560 Prof. J. W. Nicholson on the High-frequency 



Thus the valencies are not in accord with experience, and 

 if the electrons in the atoms are to be in one plane, we mnst 

 either abandon this method of calculating valency, — and on 

 any other method there would apparently be difficulty con- 

 cerning the valency of helium, — or give up van den Broek's 

 hypothesis, and suppose that the order in the Periodic Table 

 does not correspond in a very simple manner with the atomic 

 number, or charge of the central nucleus. 



But the possibility of non-coplanar rings must be considered ,. 

 and for this purpose we take an 



atom with nucleus Ae and four Q P 



electrons, arranged in two parallel 

 rings, as in the figure. The 



rings are of equal radius, so that I |_| ^IBd. 



the condition of angular mo- | n 6 



mentum is secured. The elec- 

 trons are at any instant at the 

 corners of a rectangle which can ~ e 



be inscribed to a circle. Let a be the radius of the circle 

 and a. the angle shown in the figure. 



If the electron P experiences no force along QP, 



?(*-*)«— 



4« 2 



sec- «, 



or 3 1 1 



cos a= 4N3I-i5 



if the atom is electrically neutral. 



For the steady rotation of the same electron, 



/ 1\ e 2 . e 2 



m(j» 2 a sin a = ( N — -j J — % sin a — — cosec 2 a, 



or or = -^-~ (N- 7 --j cosec 3 « ) = -i— (15 — cosec 3 a) 

 ma \ 4 .4 / 4ma 3 v 



where a is the radius of the atom. 



We find cosec 3 a = l-30923, and therefore 



o> 2 = (3-42269) -A 



v y ma 



gives the angular velocity of the system in steady motion. 



The angular momentum of each electron round the*axis of 

 the atom is 



ma 2 co sin 2 a = 7*/27r 



where h is Planck's constant, and the kinetic energy of the 

 configuration is 



W=2maW 



