Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 35A / i

Solution

V

d N = 2 \frac{V}{( 2 π ) ^{3}} d^{3} k = \frac{V}{π ^{2}} k^{2} d k .

ℏ ω = ℏ c k

(e^{ℏ c k / T} - 1)^{- 1}

. Omitting the temperature-independent zero-point energy, the total energy is therefore

E = \frac{V ℏ c}{π ^{2}} \int_{0}^{\infty} \frac{k ^{3}}{e ^{ℏ c k / T} - 1} d k .

x = ℏ c k / T

E = \frac{V T ^{4}}{π ^{2} ℏ ^{3} c ^{3}} \int_{0}^{\infty} \frac{x ^{3}}{e ^{x} - 1} d x .

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