Page:Radio-activity.djvu/148

 Let m_{0} = mass of electron for slow speeds; m    = apparent mass of electron at any speed; u    = velocity of electron; V    = velocity of light.

Let β = u/V; then it can be shown that

m/m_{0} = (3/4)ψ(β)     (1),

where     ψ(β) = (1/β^2)[(1 + β^2)/(2β) log ((1 + β)/(1 - β)) - 1]      (2).

The experimental method employed to determine e/m and u is similar to the method of crossed spectra. Some strongly active radium was placed at the bottom of a brass box. The rays from this passed between two brass plates insulated and about 1·2 mm. apart. These rays fell on a platinum diaphragm, containing a small tube about 0·2 mm. in diameter, which allowed a narrow bundle of rays to pass. The rays then struck a photographic plate enveloped in a thin layer of aluminium.

In the experiments the diaphragm was about 2 cms. from the active material and at the same distance from the photographic plate. When the whole apparatus was placed in a vacuum, a of from 2000 to 5000 volts could be applied between the plates without a spark. The rays were deflected in their passage through the electric field, and produced what may be termed an electric spectrum on the plate.

Fig. 28.

If a magnetic field is superimposed parallel to the electric field by means of an electromagnet, a magnetic spectrum is obtained perpendicular to the electric spectrum. The combination of the two spectra gives rise to a curved line on the plate. The double trace obtained on the photographic plate with reversal of the magnetic field is shown in Fig. 28. Disregarding some small corrections, it can readily be shown that if y and z are the electric and magnetic deviations respectively,

β = κ_{1}(z/y)     (3), and     e/m = κ(z^2/y)      (4).