# x-ray interaction with atmosphere

Why x-ray are stopped by atmosphere while they are more energetic than UV or IR?

They certainly interact with atmosphere but I can't understand which phenomenon stop them.

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X rays have a short wavelength. Imagine a road covered with pingpong balls. Try to roll a marble across, and it will most probably bounce back and forth amongst the balls, and eventually stop without getting to the other side. Now try a soccer ball. These will cross almost all the time. And now a truck. I doubt that you will find a situation where these don't cross..

Similarly, the atmosphere can be thought as the road with ping pong balls(molecules), and EM waves are various objects you try to roll. Radio waves have large (1 meter &c) wavelengths. Nobody stops these. Light waves have much smaller wavelengths (in $\mu\text{ m}$), but those are still large as compares to atoms (these are to the order $10^{-10}\text{ m}$). X rays are small enough to be obstructed by everyone, thus (thankfully) they do not reach the crust and fry us.

An interesting side note: The reason that makes radio waves abundant is the same reason why they are harder to use than light/xrays in astronomy. Because of their size, they do not give an accurate reading unless we employ large telescopes. Think back to the truck. If an ant hypothetically throws a truck at you, you really can't be sure where the ant is. But if the ant throws a marble or a sesame seed (assuming you catch the seed and determine its velocity), you can calculate the trajectory and find the ant with accuracy, and squash him for throwing things at you. The same principle applies when searching for a radio source..

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Ok I agree with this. Maybe what I don't understand is this: if I want to stop light I just have to use a wall maje of wood ; if I want to stop xray I have to use a large wall in lead ; what I will use to stop radio waves ? – PanAkry Feb 7 '12 at 20:23
Ah. Here you are confusing momentum with the wavelength. Unfortunately, in my analogy, I have chosen projectiles of whose momentum is directly proportional to their size, whereas for EM waves, $p\propto\frac{1}{\lambda}$. Here, the wall is infinitely large. Now it is note dependant so much on the size of the projectile as it is dependent on the mass. If you want, replace the marble with an extremely dense lead pellet; replace the soccer ball with an ice ball, and replace the truck with a huge softball. – Manishearth Feb 8 '12 at 1:02
The main point is, this analogy only goes so far. What actually happens comes from quantum mechanics. Light is both a particle and a wave, and has a quantum wavefunction. This wavefunction is a probability function of where the light "particle" (photon) can be, and has the same wavelength as the light. Thus, it is more spread out. If such a spread out wave comes across some atoms, it simultaneously passes between the gaps of the atoms, as well as being absorbed by the atoms (sounds weird, but it's due to quantum mechanics). So it has a probability of being absorbed. – Manishearth Feb 8 '12 at 1:10
Whereas the xray is inevitably absorbed, as its wavelength is smaller than an atom. So it is not spread out enough to have the choice. Aside from this, xrays have a greater tendency to be absorbed by atoms as they are ionizing radiation. Radio waved do not have enough energy to make changes to an atom, so these pass through. (This also comes from the quantized nature of light. Sending a lots of radio wave-photons is not the same as sending an xray-photon) – Manishearth Feb 8 '12 at 1:13
It sounds to me as if you are describing Thompson scattering (aka Rayleigh scattering) but this is not a major component of the X-Ray scattering in the atmosphere. See for example adsabs.harvard.edu/full/1988A%26A...193..345F where they calcuate the Thompson scattering and find it to be a factor of 100 lower than photoelectron production. – John Rennie Feb 8 '12 at 10:56

The Chandra web site has a nice description of this. see http://chandra.harvard.edu/xray_astro/absorption.html for the details. In brief, X-Rays have enough energy to ionise molecules in the atmosphere and they are absorbed in the process.

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By this approach, xrays and higher should have the same penetrating power, though we know that penetrating power varies smoothly as frequency. Comments? – Manishearth Feb 7 '12 at 19:02
In solids X-Rays are absorbed by several mechanisms that depend on the energy of the X-Rays. See en.wikipedia.org/wiki/Mass_attenuation_coefficient for details. – John Rennie Feb 8 '12 at 10:29

I would think the simple explanation would go like this. If an X -ray photon travels along a path between air molecules (no molecules along the path/ all molecules are to the side of the path) - then the photon isn't absorbed. If the photon is not on a molecule-free path and it hits a molecule, it is absorbed. A column of 14.7 pounds of air sits on every square inch of the earth's surface. There are no "molecule-free paths" from the "top of the atmosphere" to the ground when say a beam of X-Rays with cross section 1 square inch tries to travel from space, through 14.7 pounds of air, to the ground, so it is all absorbed. In the hospital, when you have an x-ray image made, a beam of X-rays with cross-section of 1 sq inch would only have to travel through something like 10 grams of air from the source to your body, then go through one pound of your muscle, and then a few more grams of air to reach the plate. Lots of photons are absorbed, but percent-wise many more make the trip than through 14.7 pounds of air. Bones are another matter - the atoms are so close together that there is no atom-free path even through a small bone (depends a bit on bone density though).

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