﻿based on Free Electrons. 1075 



and x is to be found from 



2 1 _ J _? LA 



The first approximation gives (l + # 2 j 3/2 = 8 or # = \/3 r 

 and for a second we have 



_2 L!/l-iU^(l-- V 



(l + ,r 2 ) 3 ' 2 4~~4.\ W 4\ 3V 73 / 



With . , 3 « d M 



., = v / 3 (l-^ = ^3-^(3^3-1), 



and 



4/. 



^ = 3 v /3-^(Sv / 3~l). 



Thus 



XT _ 3 r 2_ 1 1/ 1M_ ^ 3 ~1 , *Va ! \ 



M ~ l H(lH-^ 32 ""4 + 4V i ~Wj~ 4 + 4 V 3v/3J 



= 3v/ ^ 1 - ^(3x/3-l) = 1-04904- -00019 



with /a=1 : 1818 ; 



while ^ = '73205 — '00034. In each case the correction 

 for jjl is quite small. 



Table I. gives values of x and N for a vacant centre, 

 Table II. values of x' and N' for a positive centre, obtained 

 by direct solution of (6 a, b). 



For the problem of two ions placed axially two variables 

 are needed : x as before, and y where 2y is the ratio of the 

 mutual distance of the ions to the radius of the negative 

 ring. • If we take the radius of this ring as 1, the nature of 

 the change in the equations may be shown by writing them 

 for n = 4. They are 



1 4// 4y 1 



if + (a?+y 2 yP ~ (1 +pp ' 



2-.ia/2 , 2 + xs/2^ 2 



(t+^-V2) 3/2 (i+,^+V2) 3 ' 2 (i+y 2 ) 3/2 l a(n 



= •9571, ( 



x 2 (2x-\/2) x 2 (2x + \/'2) 2x __ 



(l + ^-V2f (l+r + V2f 2 (*" + irf 2 



-•9571 =N. 



