 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s16/2/3/solution

For a nonrelativistic, three-dimensional ideal electron gas with spin degeneracy two,

E_{F} = \frac{ℏ ^{2}}{2 m _{e}} (3 π^{2} n_{e})^{2/3}, ω_{pe} = \frac{n _{e} e ^{2}}{m _{e} ϵ _{0}} .

Dividing gives

\frac{ℏ ω _{pe}}{E _{F}} = \frac{4 e ^{2} m _{e}}{ℏ ^{2} ϵ _{0}} (3 π^{2})^{- 2/3} n_{e}^{- 1/6} .

Write the density-independent coefficient as

K

. From

ℏ ω_{pe} / E_{F} = 1.384

n_{e} = (K /1.384)^{6}

n_{e} ≃ 1.59 \times 1 0^{29} m^{- 3}, E_{F} ≃ 10.69 eV, ℏ ω_{pe} ≃ 14.79 eV .

The result uses the stipulated energy ratio, rather than substituting a separately tabulated aluminium density or plasmon energy.

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