PHYSICS: F. C. BLAKE 
239 
n , sin 0 . »s 
D' = log, 
M 2sin0cos0£-J^ (l-e 2sm7co*o) ^J. ( 2 ) 
and for the X-rays contained in the parallelogram BK 00 00 OJB, 
_ I y e 2sin0cos0_ e 2 sin ^ cos ^ ^xj 
sin 0 . ps (3) 
log. 
/ - ^ \ ri c m 
2 Sill 0 COS 0 \1 — 6 2sin0cos0y _ | g sin 0 cos 0 . ^ 
Now Blake and Duane, in determining the maximum positions of the ioni- 
zation chamber corresponding to a given value of 0 had in one case made 0 
equal to 4°18', which corresponded to a wave-length X equal to 0.454 A, Us- 
ing Duane's formula for the absorption, viz., m/pai= 14.9 X 3 we get m/pai = 
1.394. Applying Bragg's formula for the atomic absorption coefficient, a, 
say, we have 
a == [ioi/£ = kN 4 , 
where k is a constant and cc the atomic weight. Thus we get 
a M = 37.79 a~\ A Ca = 211.7 <z c = 1-71 ^ a 0 = 16.26 yl" 1 , 
where is the number of molecules in a gram molecule, viz. 6.062 X 10 28 . 
If now we assume the molecular absorption, m say, to be equal to the sum of 
the atomic absorptions, we have 
w Ca co 3 = «ca + #c +' 3a 0 = 229.67 A- 1 , 
whence /x: p Ca co 3 = 229.67: 100.07 = 2.295; and if we take p for calcite to be 
2.712, p. comes but equal to 6.22. 
Now W. L. Bragg has shown 2 that for calcite there is only half a molecule 
to each elementary cell. In other words, the value of p we require is 6.22: V2 
= 4.94. 
Performing the integrations indicated in (2) and (3) we get 
_ sin 0 ^ jms 
M 2 sin 0 cos 0 J"— |ps + sin 0 cos 0 (4c 2 sin^ cose - e *too™~0 - 3) j J* 
