same energy level.
• Such states are considered to have a degeneracy of 3, meaning there are three different ways to arrange the quantum numbers that result in the same energy.

The concept of degenerate states becomes particularly important when analyzing quantum systems as it relates to the potential energy and behavior of particles within fields or under various potential influences. By understanding degeneracy, we can predict how particles might behave in more complex environmental situations.