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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?

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Related: physics.stackexchange.com/q/17/2451 –  Qmechanic May 21 '12 at 23:27
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I assume you are imagining the scattering process in terms of billiard balls bouncing on some obstacle and wondering why the high momentum one is deflected more or some similar picture. Alas that is not---in general---a good metaphor for scattering processes. –  dmckee May 21 '12 at 23:59
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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.

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I assume you are thinking along the lines of why is the sky blue?

Its because of the interaction of light with matter. In this case the interaction is called Rayleigh scattering.

The intensity of Rayleigh scattering is proportional to $$(Energy\;of\; the\; Photon)^4$$ or more completely

$$I = I_0\frac{8 \pi ^4 \alpha ^2}{\lambda ^4 R^2}\left(1+ cos^2\theta \right)$$

See the wikipedia article on Rayleigh Scattering for More info:

http://en.wikipedia.org/wiki/Rayleigh_scattering

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