radiated by the [Greek: alpha] and β rays by measuring the ratio of the total number of ions produced by them. If λ is the coefficient of absorption of the β rays in air, the rate of production of ions per unit volume at a distance x from the source is q_{0}e^{-λx} where q_{0} is the rate of ionization at the source.
The total number of ions produced by complete absorption of the rays is
[integral]_{0}^[infinity] q_{0}e^{-λx}dx = q_{0}/λ.
Now λ is difficult to measure experimentally for air, but an approximate estimate can be made of its value from the known fact that the absorption of β rays is approximately proportional to the density of any given substance.
For β rays from uranium the value of λ for aluminium is about 14, and λ divided by the density is 5·4. Taking the density of air as ·0012, we find that for air
λ = ·0065.
The total number of ions produced in air is thus 154q_{0} when the rays are completely absorbed.
Now from the above table the ionization due to the β rays is ·0074 of that produced by [Greek: alpha] rays, when the β rays passed through a distance of 5·7 cms. of air.
Thus we have approximately
(Total number of ions produced by β rays)/(Total number of ions produced by [Greek: alpha] rays) = ·0074/5·7 × 154 = 0·20.
Therefore about 1/6 of the total energy radiated into air by a thin layer of uranium is carried by the β rays or electrons. The ratio for thorium is about 1/22 and for radium about 1/14, assuming the rays to have about the same average value of λ.
This calculation takes into account only the energy which is radiated out into the surrounding gas; but on account of the ease with which the [Greek: alpha] rays are absorbed, even with a thin layer, the greater proportion of the radiation is absorbed by the radio-active substance itself. This is seen to be the case when it is recalled that the [Greek: alpha] radiation of thorium or radium is reduced to half value after passing through a thickness of about 0·0005 cm. of
