 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2016/paper-2/section-c/c12/c/solution

The energy average must be taken over the atoms that actually arrive, not over the bulk population. Weighting the collision flux by

m v^{2} /2

gives

⟨ K ⟩_{hit} = \frac{( m /2 ) n A π I _{5}}{n A π I _{3}} = \frac{m}{2} \frac{I _{5}}{I _{3}}, ⟨ K ⟩_{hit} = 2 k_{B} T .

v^{3} e^{- α v^{2}}

rather than the bulk

v^{2} e^{- α v^{2}}

. Faster atoms are overrepresented among arrivals. Equivalently, the mean normal kinetic energy of the arriving flux is

k_{B} T

, while each tangential component still contributes

k_{B} T /2

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