222 Sir J. J. Thomson : Further 



Further consideration will show, however, that this view 

 is not tenable, at any rate in a large number of cases. We 

 find that the heads of the negative parabolas, like those o£ the 

 positive, are arranged in a vertical line and that the distance 

 of this line from the vertical line through the origin is the 

 same as for the corresponding distance for the positive 

 parabolas. It follows from this that the maximum value of 

 ■$>nv 2 /e is the same for the negative as for the positive 

 particles, this quantity probably being the potential fall, V, 

 between the negative glow and the cathode. Suppose now, 

 to take an example, that the negatively charged hydrogen 

 atom OA'ed its charge to hiving been in chemical combination 

 before it got through the cathode with an atom of carbon, 

 the molecule of the compound being positively charged in the 

 discharge-tube and acquiring a high velocity under the 

 electric field : after passing through the cathode the molecule 

 gets neutralised, and then dissociates into a negatively 

 charged hydrogen atom and a positively charged carbon one. 

 The kinetic energv acquired by the molecule OH, if it had 

 one charge of electricity, would be that corresponding to the 

 fall through the potential V, Since the mass of the carbon 

 atom is 12 times that of the hydrogen, the kinetic energy of 

 the hydrogen atom will be that due to the fall of unit charge 

 through a potential V/13 ; so that if this atom went through 

 the same electric field on its way to the camera as a 

 positively charged hydrogen atom which had been exposed 

 in its atomic state to the electric field in the discharge-tube, 

 it would suffer thirteen times the electrostatic deflexion of 

 the positive one ; the photographs show, however, that they 

 suffer the same deflexion. To explain the negative electrifi- 

 cation of particles in this way, it would be necessary to 

 suppose that the element with which they are combined 

 before they are dissociated is one of very much smaller atomic 

 weight. Hydrogen is the only element that would satisfy 

 this condition ; and this explanation would not account for 

 the negatively electrified particles of hydrogen itself, which 

 are a very marked feature in the discharge. Again, if the 

 negative electrification of oxygen were due to the dissociation 

 of a hydrogen compound of oxygen, then for every negative 

 oxygen atom we should have a positively electrified hydrogen 

 one ; so that if the negative oxygen line were very strong, we 

 should expect the positive hydrogen line to be strong also : 

 the positives produced in this way would, however, be moving 

 with much smaller velocities than those which had not been in 

 combination and would therefore experience a much greater 



