= Solution
Choose a flat wall and let $\theta$ measure the angle from the direction towards it. <Particles> of <speed> $v$ in <solid angle> $d\Omega$ occupy fraction $f(v)dv\,d\Omega/(4\pi)$ of an isotropic <gas>. In <time> $dt$, those within normal distance $v\cos\theta\,dt$ reach the wall. Their number per <area> per <time> is therefore $n f(v)dv\,v\cos\theta\,d\Omega/(4\pi)$.
Only the incoming <hemisphere> contributes. Integrating gives the one-way <collision> rate per unit <area>,
$$
J_N=n\int_0^\infty vf(v)\,dv\,
\frac1{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta
=\boxed{\frac14n\overline v.}
$$
Its units are $\mathrm{m^{-2}s^{-1}}$. This is not $n\overline v/2$: selecting the incoming half of the <particles> must also account for their <mean> normal <velocity>. For <effusion> through a thin hole whose diameter is much less than the <mean free path>, the same one-way rate applies because <collisions> within the opening can be neglected.
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