Final answer:
The student's physics question concerns finding the lowest degenerate energy levels for a particle in a 2D box. The correct answer is that both the ground state and the first excited state have the same energy value, resulting in E1 and E2 both being -ħ²/8mL², making option (b) the correct choice.
Step-by-step explanation:
The student is asking about the quantum particle in a two-dimensional box and the two lowest degenerate energy levels. Given the energy values equation E = (-ħ²/2mL²)(n²x n²y/4), we must identify the quantum numbers that correspond to the lowest energy states and check if these states have the same energy, thus being degenerate.
Energy levels are found by inserting different quantum numbers (nx and ny) into the equation. The quantum numbers must be positive integers starting from 1 (for the lowest energy state, also known as the ground state). Degeneracy occurs when different combinations of quantum numbers result in the same energy level.
For the ground state (n1 = 1, n2 = 1), the energy is E1 = (-ħ²/2mL²)(1² + 1²/4) = -ħ²/8mL². The first excited state can have quantum numbers (n1 = 2, n2 = 1) or (n1 = 1, n2 = 2), giving us E2 = (-ħ²/2mL²)(2² + 1²/4) = -ħ²/8mL². Therefore, E1 is degenerate with E2. So the correct answer demonstrating the two lowest degenerate levels, based on the options provided, is option (b): E1 = -ħ²/8mL², E2 = -ħ²/8mL².