Suppose you put five electrons into an infinite square well of width \(L\). Find an expression for the minimum energy of this system, consistent with the exclusion principle.

Short Answer

Expert verified
The minimum energy of the system, consistent with the Pauli Exclusion Principle, is given by \(E_{min} = \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}} \cdot 55\).

Step by step solution

01

Calculating the energy for the first electron

The first electron will be in the lowest energy state, which corresponds to \(n = 1\). Thus, the energy of the first electron is \(E_1 = \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}}\).
02

Calculating the energy for the next electrons

Following the Pauli Exclusion Principle, the second electron cannot have the same quantum number as the first electron. That means, we'll have to put it to the next energy level, corresponding to \(n = 2\). Hence, the energy of the second electron is \(E_2 = 4 \cdot \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}}\). For the five-electron system, electrons will occupy states \(n = 1, 2, 3, 4, 5\).
03

Summing up the energy

The total energy of the system will be the sum of the energies of the individual electrons. Therefore, the minimal energy of the system \(E_{min}\) will be \(E_{min} = \sum_{n=1}^{5} n^2 \cdot \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}} = \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}} \cdot (1^2 + 2^2 + 3^2 + 4^2 + 5^2) = \frac{{\pi^2 \cdot h^2}}{{2 \cdot m \cdot L^2}} \cdot 55\).

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