# If blue light has a higher energy than red light, why does it scatter more?

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?

• – Qmechanic May 21 '12 at 23:27
• 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 --- ex-moderator kitten May 21 '12 at 23:59

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.

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

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

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

Scattering may be angular interaction rather diffusion varied as a function of temperature when becomes very hot blue rays are emitted during evening temperature low orange rays are emitted being very low dark rays are emitted depending upon thermal temperature involved.

This therefore means that the blueness of the sky is from the blue light that is scattered from the sunlight in the atmosphere which, then enters our eyes from all regions of the sky. The black sky at night that shows you the moon, planets and stars is due to the absence of the Sun's light.

The light interacts with the molecules and particles and the blue light is scattered. In space, there are no air molecules or dust particles to scatter the different colors in light, so the Sun looks white and the 'sky' looks black. The effect therefore should have actually been the Tyndall effect, but it is more commonly known to physicists as Rayleigh scattering — named after Lord Rayleigh ... If you were on the Moon, which has no atmosphere, the sky would be black both night and day. You can...

For molecules, two types of scattering can occur. The second type of scattering, Raman scattering, is an inelastic scattering process in which the light scattered by a molecule emerges having an energy that is slightly different (more or less) than the incident light.

Rayleigh scattering is in the main elastic scattering from small particles whose size is less than that of the wavelength of the photon. The scattering can occur of atoms or molecules and for molecules the scattering can be inelastic with a change of rotational energy of the molecule.

When photons are scattered from an atom or molecule, most of them are elastically scattered (Rayleigh scattering), such that the scattered photons have the same energy (frequency and wavelength) as the incident photons. A small fraction of the scattered photons (approximately 1 in 10 million) are scattered inelastically by Raman Scattering.

Both Rayleigh scattering and Raman scattering are second order processes the difference is only that the final state of the atom/molecule coincides with the initial state of the atom/molecule in Rayleigh scattering, while in Raman scattering, the final state is different from the initial state. since there are many final

The molecule then relaxes back to either a lower or higher energy level (relative its original state) by emitting another photon. going photon of the same energy, frequency, and wavelength can be emitted in any direction Effectively, photons are being scattered. Although this process can ... spectrum and more. (Reddish colors that we see toward the horizon at sunrise and sunset are caused by different optical effects, not Rayleigh scattering.) but by Raman scattering. 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 ... As E=hf=hcλ, blue light - with a smaller wavelength - should have a higher energy. However, it is the case that blue light scatters the 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.most. Why is it that higher energy rays scatter more?

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 independent.

If "scattering" can be extended to include diffraction you find the wavelength dependence is inverted. Red light is diffracted more strongly than blue light.

• Hi S.N., welcome to PSE! Did you actually write this, or did you copy it from somewhere else? Note that including external content is allowed here, but you always have to properly reference the sources (as usual in science). Thank you for your collaboration. – AccidentalFourierTransform Jan 14 '18 at 2:33