Why do X-rays go through things?

Question

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?

Answer 1

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.

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 $.

Answer 2

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.