As $E=hf=\frac{hc}{\lambda}$, blue light – with a smaller wavelength – should have a higher energy. However, it is the case that blue light scatters the most. Why is it that higher energy rays scatter more?
[Physics] If blue light has a higher energy than red light, why does it scatter more
atmospheric scienceopticsscatteringvisible-light
Related Solutions
EDIT: This question discusses why the sky is blue with regards to classical rayleigh scattering.
Light does indeed travel in straight lines. However, this just indicates the direction of the movement of the wave as a whole. A light wave is an electromagnetic oscillation, and therefore higher frequencies will have quicker oscillations. The wave itself will be moving forward at the same speed, but there will be "more waviness" per unit length for a higher frequency wave than for a lower frequency wave.
If you switch from the so called "wave picture" to the so called "particle picture" (the shabbily named wave-particle duality) then the frequency $\nu$ of a photon is related to it's energy as given by $$ E = h \nu$$
These photons have an intrinsic position-momentum uncertainty wherein their position at any given moment cannot be described to an arbitrarily small accuracy. It is the probability of finding the photon at a given place that you have referred to as the quantum amplitude of a photon.
The square of the quantum amplitude describes the probability of the photon (neglecting time dependence for the moment) of being in a particular place in space, or possessing a particular momentum etc., depending on what basis you are describing the system in. So for example, if $\psi(x)$ is a function describing the quantum amplitude of a particle at every point in space, the probability that the particle lies somewhere between $0$ and $x$ is given by $$ P(x) = \int_{0}^{x} |\psi(x)|^2 dx $$
One cannot predict it's position more definitively than this probability.
Now that I've clarified what I mean by a quantum amplitude - There does exist scattering theory for quantum mechanics. This scattering theory deals with the problem of how these quantum amplitudes change when confronted by an external "potential". A potential can also be a potential as created by another particle, so it covers that case as well. A frequently used approximation to help ease of calculation is the Born approximation. It just so happens that even without the Born approximation, for a coloumb potential (which is any regular atom), the scattering problem is exactly solvable. And that solution yields the same results as Rutherford scattering. So quantum mechanically as well, everything is quite well explained.
EDIT 2: I'd earlier said that Rayleigh scattering was exactly solvable, but it was actually Rutherford scattering. Rayleigh scattering is a limit of Mie scattering which is valid for EM waves scattering of particles. However, I suspect that if you consider photons as particles that interact with the coulomb potential, rutherford scattering will be a valid quantum mechanical description of the phenomenon. I haven't been able to find any links with the relevant details worked out, but once I do I'll link them here.
The colour you're seeing is from the very small fraction of light that the panels are reflecting. The vast majority of light is being absorbed to generate electricity.
Why some of the panels appear slightly blue while others don't I don't know. Presumably there must be small differences in the manufacturing process. The absorptance of solar panels does fall off at the extreme blue end of the spectrum, so you would expect the reflected light to have a blue tinge. A quick Google found this article that includes a typical absorption spectrum:
Best Answer
In general, the scattering of light from some object depends on the how close the wavelength of light is to the size of the object.
To make an analogy, if a tidal wave with a wavelength of several kilometers hits a telegraph pole with a radius of 15 cm it isn't going to scatter very much. On the other hand, waves with a wavelength of a few cm, e.g. generated by you throwing a stone into the water, are going to be strongly scattered.
As you've said in your question, blue light has a smaller wavelength than red light. Assuming you are talking about the sky, the scattering is from particles much smaller than the wavelength of light. That means you'd expect light with the smaller wavelength to be scattered more strongly because it's nearer to the size of the objects doing the scattering.
The formula you quote is for the energy of a photon, but this is not relevant for Rayleigh scattering.
To expand the discussion a bit, when the particle size approaches or exceeds the wavelength of light the difference in the wavelengths disappears. If you look at scattering from e.g. a colloidal suspension with a one micron particle size the scattering is (mostly) wavelength independant.
If "scattering" can be extended to include diffraction you find the wavelength dependance is inverted. Red light is diffracted more strongly than blue light.