Question

A system of five identical, non-interacting particles with mass \( m \) and spin \( \frac{3}{2} \) is confined to a one-dimensional potential well of length \( L \). If the lowest energy of the system is \[ E_{{min}} = \frac{N \pi^2 \hbar^2}{2m L^2}, \] the value of \( N \) (in integer) is:

Hint:

In systems of fermions in potential wells, the Pauli exclusion principle and spin configuration must be taken into account when determining the energy states and quantum numbers.

Answer 1

1. System Description:
The system consists of five identical, non-interacting particles with mass \( m \) and spin \( \frac{3}{2} \), confined to a one-dimensional potential well of length \( L \). The energy levels of the system depend on the quantum numbers and the spin configuration.

2. Energy Levels of Non-Interacting Particles in a Potential Well:
For a particle in a one-dimensional infinite potential well, the energy levels are quantized and given by:

\[ E_n = \frac{n^2 \pi^2 \hbar^2}{2m L^2}, \]

where \( n \in \mathbb{Z}^+ \) is the quantum number. Since the particles are non-interacting, the total ground state energy is obtained by occupying the lowest available states while obeying the Pauli exclusion principle.

3. Spin and Energy Configuration:
Each particle has spin \( \frac{3}{2} \), giving rise to four distinct spin projections: \( m_s = \frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2} \). As these are fermions, the Pauli exclusion principle forbids any two particles from sharing the same quantum state (i.e., same \( n \) and spin projection).

4. Finding the Lowest Energy State:
Each spatial level \( n \) can accommodate 4 fermions (due to the 4 spin states). So:

• Level \( n = 1 \): 4 particles
• Level \( n = 2 \): 1 particle

Total energy: \[ E_{\text{min}} = 4E_1 + 1E_2 = 4\left( \frac{\pi^2 \hbar^2}{2m L^2} \right) + 1\left( \frac{4\pi^2 \hbar^2}{2m L^2} \right) = \frac{(4 + 4) \pi^2 \hbar^2}{2m L^2} \] \[ \Rightarrow N = \boxed{8} \]