7f>6 Sir J. J. Thomson on 



deflexion of 3 5 millimetres. This gives 



•41=— xl-OlxlO 4 , 

 mo 



e 2 x 10 10 

 •35= — ., x -.,- — x'o'2-2; 

 mv 3 



or r = 2-Gb" x 10 s , 



f/w=ri6xi«*. 



The value of i-> for this spot depends upon the pressure in 

 the tube. The spot d had the same electrostatic deflexion as 

 e, so that the values of e/m for the spots d and e will be as 

 ili • squares of the magnetic deflexions. 



The corresponding magnetic deflexions for d and e and 

 the square of their ratio is given in the following table : 



Deflexion of e. Deflexion of d Square of ratio. 



3-5 2o 1-96 



5-3 3-7 206 



6-8 4-7 2-09 



iW> 4 2-25 



7-u 5 l-in; 



Thus the value of e/m for d is half that for e ; hence if the 

 charges are the same, the mass of the carriers producing the 

 spot d is twice that of those producing <?, hence we ascribe d 

 to the hydrogen molecule. 



The spot c i.> the helium spot, and the value of e/m as I 

 showed in my earlier paper is \ that of the spot e. 



AVe can compare the mass of the carriers for the spot h 

 with those of d by comparing the magnetic deflexions, the 

 following are corresponding values : 



Spot d. 



Spot b. 



Square of ratio. 



90 



33 



74 



47 



1-8 



G-8 



9*2 



35 



G9 



9-5 



2-4 



7-8 



8-0 



3 



7-1 



Thus if the charges are the same, the mass of the carriers 

 of b is about seven times that of d ; if, as we supposed, the 

 carrier of d is the hydrogen molecule, then the carrier of b 

 will be an atom either of nitrogen or oxygen. I am in- 

 clined to think that this is a double spot and will be resolved 

 by the application of stronger magnetic fields. 



When the air in the tube was replaced by CO there was a 

 spot in approximately the same position as b } on increasing 



