Energy Levels in a Crystalline Solid — Question 9

Section 41-2 · Physics pages 6828–6829

Question 9 Pauli occupancy changes over an energy range set by thermal energy Use the Fermi function f(E) = 1/[1 + e(E−µ)/(kBT)]. Define E90 and E10 where f equals .90 and .10 respectively. E f Schematic; labels, not drawing lengths, specify the data. (a) Find these energies relative to the chemical potential. (b) Find their separation at 300 K using kB = 8.617 × 10−5 eV/K.
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Question 9 - Solution Pauli occupancy changes over an energy range set by thermal energy E f (a) Method and calculation f = .90 ⇒e(E90−µ)/(kBT) = 1/9 ⇒E90 = µ −kBT ln 9 . f = .10 ⇒E10 = µ + kBT ln 9 . (b) Interpretation and further calculation ∆E = E10 −E90 = 2kBT ln 9. ∆E = 2(8.617 × 10−5)(300) ln 9 = .11360 eV . Physical check The probability is one-half at the chemical potential. These are occupation probabilities per available spin state, not the density of states.

Question 9

Pauli occupancy changes over an energy range set by thermal energy

Use the Fermi function f(E)=1/[1+e(E−μ)/(kBT)]f(E)=1/[1+e^{(E-\mu)/(k_BT)}]. Define E90E_{90} and E10E_{10} where f equals .90 and .10 respectively.

Diagram shown in the original SVG worksheet.

  1. Find these energies relative to the chemical potential.

  2. Find their separation at 300 K using kB=8.617×10−5eV/Kk_B=8.617\times10^{-5}\,\mathrm{eV/K}.

Question 9 - Solution

Pauli occupancy changes over an energy range set by thermal energy

Diagram shown in the original SVG worksheet.

(a) Method and calculation

f=.90⇒e(E90−μ)/(kBT)=1/9⇒E90=μ−kBTln⁡9.f=.90\Rightarrow e^{(E_{90}-\mu)/(k_BT)}=1/9\Rightarrow\boxed{E_{90}=\mu-k_BT\ln9}.f=.10⇒E10=μ+kBTln⁡9.f=.10\Rightarrow\boxed{E_{10}=\mu+k_BT\ln9}.

(b) Interpretation and further calculation

ΔE=E10−E90=2kBTln⁡9.\Delta E=E_{10}-E_{90}=2k_BT\ln9.ΔE=2(8.617×10−5)(300)ln⁡9=.11360eV.\boxed{\Delta E=2(8.617\times10^{-5})(300)\ln9=.11360\,\mathrm{eV}}.

Physical check

The probability is one-half at the chemical potential. These are occupation probabilities per available spin state, not the density of states.