Five electrons are in a two-dimensional square potential energy well with sides of length L. The potential energy is infinite at the sides and zero inside. The single-particle energies are given by where n_x and n_y are integers. In units of (h²/8mL²) the energy of the ground state of the system is A) 0 B) 10 C) 19 D) 24 E) 48

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The energy of the ground state can be calculated by filling the lowest energy levels with electrons until all electrons have been accounted for. The lowest energy level for each direction has n=1, giving an energy of: E = (1^2 + 1^2)h^2/(8mL^2) = (h^2/(8mL^2))(2). There are two possible states for each of these energy levels, corresponding to the two different possible spin directions of the electrons. Therefore, there are a total of four electrons that can occupy the lowest energy level. The next energy level for each direction has n=2, giving an energy of: E = (2^2 + 2^2)h^2/(8mL^2) = (h^2/(8mL^2))(8). There are four possible states for each of these energy levels (two spin directions for each value of n), but only one of these states is available since we have already filled the four lowest energy states with electrons. Therefore, there are a total of four electrons that can occupy the second energy level. The total energy of the ground state is thus: 4(2h^2/(8mL^2)) + 4(8h^2/(8mL^2)) = 2h^2/(mL^2). Dividing by (h²/8mL²) gives 19 as the answer.

Five Electrons Are in a Two-Dimensional Square Potential Energy Well