Discussion
Minimum energy of a collection of six non-interacting electrons of spin \(\frac{1}{2}\) and mass m, placed in a one-dimensional infinite potential well of width L is :
Answer: B
The minimum energy is \(\frac{14\pi^{2}\hbar^{2}}{mL^{2}}\).
The levels of the well are \(E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\). By the Pauli principle each level holds two electrons of opposite spin, so in the ground state the six electrons fill n = 1, 2 and 3:
\[ \begin{aligned} E &= 2(1^{2} + 2^{2} + 3^{2})\frac{\pi^{2}\hbar^{2}}{2mL^{2}} \\ &= 28\frac{\pi^{2}\hbar^{2}}{2mL^{2}} = \frac{14\pi^{2}\hbar^{2}}{mL^{2}} \end{aligned} \]
Putting all six in the lowest level, as for bosons, gives \(3\pi^{2}\hbar^{2}/mL^{2}\). Placing one electron in each of the six lowest levels, ignoring spin, gives a sum of squares of 91. Seven is half the correct coefficient.
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