1 00 Prof. Rutherford on Excited Radioactivity and 



The greatest distances * x^ x 2 passed over by the positive 

 carrier during two succeeding half alternations is thus 

 given by 



a? 3 = — ^H — - & • I, and # 2 = — — -, — L K . T. 

 a d 



Suppose the positive carriers are produced at a uniform 

 rate of g per second for unit distance between the plates. 

 The number of positive carriers which reach B during a half 

 alternation may be divided into two parts : 



(1) One half of those carriers which are produced within 

 the distance x of the plate B. This number is equal to 



(2) All the carriers which are left within the distance x x 

 from B at the end of the previous half alternation. The 

 number of these can be readily shown to be 



Now all the rest of the carriers produced between A and 

 B during a complete alternation will reach the other plate A 

 in the course of succeeding alternations, provided no appre- 

 ciable recombination takes place. This must obviously be 

 the case, since the positive carriers travel further in a half 

 alternation towards A than they return towards B during the 

 next half alternation. The carriers thus move backwards 

 and forwards in the changing electric field, but on the whole 

 move towards the plate A. 



The total number of positive carriers produced between the 

 plates during a complete alternation is 2dqT. The ratio p of 

 the number which reach B to the total number produced is 

 thus given by 



i^'o J- X\ Xy-\~X^ 



Substituting the value of x l and x 2 we obtain 



2(E +E,) cP 

 1V_ EoCEo-EO " T - p - 



In the experiments the values of E , E l5 d, and T were 

 varied, and the results obtained w T ere in general agreement 

 with the above equation. 



* In the equations that follow it is assumed that :i\ is less than the 

 distance between the plates. If :i\>>d the equations have to be modified. 



