Page:Popular Science Monthly Volume 59.djvu/334

324 too, by the results of experiments on electrolysis, that to carry the unit charge of electricity requires a collection of atoms of hydrogen which together weigh about of a milligram; hence if we can measure the charge of electricity on an atom of hydrogen we see that  of this charge will be the weight in milligrams of the atom of hydrogen. This result is for the case when electricity passes through a liquid electrolyte. I will now explain how we can measure the mass of the carriers of electricity required to convey a given charge of electricity through a rarefied gas. In this case the direct methods which are applicable to liquid electrolytes cannot be used, but there are other, if more indirect, methods, by which we can solve the problem. The first case of conduction of electricity through gases we shall consider is that of the so-called cathode rays, those streamers from the negative electrode in a vacuum tube which produce the well-known green phosphorescence on the glass of the tube. These rays are now known to consist of negatively electrified particles moving with great rapidity. Let us see how we can determine the electric charge carried by a given mass of these particles. We can do this by measuring the effect of electric and magnetic forces on the particles. If these are charged with electricity they ought to be deflected when they are acted on by an electric force. It was some time, however, before such a deflection was observed, and many attempts to obtain this deflection were unsuccessful. The want of success was due to the fact that the rapidly moving electrified particles which constitute the cathode rays make the gas through which they pass a conductor of electricity; the particles are thus as it were moving inside conducting tubes which screen them off from an external electric field; by reducing the pressure of the gas inside the tube to such an extent that there was very little gas left to conduct, I was able to get rid of this screening effect and obtain the deflection of the rays by an electrostatic field. The cathode rays are also deflected by a magnet, the force exerted on them by the magnetic field is at right angles to the magnetic force, at right angles also to the velocity of the particle and equal to $$Hev\ \sin\theta$$ where $$H$$ is the magnetic force, $$e$$ the charge on the particle and $$\theta$$ the angle between $$H$$ and $$v$$. Sir George Stokes showed long ago that, if the magnetic force was at right angles to the velocity of the particle, the latter would describe a circle whose radius is $$mv/eH$$ (if $$m$$ is the mass of the particle); we can measure the radius of this circle and thus find $$m/ve$$. To find $$v$$ let an electric force $$F$$ and a magnetic force $$H$$ act simultaneously on the particle, the electric and magnetic forces being both at right angles to the path of the particle and also at right angles to each other. Let us adjust these forces so that the effect of the electric force which is equal to $$Fe$$ just balances that of the magnetic force which is equal to $$Hev$$; when this is the case $$Fe = Hev$$ or $$v = F/H$$. We can thus find $$v$$, and knowing