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I always heard that the smaller the wavelength, the more interactions take place. The sky is blue because the blue light scatters. So why is this not true for X-rays, which go through objects so readily that we need often use lead to absorb it?

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    $\begingroup$ Why do visible light pass through glass? $\endgroup$ – Thorbjørn Ravn Andersen Apr 21 '16 at 10:46
  • $\begingroup$ Also, be aware that x-rays do interact with the matter. A significant amount of photons will diffract from their "straight path" die to "collisions" with matter. Photon scattering is one of the main sources of noise in X-ray imaging. There is even a tomography modallity called "X-ray diffraction tomography" that uses the information from the photons that "do not go through things" $\endgroup$ – Ander Biguri Apr 21 '16 at 11:32
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    $\begingroup$ Possible duplicate of this pair of questions: Why are objects opaque? and Why aren't all objects transparent? $\endgroup$ – Stop Harming Monica Apr 21 '16 at 12:53
  • $\begingroup$ It should be noted that lead is not required; bone is also opaque to x-rays. $\endgroup$ – Stop Harming Monica Apr 21 '16 at 12:54
  • $\begingroup$ @OrangeDog: Hmm, if everything would be transparent, we couldn't see anything - interesting thought. $\endgroup$ – phresnel Apr 21 '16 at 14:21
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You have to distinguish, which interactions take place, when electromagnetic radiation passes through a solid and interacts with it.

There is a nice plot on Wikipedia, showing the dielectric response of solids for different wavelengths/frequencies.

Dielectric response

Basically, as the frequency gets higher, the wavelength becomes shorter, and the molecules or atoms are no longer able to follow the driving force that is transferred by the electromagnetic wave. Therefore in this picture the real part of the refractive index goes to $ 1 $, while the imaginary part, which leads to optical losses or absorption, goes to $ 0 $.

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    $\begingroup$ Good answer. But there are things smaller than atoms (e.g. electrons, nuclei). Since the question is about X-rays and this answer is about distinguishing different interactions - a plot which distinguishes the different interactions of X-rays would also be helpful. See Fig. 24.3 here also. Good illustration/explanation here too. $\endgroup$ – uhoh Apr 21 '16 at 11:19
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Light is composed out of a large ensemble of photons, and photons are quantum mechanical elementary particles. Matter is composed out of atoms and molecules , which have small dimensions and are in the quantum mechanical range.

The quantum mechanical "size of interaction region" is given by the Heisenberg uncertainty relation. Even though a photon is an elementary zero mass particle it has a momentum given by

$$p = \hslash k= \frac{h\nu}{c}= \frac{h}{\lambda}\,.$$

As the electromagnetic wave impinges on a solid, each individual photon will interact/scatter with an atom or molecule on its path.

The Heisenberg uncertainty principle says that if the photon has momentum p

\begin{align}\sf \Delta x\Delta p &\gt \frac{\hslash}{2}\\\end{align}

its position x is uncertain by a volume bounded by the HUP.

The uncertainty in the position of the photon, is inversely proportional to the wavelength. If $\lambda$ is large the photon has the probability to exist in a large x dimension in order for the HUP to be fulfilled.

One can think of the volume defined by the HUP as the measure of how "large" the photon is. The smaller the wavelength the more "point like" the interactions of the photon will be.

For optical frequencies, large $\lambda$ s, this distance is composed of a huge number of atoms and molecules on its way, and the probability that the photon, and therefore the electromagnetic wave built up by photons, will interact, is practically 1.

For x-rays the (HUP limit) $\sf \Delta x$ becomes smaller than the distances between the lattice distances of atoms and molecules, and the photon will interact only if it meets them on its path, because most of the volume is empty of targets for the x-ray wavelengths of the photon.

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  • $\begingroup$ So why blue light is scattered more than red light, according to your argument? $\endgroup$ – velut luna Apr 21 '16 at 5:41
  • $\begingroup$ @Mathaholic When in optical frequencies one has to look at the specific situations for the solutions of the scattering . I am answering about x-rays with a general HUP argument. See en.wikipedia.org/wiki/Rayleigh_scattering when wavelength is very much larger than atomic distances $\endgroup$ – anna v Apr 21 '16 at 6:04

protected by Qmechanic Apr 21 '16 at 6:40

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