λ_{1} and λ_{2} are difficult to determine directly for the radio-active substance itself, but it is probable that the ratio λ_{1}/λ_{2} is not very different from the ratio for the absorption coefficients for another substance like aluminium. This follows from the general result that the absorption of both [Greek: alpha] and β rays is proportional to the density of the substance; for it has already been shown in the case of the β rays from uranium that the absorption of the rays in the radio-active material is about the same as for non-radio-active matter of the same density.
With a thick layer of uranium oxide spread over an area of 22 sq. cms., it was found that the saturation current between parallel plates 6·1 cms. apart, due to the [Greek: alpha] rays, was 12·7 times as great as the current due to the β rays. Since the [Greek: alpha] rays were entirely absorbed between the plates and the total ionization produced by the β rays is 154 times the value at the surface of the plates,
W_{1}/W_{2} = (total number of ions due to [Greek: alpha] rays)/(total number of ions due to β rays)
= (12·7 × 6·1)/154 = 0·5 approximately.
Now the value ofλ_{1} for aluminium is 2740 and of λ_{2} for the same metal 14, thus
E_{1}/E_{2} = λ_{1}W_{1}/(λ_{2}W_{2}) = 100 approximately.
This shows that the energy radiated from a thick layer of material by the β rays is only about 1 per cent. of the energy radiated in the form of [Greek: alpha] rays.
This estimate is confirmed by calculations based on independent data. Let m_{1}, m_{2} be the masses of the [Greek: alpha] and β particles respectively and v_{1}, v_{2} their velocities.
(Energy of one [Greek: alpha] particle)/(Energy of one β particle) = m_{1}v_{1}^2/(m_{2}v_{2}^2) = ((m_{1}/e)v_{1}^2)/((m_{2}/e)v_{2}^2).
Now it has been shown that for the [Greek: alpha] rays of radium
v_{1} = 2·5 × 10^9,
e/m_{1} = 6 × 10^3.
