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For a fixed sphere and fixed refractive index, both the Rayleigh scattering cross-section and scattering efficiency are proportional to ${\lambda }^{-4}$. Halving wavelength multiplies either by $16$, as long as the shorter wavelength still satisfies $ka\ll 1$. If optical dispersion changes $m$, the material factor changes too; the pure fourth-power comparison then needs correction.

For equal incident light intensities at blue wavelength $450\mathrm{nm}$ and red wavelength $650\mathrm{nm}$, the same toy sphere givesBlue light therefore produces about $4.35$ times the scattered power. This compares the particle's response at two wavelengths; it does not assert that an actual sky has this ratio of observed blue and red light intensities.

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 $k$,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 $ka=2\pi (10/550)\simeq 0.1142$ and $(m^2-1)/(m^2+2)=5/17$. Substitution gives Rayleigh scattering cross-section $\sigma \simeq 1.234\times 10^{-20}{\mathrm{m}}^2$ and scattering efficiency $Q_{\mathrm{sca}}\simeq 3.929\times 10^{-5}$. At incident light intensity $1\mathrm{W}{\mathrm{m}}^{-2}$, the scattered power is approximately $1.234\times 10^{-20}\mathrm{W}$. These are leading small-particle estimates, not a measured air-molecule cross-section.