= Solution
Wall arrivals are not an unbiased sample of the <gas>: faster atoms arrive more frequently. The normalized <speed distribution> among incident atoms is
$$
f_{\rm hit}(v)=\frac{vf(v)}{\overline v}.
$$
The angular factor in the arrival rate cancels when calculating the <mean> <kinetic energy>. Thus
$$
\langle K\rangle_{\rm hit}
=\frac m2\frac{\int_0^\infty v^3f(v)\,dv}{\int_0^\infty vf(v)\,dv}
=\frac m2\frac{I_5}{I_3}.
$$
Using $I_5=2I_3/\alpha$ gives
$$
\boxed{\langle K\rangle_{\rm hit}=\frac m\alpha=2k_BT.}
$$
By contrast, an atom chosen randomly from the bulk has <mean> <kinetic energy> $3k_BT/2$. The arrival weighting raises the normal-direction contribution to $k_BT$ while each tangential direction still contributes $k_BT/2$. This is also the <mean> <energy> carried by atoms effusing through a thin aperture from an equilibrated reservoir.
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