Expert: Philip A. Stahl - 6/13/2010

Question
can electricity be generated from cosmic rays and if so how? and what energy are they at ground level

Answer
Hello,

The erroneous perception that cosmic rays might have an association with electricity probably originated with the original (incorrect) theory of their formation: from atmospheric electricity (due to Victor Hess). Thus, if atmospheric electrical currents might give rise to these cosmic rays, perhaps it was feasible that reversing the process might yield electricity.

Later, Robert Millikan (famed for his "oil drop" experiment to measure the electronic charge) showed that the cosmic rays were extraterrestrial in origin and not formed from atmospheric electricity. Millikan's conception of them was incorrect, however, because he viewed them mainly as photons, together with some secondary electrons.

Today we are aware, from many sophisticated detector analyses, that cosmic rays are nearly 90% protons, and about 1% electrons.(About 9% are alpha particles, or helium nuclei).

Another reason for the electricity misconception, or assumption cosmic rays can yield such, is the term "rays", which is a misnomer. Cosmic rays do not arrive as any kind of beam or stream of particles, as the name implies, but *individually*.

Indeed, the energy spectrum of cosmic rays shows an interaction range from 1 per square meter per second (for the most interactive), to 1 per square meter per year, to 1 per square *kilometer* per year. With these sort of sparse concentrations, even at the highest energy level, developing any kind of sustained electric current would be impossible.

Re: impact at ground level, one must reckon in attentuation by the Earth's atmosphere. Thus, primary cosmic rays entering the atmosphere at say an energy of 10^16 eV (where 1 eV = 1.6 x 10^-19 J) will become "secondary cosmic rays" at ground level with diminished energy.

The ground effect(for the energy absorbed), of cosmic rays, is ~ 0.3 mSv (milli-Sieverts), compared to a whole radiation effect (from all sources) of ~ 2.4 mSv. Since, 1 Sv = 1 J/kg (joule per kilogram) then for 0.3 milli-Sievert: 0.3 mSv ~  0.0003 J/kg


Hope this sheds some light!