Perkins — Rectification Effect in a Vacuum Tube. 489 



the previous half cycle, failed to give up their charges, and 

 are returning under the electrostatic attraction of the ring. 

 To account for their effect 



Let n = number of ions in the tube per cm. length 



Then K« 2 .— number of recombinations per second per cm. 



length 



" TKtt 2 = total number during half period 



„ . . 2md . 2md I 



but n, in lon» segment, == v -j- velocity = =• X — r^ 



& h l + d * l + d vV 



.'. total number of recombinations in the time T = 



TK4mT<f ^ 72 72 . _ 4TKm' 



KPd 2 where K, = 



v 2 v\i+dy i x v*\ 7 \i+dy 



and the number uncombined at the end B 



= K 3 (l-KAn(T-K/)^ (1) 



where K„ = =- and K„ 



Vv 3 l + d 



Similarly the total number arriving at A during the same 

 time is 



= K S (1-K/Vf) (T-K/P)£ (2) 



And the negative ions arriving at B during the second half 

 cycle are 



= K 3 (l-K 1 - ?7 j(T-^-|<* (3) 



And the negative ions arriving at A are 

 = K„(l-K 



(,-*,!£)(t-?J*), (4) 



The sum of (2) + (3) — { (1) + (4) \ gives us the number of 

 outstanding 4- ions at A at the end of a complete cycle. This 

 is equal to 



N= -^ \ld(l-d) [K/ (r-l)-TE/*(/-l)-E I K 2 rrf 2 0' 3 -l)l [ 



But K, is necessarily small with the short period used, so in 

 order to obtain an approximate form of the theoretical curve, 

 we may drop all terms multiplied by K,, and substituting for K 2 

 and K 3 we have 



TVT 2}U 0' —1 ) ,77/7 7\) 



N = ^—t — jr - ; ld{l—d) 



Vv(l + d) r ( v .' 



This curve is plotted in fig. 7, assuming Z + e?=20 and 

 letting the constant part of the expression equal unity. The 

 result satisfactorily accounts for the. two maxima, the polarity 



