Number  of  Corpuscles  in  an  Atom.  773 
Bat  i£  n  is  the  number  of  molecules  per  c.  c., 
ne  =  '4 ; 
hence  N  =  25w. 
From  this  we  deduce  that  there  are  25  corpuscles  in  each 
molecule  o£  air,  and  this  indicates  that  the  number  o£  cor- 
puscles in  the  atom  is  equal  to  the  atomic  weight  ;  for  the 
scattering  by  air  is  very  nearly  the  same  as  that  by  nitrogen, 
and  25,  the  number  o£  corpuscles  in  the  molecule  deduced 
from  Barkla's  experiment,  is  near  to  28,  the  number  in  each 
molecule  if  the  number  in  the  atom  were  equal  to  the  atomic 
weight. 
?>rd  Method.  Absorption  of  j3  Rays. — We  regard  the 
absorption  of  the  {3  rays  as  due  to  the  effect  of  the  collisions 
between  these  rays  and  the  corpuscles  which  they  meet  with 
in  their  path  through  the  absorbing  substance.  If  X  is  the 
coefficient  of  absorption,  it  is  shown  in  the  latter  part  of  the 
paper  that 
where  N  is  the  number  of  corpuscles  per  cubic  centimetre, 
V  the  velocity  of  the  j3  particles,  Y0  the  velocity  of  light, 
e  the  charge  on  a  corpuscle  in  electromagnetic  measure,  m  the 
mass  of  a  corpuscle,  and  a  a  length  comparable  with  the 
distance  between  the  corpuscles  in  an  atom. 
If  8  is  the  density  of  the  absorbing  substance,  M  the  mass 
of  an  atom,  n  the  number  of  corpuscles  in  the  atom,  we  have 
-M=8;  so  that 
n 
S2  en      Vo4,      /l  aV2  m 
,     *■  en       V04,      /l»V!n       A 
Now  \/S  is  approximately  constant  whatever  be  the  nature 
of  the  absorbing  substance;  hence  since  the  logarithmic  term 
only  varies  slowly,  we  conclude  that  n  must  be  proportional 
to  M,  i.  e.  that  the  number  of  corpuscles  in  an  atom  is  pro- 
portional to  the  atomic  weight.  To  find  the  number  of 
corpuscles  in  an  atom,  let  us  apply  the  formula  to  the  case  of 
the  j3  particles  from  uranium,  for  which,  as  Becquerel  has 
shown,  V  =  16  X  1010;  and  Rutherford  finds  that  for  copper  and 
silver  X/S=7.  Putting*/™  =  1-7  X  1010,  *=10-20,V0  =  3  x  1010, 
we  get 
™?_=        l-4xl04 
M 
