Rayleigh scattering problem We read everywhere that the blue sky is due to the white light from the sun interacting with the air molecules. But the blue wavelength is around 420 nm, the O2 and N2 molecules around 0.3 nm that is roughly 1000 times smaller. How can the sun light interact with air ? So what is the explanation of the blue sky ?
 A: tl;dr Although the cross section of molecules to optical light is very small, there are a lot of molecules. Moreover, the cross section is very dependent on wavelength ($\sigma\propto\lambda^{-4}$), so even if blue light doesn't scatter much, red light scatters even less.

It's true that the cross section of oxygen and nitrogen is immensely small. In fact, because the ratio $r/\lambda$ is of the order of $3\times10^{-4}$ for blue light, i.e. $\ll1$, we're so far out in the Rayleigh regime that, due to quantum mechanical effects, the cross section of O2 and N2 are much, much smaller than the geometrical cross section (the standard, $\sigma = 2\pi r^2$, cross section that you might think the molecules would have), roughly a factor $10^{-12}$ smaller. Thus, to a blue photon at $\lambda = 420\,\mathrm{nm}$, an oxygen/nitrogen molecule has a cross section of some $7\times10^{-27}\,\mathrm{cm}^2$, if I've calculated correctly.
On the other hand, there are a lot of molecules. From the ideal gas law, at sea level we have a molecular number density of
$$
\begin{array}{rcl}
n & =      & \frac{P}{k_\mathrm{B} T} \\
  & \simeq & \frac{1\,\mathrm{atm}}{k_\mathrm{B} \times 300\,\mathrm{K}} \\
  & \simeq & 2\times10^{19} \,\mathrm{cm}^{-3},
\end{array}
$$
where $k_\mathrm{B}$ is Boltzmann's constant.
Hence, the mean free path of a blue photon at sea level is of $\ell = 1 / n\sigma \sim 60\,\mathrm{km}$. The atmospheric density decreases as you go to higher altitudes with a scale height of $\sim8\,\mathrm{km}$, but integrating density through the entire atmosphere, you get that, above each $\mathrm{cm}^2$, you have $N\sim2.5\times10^{25}$ molecules (this is called the column density).
Hence, the probability of scattering for a blue photon across the atmosphere is
$$
P_\mathrm{abs, blue} = 1 - e^{-N\sigma} \simeq 0.16,
$$
meaning that $\sim84$% of the photons don't scatter.
However, in the Rayleigh regime, the scattering cross section goes a $\sigma \propto \lambda^{-4}$. That means that the cross section as seen by a red photon with $\lambda = 700\,\mathrm{nm}$ is $(700/420)^4 \simeq 7.7$ times smaller, so now the probability is only
$$
\begin{array}{rcl}
P_\mathrm{abs, red} & = & 1 - e^{-N\sigma_\mathrm{red}} \\
                    & = & 1 - e^{-N\sigma_\mathrm{blue}/7.7} \\
                    & \simeq & 0.02.
\end{array}
$$
i.e. $\sim98$% of the red photons don't scatter.
A: The wavelength does not to be smaller than the radius of the particle. Rayleigh scattering happens because the E-field from the incoming light wave polarizes these particles, converting it into a dipole. You could benefit from reading the Wikipedia article 1.
