air is small compared with that of water. If, then, u is measured, the radius of the drop can be calculated..

Now it is easy to deduce the total mass of water precipitated if the expansion of the gas by which the cloud was formed took place adiabatically, that is to say in such a way that no heat is either gained or lost during the process of expansion. Hence, if we know the mass of each drop we can calculate the number of drops present, and on the assumption that every negative ion acts as a condensation nucleus we can calculate the charge carried by a single ion. The details of the calculation are somewhat complicated and need not here be considered, but without entering into details the general method will be understood from the above account.

Proceeding in this way, J. J. Thomson found 34 × 10-10 electrostatic units for the charge carried by the negative ion produced by Röntgen rays and ultra-violet light. H. A. Wilson, by a slightly different method, gives the value as 3.1 × 10-10. Considering the difficulties of the experiments both these values

agree sufficiently nearly with that obtained for the charge carried by the hydrogen ion in electrolysis, and it may therefore be concluded that these two charges are, in fact, equal.

THE RATIO OF THE CHARGE TO THE MASS OF THE CATHODE RAYS

When a current of electricity is passed through a rarefied gas at a pressure of less than 1 millimetre of mercury a complicated series of phenomena can be seen as the discharge passes. With the various features of the discharge and their significance we are not here concerned, but there are certain phenomena to which we shall have occasion to refer later and which we will consider briefly.

If a current of electricity is passed between two electrodes through the gas contained in a glass vessel which can be evacuated by means of a pump, the character of the discharge gradually changes as the pressure of the gas is reduced. When the vacuum is sufficiently high, a faint stream of light is seen proceeding in straight lines from the negative electrode or cathode which frequently produces a

green phosphorescence on the walls of the glass at places at which the stream impinges. On account of the fact that the luminosity appears to emanate from the cathode it is usually spoken of as the cathode stream or radiation. Now these rays have been the subject of much investigation and discussion, and their nature was for a long time not understood, but they have ultimately been shown to consist of material particles projected from the cathode with very high velocities; moreover, they have been shown to carry with them a negative charge of electricity; for if the rays are caused to pass through a strong electric field, the stream which previously proceeded in a straight line from the cathode is now bent towards the positive electrode into a curve, as would be expected if it consisted of negatively charged particles projected from the cathode. Furthermore they are deflected by a magnetic field applied at right angles to the direction in which they are being propagated, which should also be the case on the view just mentioned as to the nature of the rays. For it has been shown by Rowland

that a charged particle in motion has the same magnetic effect as a current of electricity, and hence we should expect that this stream of particles, if charged, should be deflected by a magnetic field acting at right angles to the direction of motion of the particles and the magnitude of the deflection to be expected for a given magnetic field can easily be calculated. For if we have a field of strength H at right angles to the direction of a current C flowing in a straight wire, then, as is well known, the force per unit length exerted on the wire is HC.

Now in our case of a moving charged body, C=ev where e is the charge carried by the particle and v is its velocity.

Since the force always acts at right angles to the path of the particle, it does not affect its velocity but only the direction of its motion. Thus if r be the radius of curvature of the path of the particle, the electro-magnetic force (ev H) is balanced by the centripetal mv2 toward the centre of curvature of

force

r

the path of the particle. We thus have

If we measure r and H this gives us a relation

between and v.

m

In order to find them

separately it is necessary to find another relation between these two quantities, and this is done by measuring the deflection of the rays

an electric field. Consider the rays passing between two charged parallel plates and let F be the electric force between the plates, then Fe is the mechanical force acting on each particle, and hence each particle moves with Fe

an acceleration towards the centre of cur

m

vature of the path of the particle, and this

acceleration we know to be equal to

Combining (1) and (2) we have

v2

r