lONIZATION BY NEGATIVE IONS 23 



or when Xi/ exceeds a certain potential V. The value of 

 a corresponding to X represents, therefore, the number 

 of free paths per centimetre which exceed the value 

 #=V/X. It is easy to express a as a function of X 

 and the mean free path. Let N be the total number of 

 collisions that an ion makes with molecules in travelling 

 through a distance of 1 centimetre in a gas at 1 milli- 

 metre pressure, so that ^ is the mean free path. Let n 



be the number of paths per centimetre which exceed the 

 distance y. In going through the element dy of these paths 

 the number of collisions will be proportional to ndy, 

 hence dn=kndy where A; is a constant. On integrating, 



fry 



this equation gives n=ct J , the constant c being 

 equal to N, which represents the number of paths which 

 exceed the value y=0. The constant k may be found 

 in terms of the mean free path, for since ydn is the 

 sum of the paths whose lengths are between y and 

 y+dy, the sum of all the paths in 1 centimetre is 



- [ydn= f 



JN J 



which gives fc=N. 

 Hence the number of paths per centimetre which exceed 



Kfy 



the distance y is Ne % which is therefore the value of 



y 

 a when y=y 



The connection, therefore, between a and X when 



NV/ a NVp\ 



p=l is a=Ne liar in general -=Ne~ x j where N, 



the maximum value of a, is the number of collisions 

 that an ion makes in travelling a distance of 1 



