* * * * * In a similar way it may be shown that the energy _W₂_ of the β rays reaching the surface is given by $$ W_2 = \frac {E_2 \sigma} {2 λ_2} $$ where _E₂_ and λ₂ are the values for the β rays corresponding to _E₁_ and λ₁ for the α rays. Thus it follows that _E₁_ λ₁_W₁_ ---- = ------ _E₂_ λ₂_W₂_ λ₁ and λ₂ are difficult to determine directly for the radio-active substance itself, but it is probable that the ratio λ₁/λ₂ 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 α 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 α rays, was 12·7 times as great as the current due to the β rays. Since the α 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, _W₁_ total number of ions due to α rays ----- = ---------------------------------- _W₂_ total number of ions due to β rays 12·7 × 6·1 = ------ = 0·5 approximately. 154 Now the value of λ₁ for aluminium is 2740 and of λ₂ for the same metal 14, thus _E₁_ λ₁_W₁_ ----- = ---------- = 100 approximately _E₂_ λ₂_W₂_ 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 α rays. This estimate is confirmed by calculations based on independent data. Let _m₁_, _m₂_ be the masses of the α and β particles respectively and _v₁_, _v₂_ their velocities. $$ \frac {Energy of one \alpha particle} {Energy of one \beta particle} = \frac {m_1 v_1^2} {m_2 v_2^2} $$ $$ = \frac {\frac {m_1} {e} v_1^2} {\frac {m^2} {e} v_2^2} $$ . Now it has been shown that for the α rays of radium _v₁_ = 2·5 × 10⁹, _e_ ---- = 6 × 10³. _m₁_ The velocity of the β rays of radium varies between wide limits. Taking for an average value _v₂_ = 1·5 × 10¹⁰, _e_ ---- = 1·8 × 10⁷, _m₂_