Multiple Electrons in Rectangular Traps — Question 4

Section 40-7 · Physics pages 6698–6699

Question 4 A filled square-box shell includes degenerate spatial states In a two-dimensional square of side L, Enxny = ϵ(n2 x + n2 y) with positive integer indices and two spins per spatial orbital. Six independent electrons occupy their zero-temperature ground state. L × L E Schematic; labels, not drawing lengths, specify the data. (a) Identify occupied spatial orbitals and calculate the total energy. (b) Find the lowest energy to promote one electron into an unoccupied spatial orbital, ignoring transition selection rules.
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Question 4 - Solution A filled square-box shell includes degenerate spatial states L × L E (a) Method and calculation (1, 1) : 2ϵ; (1, 2), (2, 1) : 5ϵ each. Eg = 2(2ϵ) + 4(5ϵ) = 24ϵ . (b) Interpretation and further calculation Lowest empty orbital (2, 2) : E = 8ϵ. ∆Emin = 8ϵ −5ϵ = 3ϵ . Physical check Degenerate orbitals remain distinct states and each has its own two spin slots. The excitation cost is an energy difference, not the absolute energy of the empty orbital.

Question 4

A filled square-box shell includes degenerate spatial states

In a two-dimensional square of side L, Enxny=ϵ(nx2+ny2)E_{n_xn_y}=\epsilon(n_x^2+n_y^2) with positive integer indices and two spins per spatial orbital. Six independent electrons occupy their zero-temperature ground state.

Diagram shown in the original SVG worksheet.

  1. Identify occupied spatial orbitals and calculate the total energy.

  2. Find the lowest energy to promote one electron into an unoccupied spatial orbital, ignoring transition selection rules.

Question 4 - Solution

A filled square-box shell includes degenerate spatial states

Diagram shown in the original SVG worksheet.

(a) Method and calculation

(1,1):2ϵ;(1,2),(2,1):5ϵ each.(1,1):2\epsilon;\quad(1,2),(2,1):5\epsilon\text{ each}.Eg=2(2ϵ)+4(5ϵ)=24ϵ.\boxed{E_g=2(2\epsilon)+4(5\epsilon)=24\epsilon}.

(b) Interpretation and further calculation

Lowest empty orbital (2,2):E=8ϵ.\text{Lowest empty orbital }(2,2):E=8\epsilon.ΔEmin=8ϵ−5ϵ=3ϵ.\boxed{\Delta E_{\min}=8\epsilon-5\epsilon=3\epsilon}.

Physical check

Degenerate orbitals remain distinct states and each has its own two spin slots. The excitation cost is an energy difference, not the absolute energy of the empty orbital.