 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-2/section-c/c12/energy-of-incident-particles/solution

Wall arrivals are not an unbiased sample of the gas: faster atoms arrive more frequently. The normalized speed distribution among incident atoms is

f_{hit} (v) = \frac{v f ( v )}{v} .

The angular factor in the arrival rate cancels when calculating the mean kinetic energy. Thus

⟨ K ⟩_{hit} = \frac{m}{2} \frac{\int _{0}^{\infty} v ^{3} f ( v ) d v}{\int _{0}^{\infty} v f ( v ) d v} = \frac{m}{2} \frac{I _{5}}{I _{3}} .

Using

I_{5} = 2 I_{3} / α

gives

⟨ K ⟩_{hit} = \frac{m}{α} = 2 k_{B} T .

By contrast, an atom chosen randomly from the bulk has mean kinetic energy

3 k_{B} T /2

. The arrival weighting raises the normal-direction contribution to

k_{B} T

while each tangential direction still contributes

k_{B} T /2

. This is also the mean energy carried by atoms effusing through a thin aperture from an equilibrated reservoir.

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