﻿910 Mr. C. G. Darwin : A Theory of the 



By division then : — 



Ax _ NVlog (1 + m/) -f(iv') 



(3) 



Ax' N/i Jog (1 + 20) — /'(«?) ' 

 Then Ax is the " air equivalent " of a film of thickness Ax' 

 of the substance. (8) shows why equivalence should depend 

 on the velocity of the a rays. By means of (8) we might 

 theoretically solve and find all the atomic constants in- 

 cluding the radius of the air atom. For if the equivalent of 

 Ax' is known at three different velocities, then two equations 

 can be formed involving a and a', and can he solved. (7) 

 then gives n and (8) n'. I have attempted such a solution 

 for gold, but the values of a and a which were obtained 

 were certainly too small to be admissible. The data used 

 were taken from some experiments by Taylor *, who records 

 the equivalence of a certain gold foil at various velocities. 

 He measured the velocity by finding the range which the 

 a rays still had to run after passing the foil. This method 

 of determining equivalence is open to the objection that it is 

 not quite certain what is the quantity that is measured. The 

 ionization falls off very rapidly at the end of the range and 

 so very accurate measurements of change of range can be 

 made. But the shape of the curve at the end depends 

 largely on the amount of the straggling among the a rays, 

 and this will vary systematically from one substance to 

 another. Hence measurements at the end of the range need 

 not give at all accurately the velocity at an earlier point, and 

 we must abandon the hope of a complete direct solution. 



By supposing a known for air we need only use two 

 measurements, and these may be taken fairly close together 

 so as to minimize the effect of change of shape o£ the 

 ionization curve. If A^, Ax 2 are the distances in air 

 equivalent to Ax' at velocities i\, v 2 , then from (8) w T e have 

 Ax, = log (1 -f V) -/«) log (1 + w 2 ) -f(w^) 

 Ax 2 log (1 + w 1 )-f(w 1 ) log (1 + w 2 ')-f(w 2 y ' \ } 

 To solve this for a' we ought, strictly speaking, to determine 

 the velocities i\. v 2 from the recorded ranges by the velocity 

 curve itself, but it is easy to see that the empirical formula 

 v 3 = V 3 (l — a.'/R) is quite accurate enough. The solution of 

 (9) determines w% and w 2 . (8) then gives n' . The solu- 

 tions all depend on the value assumed for a the radius of 

 the air atom. Table I. gives the solutions for the sub- 

 stances which Taylor f examined. The numbers in each 

 column all depend on the value taken for the first row. For 



* T. S. Taylor, Phil. Mag. vol. xviii. p. 604 (1909). The numbers 

 used were extracted from Tables I. and II. 

 t Taylor, he. cit. 



