 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2/36a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

E (v) = m ∣ v ∣^{2} /2

and hence probability density proportional to

e^{- βE}

, where

β = (k_{B} T)^{- 1}

f (v) = (\frac{m}{2 π k _{B} T})^{3/2} exp (- \frac{m ∣ v ∣ ^{2}}{2 k _{B} T}) .

v

4 π v^{2}

g (v) = 4 π v^{2} (\frac{m}{2 π k _{B} T})^{3/2} e^{- m v^{2} / (2 k_{B} T)}, v \geq 0.

Since

\frac{d}{d v} lo g g (v) = \frac{2}{v} - \frac{m v}{k _{B} T},

v_{mp} = \frac{2 k _{B} T}{m} .

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