On separating Mercury into its Isotopic Forms. 819 



Let ni = number of molecules per c.c. of isotope A at 

 this depth, 

 w 2 = number of molecules per c.c. of isotope B at 



this depth, 

 mx=mass of molecules A, m 2 = mass of molecules B. 

 Then, since we have assumed that mercury is incompres- 

 sible, 



?ii + n 2 = 3i constant, 



dn x dn 2 _ n 

 dx dx 



Also the density in the \&yzx=.n x mi-Vn 2 m 2 . Consider an 

 A molecule- It is acted on by a vertical force downwards 

 due to its weight of m^g, and a vertical force upwards due to 



flotation of — — - — (nim 1 + n 2 m 2 )g since the volume of a 



molecule = 



n 1 + n 2 



Hence there will be a net downward force on an A mole- 

 cule of 



L Ui -\-n 2 j 



n 2 (m l —m 2 )g 



?ii + n 2 



If S is the area of the column, the total force on all the 

 A molecules in a layer of thickness dx due to gravitational 

 effects 



= — — — (mi- m 2 ) . S . dx. 

 ni + n 2 y l 2J 



Now we have assumed that the mercury is in equilibrium 

 hence there must be a force acting which will counterbalance 

 this effect. Since we have also assumed that the two isotopes 

 A and B are exactly alike except as regards mass, and that 

 the total number of molecules per c.c. is constant, it is plain 

 that the attractive force on a molecule due to the sur- 

 rounding molecules will on the average be zero, and no 

 re-distribution of the two classes of molecules will affect this 

 result. 



It would thus appear that the only force which could 

 counterbalance the tendency of the heavier molecules to 

 move downwards would be due to the increase of the osmotic 

 pressure of the latter owing to an increase in the number of 



