﻿210 Mr. L. St. C. Broughall on Theoretical 



Let us now consider the energy of an electron in the 

 outer shell when the diameter of the shell = 2^. 



Let the angular velocities about the axes YY' and ZZ' 



=w 1 . 



Let the angular velocity about axis XX / = W 3 . 



Since the diameter of the shell is equal to ' 2c 1? ' it follows 

 that the radius of the electron orbit =Sj where Sx = C! v 6/3. 



Using the above notation and remembering farther that 

 the kinetic energy of a particle describing a circular path 

 of radius ' R ' with an angular velocity 'W is equal to 

 ^MR 2 W 2 , where M is the mass of the particle, we find that 

 the kinetic energy due to rotation about the axis XX' 



= ±mS{ 2 W 



and the kinetic energy due to rotation about the axis YY' 

 plus that due to rotation about ZZ' 



= mS 1 2 W 1 2 since Wi = W 2 . 



Therefore the total kinetic energy of the particle is 

 equal to 



>S 1 2 {2W 1 2 + W 3 2 }. 



In the case of an electron in the inner shell where the 

 radius of the orbit =Rj, we have the kinetic energy of 

 the electron due to its rotation about the axes YY' and 

 ZZ'=E X where 



E 1 =wR 1 2 W 1 2 . 



Now let the inner shell expand until it occupies the space 

 previously occupied by the outer shell — that is to say, until 

 Ri = Ci, then kinetic energy in new orbit = E 2 » 



Where E 2 = ?nC 1 2 W 11 2 , 'W n ' being the new angular 

 velocity about the axes YY' and ZZ', the change of energy, 

 = E 1 -E» 



= m(R 1 2 W 1 2 -C 1 2 W 11 2 ). 



Meanwhile the outer electrons have moved further away 

 from the nucleus, and now the outer shell has a radius = G 2 , 

 and the orbit of the electrons is now S 2 . 



Therefore the energy in the new position is equal to 



*mS 1 *[2W 11 J + W 1 ,»]. 

 Where ' Wi 3 ' is the new angular velocity about the axis 



