Cathode and Anode in a Geissler Tube. 459 



per second; e is the charge per ion (4*65 x 10" 10 e.s.). This 

 gives, for i = 0*00197 amp. (the current producing an excess 

 temperature of 13 0, 5 C.) 



71 = 6-36 xlO 15 , 



so that the mean energy given to the cathode per ion 



= 6-5xl0- 10 erg. 



Now the cathode fall of potential is insufficient to supply 

 this energy. The cathode fall for copper in air is 252 volts*, 

 which gives the ion passing through it only 



3-9 x 10- 10 erg. 



This leads to the conclusion that all the positive ions do not 

 ionize, and that there is an excess of positive over negative 

 carriers taking part in the discharge. The electron leaving 

 the cathode gains energy sufficient to ionize many times in 

 passing through the cathode dark space, and the positive 

 ions produced inside the cathode dark space will give up 

 their energy wholly as heat. Recent experiments by Sir 

 J. J. Thomson | indicate that ionization takes place inside 

 the cathode dark space. Farther, owing to collision, some 

 of the positive ions originating outside the cathode dark 

 space may lose energy by collision and also cease to be 

 ionizers. 



Now if n positive ions reach the cathode per second and 

 ionize, and m per second give up their energy to the cathode 

 as heat, then 



e(2n-\-m) = i 

 2n + m = 12-72 xlO 15 . 



A minimum value for m can be found if we further 

 assume that the n ions which ionize do not impart any heat 

 energy to the cathode. If the average energy of the m 

 carriers is E, then 



??2E = 4-lxl0 6 ergs. 



Now E is less than the energy gained by falling through 

 the cathode fall (i. e. 252 volts), so that 



m > 1-04 xlO 16 

 and 



™<0'12xl0 16 . 



Thus only -j 1 -^ or T ] ^ of the current at the surface of the 

 cathode is carried by electrons. 



* Rottg-ardt, Ann. d. Phys. xxxiii. p. 1193 (1910). 

 t Sir J. J. Thomson. Phil Mag-. August 1912, p. 225. 



