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 [Greek: alpha] 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 λ_{1} be the average coefficient of absorption of the [Greek: alpha] rays in the radio-active substance itself and σ the specific gravity of the substance. Let E_{1} 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
(1/2)E_{1}σe^{-λ_{1}x}dx.
The total energy W_{1} per unit area radiated to the surface per sec. by a thickness d is given by
W_{1} = (1/2)[integral]_{0}^d E_{1}σe^{-λ_{1}x}dx
= (E_{1}σ/(2λ_{1}))(1 - e^{-λ_{1}d}) = E_{1}σ/(2λ_{1})
if λ_{1}d is large.
In a similar way it may be shown that the energy W_{2} of the β rays reaching the surface is given by W_{2} = E_{2}σ/(2λ_{2}) where E_{2} and λ_{2} are the values for the β rays corresponding to E_{1} and λ_{1} for the [Greek: alpha] rays. Thus it follows that
E_{1}/E_{2} = λ_{1}W_{1}/(λ_{2}W_{2});
