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}

. In this free-electron model,

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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