= Scattering efficiency
{title2=$Q_{\rm sca}=\sigma/(\pi a^2)$}
For a <sphere> of <radius> $a$, the <scattering efficiency> is its effective scattering <area> divided by the projected <geometric cross-section> $\pi a^2$. It is a <dimensionless quantity>. For <Rayleigh scattering by a small sphere> with relative <refractive index> $m$ and incident <wavenumber> $k$,
$$
Q_{\rm sca}=\frac{\sigma}{\pi a^2}=\frac83(ka)^4\left(\frac{m^2-1}{m^2+2}\right)^2.
$$
At fixed <wavelength> and fixed <refractive index>, the <Rayleigh scattering cross-section> scales as $a^6$, whereas the <scattering efficiency> scales as $a^4$. Doubling <radius> multiplies the former by $64$ and the latter by $16$. An efficiency here is an effective scattering <area> relative to a reference <area>, rather than a fraction of all light in an arbitrary beam.
For a worked toy example, take $a=10\,\mathrm{nm}$, $m=1.5$, and $\lambda=550\,\mathrm{nm}$. Then $ka=2\pi(10/550)\simeq0.1142$ and $(m^2-1)/(m^2+2)=5/17$. Substitution gives <Rayleigh scattering cross-section> $\sigma\simeq1.234\times10^{-20}\,\mathrm{m}^2$ and <scattering efficiency> $Q_{\rm sca}\simeq3.929\times10^{-5}$. At incident <light intensity> $1\,\mathrm{W\,m^{-2}}$, the scattered <power> is approximately $1.234\times10^{-20}\,\mathrm W$. These are leading small-particle estimates, not a measured air-<molecule> cross-section.
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