of Electricity through Gdses exposed to Rdntgen Rays. 395 



this property is produced by the Rontgen rays. If we regard 

 the gas as an electrolyte, then the passage of a quantity e of 

 electricity will destroy e/e of the conducting particles, where 

 e is the charge carried by one of these particles. Let n be the 

 number of conducting particles in unit volume of the gas, 

 q the r.ite at which these are produced by the rays, an 2 the 

 rate at which these disappear independently of the passage of 

 the current, i the current through unit area of the gas, I the 

 distance between the electrodes. Then we have 



dn 2 i /1N 



w=i-? n -h ; W 



so that when the state of the gas is steady, 



= q-«v?-j e (2) 



When the current is small this equation gives 



n 2 = q/a ; 



and as the number of conducting particles is independent of 

 the current, the current will be proportional to the E.M.F. 

 This corresponds to the straight part of the curve. 



In the general case the current is proportional to the pro- 

 duct of >i, the number of conducting molecules, and the 

 potential gradient, If E is the difference of potential between 

 the plates, U the sum of the velocities of the positively and 

 negatively electrified particles when the potential gradient is 

 unity, we have 



t = neUE// or n= TTr v 

 eUE 



Substituting this value of n in equation (2), we get 



°^-7WW^h (3) 



We see from this that c approaches the limit qel. Thus the 

 limiting current is proportional to the distance between the 

 electrodes ; so that when we approach saturation the current 

 will increase as the distance between the electrodes increases, 

 and we get what is at first sight the paradoxical result that a 

 thin layer of air offers a greater resistance to the passage of a 

 current than a thicker one. This is, however, easily accounted 

 for if we remember that the current destroys the conducting- 

 power, and that as in a thicker layer there are more con- 

 ducting particles than in a thinner one the current required 

 to destroy them all will be greater. 



