Multiple Electrons in Rectangular Traps — Question 10

Section 40-7 · Physics pages 6710–6711

Question 10 Finite-temperature occupations smooth the highest filled shell A supplied independent-electron reservoir model gives mean occupation per spin orbital f(E) = 1/[e(E−µ)/(kBT) + 1]. Two spatial orbitals have energies µ −∆and µ + ∆, each with two spins. Let ∆/(kBT) = ln 9. µ −∆ µ + ∆ ∆E Schematic; labels, not drawing lengths, specify the data. (a) Find both occupations and the total mean electron number in these four spin orbitals. (b) Find the mean excitation energy relative to two electrons occupying only the lower spatial orbital.
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Question 10 - Solution Finite-temperature occupations smooth the highest filled shell µ −∆ µ + ∆ ∆E (a) Method and calculation fl = 1/(1 + 1/9) = .9, fu = 1/(1 + 9) = .1 . ⟨N⟩= 2fl + 2fu = 2 . (b) Interpretation and further calculation ⟨E⟩= 2(.9)(µ −∆) + 2(.1)(µ + ∆) = 2µ −1.6∆. Eref = 2µ −2∆⇒⟨E⟩−Eref = .4∆. Physical check This model permits number fluctuations while fixing the mean through the chemical potential. It is not the exact canonical distribution of an isolated two-electron system.

Question 10

Finite-temperature occupations smooth the highest filled shell

A supplied independent-electron reservoir model gives mean occupation per spin orbital f(E)=1/[e(E−μ)/(kBT)+1]f(E)=1/[e^{(E-\mu)/(k_BT)}+1]. Two spatial orbitals have energies μ−Δ\mu-\Delta and μ+Δ\mu+\Delta, each with two spins. Let Δ/(kBT)=ln⁡9\Delta/(k_BT)=\ln9.

Diagram shown in the original SVG worksheet.

  1. Find both occupations and the total mean electron number in these four spin orbitals.

  2. Find the mean excitation energy relative to two electrons occupying only the lower spatial orbital.

Question 10 - Solution

Finite-temperature occupations smooth the highest filled shell

Diagram shown in the original SVG worksheet.

(a) Method and calculation

fl=1/(1+1/9)=.9,fu=1/(1+9)=.1.\boxed{f_l=1/(1+1/9)=.9,\quad f_u=1/(1+9)=.1}.⟨N⟩=2fl+2fu=2.\boxed{\langle N\rangle=2f_l+2f_u=2}.

(b) Interpretation and further calculation

⟨E⟩=2(.9)(μ−Δ)+2(.1)(μ+Δ)=2μ−1.6Δ.\langle E\rangle=2(.9)(\mu-\Delta)+2(.1)(\mu+\Delta)=2\mu-1.6\Delta.Eref=2μ−2Δ⇒⟨E⟩−Eref=.4Δ.E_{\mathrm{ref}}=2\mu-2\Delta\Rightarrow\boxed{\langle E\rangle-E_{\mathrm{ref}}=.4\Delta}.

Physical check

This model permits number fluctuations while fixing the mean through the chemical potential. It is not the exact canonical distribution of an isolated two-electron system.