Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can gamma rays of high enough energy entering our planet's atmosphere reach the surface (50% probability)?

Or, in other words, is there a window for extremely high-energy gamma rays like for the visible spectrum and radio?

This figure, from Electromagnetic Spectrum, shows that the penetration of gamma rays increases with increasing energy, but it seems to level out at about 25 km altitude:

Penetration depth of electromagnetic radiation as a function of frequency

There are no units on the X-axis, and thus it does not show the energy for the highest energy gamma rays for this figure. This does not rule out a window at even higher energies.

share|cite|improve this question
This question needs more tags. Some ideas: electromagnetic-radiation, photon-matter-interaction – Peter Mortensen Nov 9 '10 at 20:00
I put in some of the subject tags that seem to be emerging as "standard" on the site, hopefully that will help. But the tags can be changed again if there are better ones ;-) – David Z Nov 9 '10 at 20:05
up vote 4 down vote accepted

I found a reference through Google Books, Very high energy gamma-ray astronomy by Trevor Weekes, which says that the atmosphere is essentially opaque to high-energy gamma rays, equivalent to a meter-thick wall of lead. We are able to do gamma-ray astronomy with ground-based telescopes by detecting the decay products of the gamma rays' interactions with atmospheric particles, but the photons themselves never (well, essentially never) reach the ground.

From page 13:

The earth's atmosphere effectively blocks all electromagnetic radiation of energies greater than $10\text{ eV}$. The total vertical thickness of atmosphere above sea level is $10^{30}\ \mathrm{g\ cm^{-2}}$, and since the radiation length is $37.1\ \mathrm{g\ cm^{-2}}$, this amounts to more than 28 radiation lengths. This is true up to the energy of the highest known cosmic rays (some of which may be gamma rays).

share|cite|improve this answer
It is at…, but how can the content be viewed? – Peter Mortensen Nov 9 '10 at 21:33
Click on "Preview this book". It might or might not let you view the relevant portion of the book, but if not, you can always try to find it in a library. – David Z Nov 9 '10 at 21:35
I see no such link, only "No preview available". I may be geographically challenged. – Peter Mortensen Nov 9 '10 at 21:47
Maybe so, I don't know what sort of geographic restrictions are available in Google Books. I would recommend having a look at the closest university library. – David Z Nov 9 '10 at 22:15
@PeterMortensen I've never read the book, but it's easy to see how that number arises. Given their short wavelengths, gamma rays don't "see" electronic structure; all they do is Compton scatter or pair produce. Thus the only thing that matters (to first order, of course) is column density, which is $10^3\ \mathrm{g/cm^2}$ for both the atmosphere and a meter of lead. – Chris White Nov 2 '13 at 1:18

Propagation of very energetic photons through medium looks like this. The photon enter the medium, at some point it scatters off an electron or creates an electron-positron pair in a coulomb field, the initial energy distributed between the two daughter particles. Each of these particles then "splits" again, and so on, and as a result an electromagnetic shower develops. The typical length at which this "doubling" occurs is called the radiation length. This length is calculated (and measured) for many materials; it is a property of a medium and does not depend on the energy of the initial photon. For air at normal pressure and temperature is it of the order of hundreds of meters (and it is increases with pressure drop). So, it means that the initial extremely energetic photon will hardly make it into the troposphere, not to mention ground.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.