An Interpretation of Cosmic-Ray Phenomena

Abstract

Schindler's data on the transition effects of the cosmic rays have been interpreted on the assumption that the equilibrium between the primary radiation and its secondary corpuscular rays is different in different media. The ionization behind any thickness of absorbing materials, assumed to be proportional to the flux of secondaries, is calculated in terms of the absorption coefficients, $\nu \mathrm{}\left(m\right)\mathrm{}$, of the primary radiation and the production and absorption coefficients, $\beta \mathrm{}\left(m\right)\mathrm{}$ and $\mu (m, m)$, respectively, of the secondary rays, these coefficients being characteristics of the media. A comparison with the data permits a determination of each of the absorption coefficients as well as the product of each of the production coefficients by the number of primary rays. The results are as follows, expressed in ${\mathrm{cm}}^{-1\mathrm{}}$ lead equivalent: $\nu \left(\mathrm{Pb}\right)=0.0064\mathrm{}$; $\nu \left(\mathrm{Fe}\right)=0.009\mathrm{}$; $\mu (\mathrm{P}\mathrm{b},\mathrm{a}\mathrm{i}\mathrm{r})$, the absorption coefficient in lead for secondaries produced in air=0.50; $\mu (\mathrm{P}\mathrm{b},\mathrm{P}\mathrm{b})=0.98\mathrm{}$; $\mu (\mathrm{F}\mathrm{e},\mathrm{a}\mathrm{i}\mathrm{r})=0.30\mathrm{}$; $\mu (\mathrm{F}\mathrm{e},\mathrm{F}\mathrm{e})=0.45\mathrm{}$; $\mu (\mathrm{P}\mathrm{b},\mathrm{F}\mathrm{e})=0.72\mathrm{}$; $\mu (\mathrm{F}\mathrm{e},\mathrm{P}\mathrm{b})=0.48\mathrm{}$; $\frac{\beta \left(\mathrm{Pb}\right)}{\beta \left(\mathrm{Fe}\right)}=2.0\mathrm{}$. This determination of the absorption coefficients of the primary and secondary radiations allows the following estimates to be made. (a) The lower limit of the average energy of the secondary radiation is about 30 million volts. (b) The average number of secondaries per primary is about 100 in iron and 230 in lead. (c) The energy of a primary cosmic ray, equal to the sum of the energies of its secondaries, is about 2×${10}^{10}$ volts.