 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2022/paper-2/section-a/a5/solution

Let

N_{u}, N_{l}

be the upper- and lower-level populations and

g_{u}, g_{l}

\frac{N _{u}}{N _{l}} = \frac{g _{u}}{g _{l}} exp (- \frac{Δ E}{k _{B} T}), \frac{Δ E}{k _{B}} = \frac{27 \times 1 0 ^{- 6} eV}{8.617333262 \times 1 0 ^{- 5} eV K ^{- 1}} = 0.313322 K .

No degeneracy factors are supplied, so the numerical answers use equal degeneracy factors. Stating the ratio as upper population divided by lower population removes an otherwise arbitrary inversion.

(a) At

T = 500 K

\frac{N _{u}}{N _{l}} = e^{- 0.000626644} = 0.999374.

The two populations are nearly equal because the level separation is much smaller than the thermal energy

k_{B} T

(b) At

T = 2 mK = 0.002 K

\frac{N _{u}}{N _{l}} = e^{- 156.661} = 9.18 \times 1 0^{- 69} .

Almost every atom occupies the lower level. If the levels have unequal degeneracy factors, each boxed ratio must be multiplied by

g_{u} / g_{l}

; their individual values cannot be inferred from the paper.

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