Multiple Electrons in Rectangular Traps — Question 2

Section 40-7 · Physics pages 6694–6695

Question 2 Adding one electron creates an even-odd pattern in the addition cost Use noninteracting electrons with En = ϵn2 and two spin states per spatial level. Define AN = Eg(N + 1) −Eg(N) as the energy cost to add one electron into the same fixed infinite well. 1 2 3 4 Schematic; labels, not drawing lengths, specify the data. (a) Derive A2K and A2K+1. (b) Find A5, A6, A7 and distinguish this cost from ionization to free space.
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Question 2 - Solution Adding one electron creates an even-odd pattern in the addition cost 1 2 3 4 (a) Method and calculation N = 2K : 1, . . . , K filled ⇒A2K = ϵ(K + 1)2 . N = 2K + 1 : one vacancy in level K + 1 ⇒A2K+1 = ϵ(K + 1)2 . (b) Interpretation and further calculation A5 = 9ϵ, A6 = 16ϵ, A7 = 16ϵ . Voutside = ∞⇒no finite escape threshold in this model . Physical check The addition cost compares different particle-number ground states in the same confinement. A physical ionization energy requires a finite exterior potential and an energy reference.

Question 2

Adding one electron creates an even-odd pattern in the addition cost

Use noninteracting electrons with En=ϵn2E_n=\epsilon n^2 and two spin states per spatial level. Define AN=Eg(N+1)−Eg(N)A_N=E_g(N+1)-E_g(N) as the energy cost to add one electron into the same fixed infinite well.

Diagram shown in the original SVG worksheet.

  1. Derive A2KA_{2K} and A2K+1A_{2K+1}.

  2. Find A5,A6,A7A_5,A_6,A_7 and distinguish this cost from ionization to free space.

Question 2 - Solution

Adding one electron creates an even-odd pattern in the addition cost

Diagram shown in the original SVG worksheet.

(a) Method and calculation

N=2K:1,…,K filled⇒A2K=ϵ(K+1)2.N=2K:\quad 1,\ldots,K\text{ filled}\Rightarrow\boxed{A_{2K}=\epsilon(K+1)^2}.N=2K+1:one vacancy in level K+1⇒A2K+1=ϵ(K+1)2.N=2K+1:\quad\text{one vacancy in level }K+1\Rightarrow\boxed{A_{2K+1}=\epsilon(K+1)^2}.

(b) Interpretation and further calculation

A5=9ϵ,A6=16ϵ,A7=16ϵ.\boxed{A_5=9\epsilon,\quad A_6=16\epsilon,\quad A_7=16\epsilon}.Voutside=∞⇒no finite escape threshold in this model.V_{\mathrm{outside}}=\infty\Rightarrow\boxed{\text{no finite escape threshold in this model}}.

Physical check

The addition cost compares different particle-number ground states in the same confinement. A physical ionization energy requires a finite exterior potential and an energy reference.