THE LINEAR FORCE OF GROWING CRYSTALS 



3 I 7 



less stable than the other owing to its inferior size, or because it is 

 strained, or because it is an inherently instable form, or for any 

 other reason ? In the first place, it is plain that a solution to be 

 in equilibrium with a less stable crystal will require to be more 

 concentrated than that in equilibrium with the more stable one; 

 consequently in the foregoing case the solution will become super- 

 saturated with respect to the more stable crystal before it is super- 

 saturated with respect to the other, and the former will begin to 

 grow before the latter, which, indeed, will not grow at all (it may 

 even dissolve) unless the degree of supersaturation is greater than 

 the growth of the more stable crystal can keep balanced. 



TABLE II 



A Loaded (i Gm.) and an Unloaded Crystal in the Same Solution. 

 Temperature = i o° C. 



Analogous cases also arise when the crystals are of equal stability 

 but the solution is inhomogeneous, and there results from the action 

 of outside forces (imperfect stirring, thermal convection, gravitative 

 adjustments) a distribution of concentration such that one crystal 

 is in contact with solution of higher concentration and grows while 

 the other cannot. 



A familiar instance may be cited. If in an unstirred saturated 

 solution in a closed vessel two identical crystals are placed, of which 

 one is suspended a few millimeters above the other, the lower crystal 

 will grow while the upper one dissolves slowly. And similarly under 

 like conditions the bottom of a very large crystal will grow at the 

 expense of the top, and the prismatic lateral faces gradually acquire 

 the' contour of a flight of steps; the reason in either case is that 

 under the action of gravity the solution tends to become more 

 concentrated in its lower layers, and hence, since it is kept 



