 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/2017/s16/1/4/solution

ρ / (9 m_{u})

n_{e} ≃ 2.48 \times 1 0^{29}

^{- 3}

. Therefore

ℏ ω_{p} ≃ 18.5

eV. The photon energy loss gives

ℏ ω = 30

k = \frac{E _{i}^{2} + E _{f}^{2} - 2 E _{i} E _{f} c o s ( π /4 )}{ℏ c} ≃ 1.14 \times 1 0^{10} m^{- 1} .

Inverting the stipulated dispersion law gives

k_{B} T = \frac{m _{e} ( ω ^{2} - ω _{p}^{2} )}{3 k ^{2}} ≃ 18.7 eV .

An angular frequency expressed “in eV” means

ℏ ω

, not the frequency itself. The temperature is a formal inference conditional on the dispersion model, whose validity is tested next.

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