782 Sir J. J. Thomson on the 



lithium, sodium, potassium, each 1 ; beryllium, magnesium, 

 calcium, and strontium, 2 ; boron, aluminium, 3 ; carbon 

 and silicon, 4; nitrogen, phosphorus, and arsenic, 5 ; oxygen, 

 sulphur, and selenium, 6 ; fluorine, chlorine, bromine, and 

 iodine, 7. 



Now let us consider how the corpuscles in these atoms 

 can be fixed. They are not fixed when the atom is by itself. 

 In this case the tube of force starting from a corpuscle in 

 the atom, returns to a positive charge in the same atom and 

 possesses considerable mobility, as the corpuscle at one end of 

 it can move freely about in the atom. The corpuscle will 

 not be fixed unless the tube of force at its end is anchored to 

 something not in the atom, i. e. it must end on another atom. 

 Thus if there are n free corpuscles in the atom, to fix these 

 and thus saturate the atom, the n tubes of force which start 

 from the n corpuscles must all end on other atoms and not 

 return to the original atom. Thus to ensure saturation from 

 every free corpuscle in an atom, a tube of force must pass 

 out of that atom and end on some other, and this must hold 

 for every atom in the molecule. When the atoms are 

 electrically neutral, i. e. have no excess of positive over 

 negative charge or vice-versa, for each tube of force which 

 passes out of an atom, another must come in: and thus each 

 atom containing n corpuscles must be the origin of n tubes 

 going to other atoms and also the termination of n tubes 

 coining from other atoms. 



Thus consider two atoms A and B each of which contains 

 one free corpuscle, denote these corpuscles by ot and ft re- 

 spectively. Then a and ft will be fixed if a tube of force 

 goes from a in A to the positive core of B and another from /5 

 to the positive core of A : in this case the molecule AB will 

 be saturated. If, however, B had contained two mobile 

 corpuscles /3 and 7, then, though a might be fixed by a tube 

 going from it to the core of B, yet since only one tube goes 

 out of A only one can come in ; thus only the tube from one 

 of the corpuscles in B can go to A ; the tube from the other 

 must return to B, and thus this corpuscle will not be fixed and 

 B will not be saturated. If, however, there is a second atom 

 of A in the neighbourhood or any other atom C which contains 

 only one free corpuscle, then this tube of force from 7 can go 

 to the positive core of C while the tube from the corpuscle 

 in comes to the core of B. With this arrangement every 

 mobile corpuscle in the system is anchored by a tube of force 

 to some other atom, and thus deprived of mobility : hence 

 the system will be saturated. 



