﻿902 
  Prof. 
  0. 
  W. 
  Richardson 
  on 
  the 
  Kinetic 
  

  

  § 
  5. 
  Second 
  Method. 
  

  

  (a) 
  Negative 
  ion*. 
  — 
  This 
  method 
  is 
  not 
  independent 
  of 
  the 
  

   first. 
  It 
  rests 
  directly 
  on 
  equation 
  (8) 
  and 
  is 
  really 
  only 
  a 
  

   more 
  accurate 
  method 
  of 
  testing 
  the 
  applicability 
  of 
  equa- 
  

   tion 
  (7). 
  According 
  to 
  equation 
  (8) 
  the 
  curve 
  obtained 
  on 
  

   plotting 
  log 
  i 
  against 
  ,r- 
  should 
  be 
  a 
  straight 
  line. 
  The 
  

   values 
  of 
  log 
  10 
  i 
  and 
  of 
  .r 
  have 
  been 
  calculated 
  for 
  the 
  points 
  

   shown 
  in 
  fig. 
  3 
  and 
  are 
  plotted 
  against 
  each 
  other 
  in 
  tig. 
  5. 
  

   It 
  will 
  be 
  seen 
  that 
  they 
  all 
  lie 
  very 
  near 
  to 
  the 
  straight 
  line 
  

   drawn 
  on 
  the 
  diagram. 
  There 
  is 
  a 
  tendency 
  for 
  the 
  points 
  

   corresponding 
  to 
  largo 
  values 
  of 
  a- 
  to 
  lie 
  above 
  the 
  line. 
  

   This 
  tendency 
  is 
  much 
  more 
  marked 
  in 
  the 
  case 
  of 
  the 
  positive 
  

   ion-, 
  and 
  the 
  probable 
  eause 
  of 
  it 
  will 
  be 
  discussed 
  when 
  

   their 
  behaviour 
  is 
  considered. 
  

  

  If 
  ij, 
  ./■]. 
  and 
  i. 
  : 
  . 
  ./'.. 
  are 
  any 
  two 
  pairs 
  of 
  corresponding- 
  

   values 
  of 
  i 
  and 
  ./■. 
  it 
  follows 
  from 
  equation 
  (8) 
  that 
  

  

  110 
  = 
  

  

  V 
  

  

  ne 
  

  

  

  9-2z* 
  1 
  

  

  og 
  /. 
  

   Fh 
  ' 
  

  

  log/, 
  

  

  (10) 
  

  

  >, 
  

  

  •4 
  

  

  \ 
  

  

  

  

  

  

  \ 
  

  

  \ 
  

  

  

  

  

  

  x\ 
  

  

  \ 
  

  

  

  

  

  

  >K 
  

  

  

  

  

  

  

  \ 
  x 
  

  

  

  

  

  

  

  \ 
  x 
  

  

  

  

  

  

  \ 
  

  

  20 
  

   /■008x/0~ 
  

  

  30 
  

  

  40 
  

  

  50 
  

  

  Referring 
  to 
  fig. 
  5, 
  we 
  find 
  Av- 
  t 
  r 
  2 
  2 
  = 
  4'48 
  x 
  10 
  cm. 
  2 
  and 
  

   log 
  10 
  i 
  2 
  — 
  log 
  10 
  2 
  X 
  = 
  1*1. 
  Substituting 
  these 
  and 
  0= 
  1500° 
  abs., 
  

  

  