164 THE QUANTUM [xn. 3 
elementary magnet of moment M, so that the strength of the 
magnetic field is M//¿a 3 . Thus at the point D, B = /¿H = M/a 3 . 
Hence 
E M 
Ka 2 a* V> 
or since v — aco, 
E M 
Ka 2 a 2 ™' 
Therefore we have 
E 
k 0J ‘ 
Thus the equivalent charge at O is given by 
E = KMco 12:6 
This relation may now be used in expressing the angular 
momentum of the rotating quantum tube. 
Angular Momentum = . - . co 
n e 
E 
e 
12:7 
If we assume that the charge E = Ze, expressing the charge in 
terms of the electron charge e, then the angular momentum 
Zx* 
n 
12:8 
The kinetic energy of the tube takes the simple form 
iE h 
. — . co 
2 e n 
= -hv, 
e 
Zhv 
since co = 2nv, 
. . . 12:9 
If, in the above results we take Z = 1, so that the equivalent 
charge equals the electron charge e, we find for the angular 
momentum of the rotating tube h/n, and for the kinetic energy 
hv. It is important to observe that these values are exactly 
double those obtained by adopting Nicholson’s quantum condition 
which postulates a natural unit of angular momentum of magni 
tude h/2n, corresponding to an amount of kinetic energy \hv. 
We have, in this case, an amount of energy corresponding to 
Planck’s quantum hv. We may compare the results we have 
obtained with those in the case of the simple Bohr atom, where 
the kinetic energy of the circulating electron is \hv and the