Multiple Electrons in Rectangular Traps — Question 8

Section 40-7 · Physics pages 6706–6707

Question 8 An elongated rectangle changes the order in which orbitals fill A rectangular well has Lx = 2L, Ly = L and energies E = ϵ(n2 x/4 + n2 y), with ϵ = π2ℏ2/(2mL2). Six noninteracting electrons occupy the ground state. 2L L Schematic; labels, not drawing lengths, specify the data. (a) Find the three lowest spatial orbitals and their energies. (b) Calculate the total ground energy and the first excitation gap.
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Question 8 - Solution An elongated rectangle changes the order in which orbitals fill 2L L (a) Method and calculation E11 = 5ϵ/4, E21 = 2ϵ, E31 = 13ϵ/4 . E12 = 17ϵ/4 < E41 = 5ϵ ⇒next empty orbital (1, 2). (b) Interpretation and further calculation Eg = 2ϵ(5/4 + 2 + 13/4) = 13ϵ . ∆Emin = E12 −E31 = ϵ . Physical check Filling requires comparing all candidate directions. Increasing the long-axis quantum number is initially cheaper, but eventually exciting the short direction becomes favorable.

Question 8

An elongated rectangle changes the order in which orbitals fill

A rectangular well has Lx=2LL_x=2L, Ly=LL_y=L and energies E=ϵ(nx2/4+ny2)E=\epsilon(n_x^2/4+n_y^2), with ϵ=π2ℏ2/(2mL2)\epsilon=\pi^2\hbar^2/(2mL^2). Six noninteracting electrons occupy the ground state.

Diagram shown in the original SVG worksheet.

  1. Find the three lowest spatial orbitals and their energies.

  2. Calculate the total ground energy and the first excitation gap.

Question 8 - Solution

An elongated rectangle changes the order in which orbitals fill

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E11=5ϵ/4,E21=2ϵ,E31=13ϵ/4.\boxed{E_{11}=5\epsilon/4,\quad E_{21}=2\epsilon,\quad E_{31}=13\epsilon/4}.E12=17ϵ/4<E41=5ϵ⇒next empty orbital (1,2).E_{12}=17\epsilon/4<E_{41}=5\epsilon\Rightarrow\text{next empty orbital }(1,2).

(b) Interpretation and further calculation

Eg=2ϵ(5/4+2+13/4)=13ϵ.\boxed{E_g=2\epsilon(5/4+2+13/4)=13\epsilon}.ΔEmin=E12−E31=ϵ.\boxed{\Delta E_{\min}=E_{12}-E_{31}=\epsilon}.

Physical check

Filling requires comparing all candidate directions. Increasing the long-axis quantum number is initially cheaper, but eventually exciting the short direction becomes favorable.