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 154_q₀_ when the rays are completely absorbed. Now from the above table the ionization due to the β rays is ·0074 of that produced by α 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 ·0074 ---------------------------------- = ----- × 154 = 0·20. Total number of ions produced by α rays 5·7 Therefore about ⅙ 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 α 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 α radiation of thorium or radium is reduced to half value after passing through a thickness of about 0·0005 cm. of aluminium. Taking into consideration the great density of the radio-active substances, it is probable that most of the radiation which escapes into the air is due to a thin skin of the powder not much more than ·0001 cm. in thickness. * * * * * An estimate, however, of the relative rate of emission of energy by the α and β rays from a thick layer of material can be made in the following way:—For simplicity suppose a thick layer of radio-active substance spread uniformly over a large plane area. There seems to be no doubt that the radiations are emitted uniformly from each portion of the mass; consequently, the radiation, which produces the ionizing action in the gas above the radio-active layer, is the sum total of all the radiation which reaches the surface of the layer. * * * * * Let λ₁ be the average coefficient of absorption of the α rays in the radio-active _substance itself_ and σ the specific gravity of the substance. Let _E₁_ be the _total_ energy radiated per sec. per unit mass of the substance when the absorption of the rays in the substance itself is disregarded. The energy per sec. radiated to the upper surface by a thickness _dx_ of a layer of unit area at a distance _x_ from the surface is given by $$ \frac {1} {2} E_1 \sigma e^{–λ_1 x} dx $$ . The total energy _W₁_ per unit area radiated to the surface per sec. by a thickness _d_ is given by $$ W_1 = \frac {1}{2} \int₀^d E_1 \sigma e^{–λ_1 x} dx $$ $$ = \frac {E_1 \sigma} {2 λ_1} (1 − e^{–λ_1 d}) $$ $$ = \frac {E_1 \sigma} {2 λ_1} $$ if λ₁_d_ is large.