758 Sir J. J. Thomson on 



line, the most deflected portion of which, B, will be due to 

 particles which were charged when they passed through the 

 cathode. Now let the magnet next the screen be excited, 

 the appearance on the screen is as in fig. 6 ; those rays 



Fig. 6. 



which were charged when they passed through the first field 

 and were deflected by it are still further deflected along the 

 line BC, but in addition to this the stream of neutral 

 particles as it passed between the two magnets has pro- 

 duced new secondary rays and these are deflected along OA. 

 Thus all the rays which were charged when they passed 

 through the cathode are found on the line BC, while OA 

 consists exclusively of those which have been produced or 

 which have acquired their charge after they left the first 

 magnet. If now we put on the electric field in the system 

 next the screen, we find that, at low pressures, the portion 

 B( \ which consists of rays chaiged when they parsed through 

 the cathode, is broken up into the two bands of which we 

 have been speaking, and which were seen when only one 

 system of electric and magnetic forces was used. On the 

 other hand, the band OA, due to ravs which were produced 

 nearer the screen than the first magnet, does not bifurcate 

 but consists of only one branch for which the maximum 

 value of e/m = 10 4 . The appearance of the phosphorescence 

 is shown in fig. 7. 



Fig. 7. 





1 have hitherto spoken only of two bands, but when the 

 pressure is low there seem with these large tubes to be para- 

 bolic bands corresponding to every gas in the tube. By using 

 very sensitive screens I have been able to detect the bands 

 corresponding to hydrogen, helium, carbon, air, oxygen, 

 neon, and mercury vapour. The appearance on the screen 

 when there are several gases in the tube is almost like a 

 spectrum, and I think this effect may furnish a valuable 

 means of analysing the gases in the tube and determining 

 their atomic weights. There is a band on the screen corre- 

 sponding to a value of e/m about \ X 10 4 , clue to the air in 



