20 



produced is proportional to the energy expended, we may say 

 that 8e = kSr/e, where e is the energy possessed by the particle 

 and A; is a constant. Hence e is proportional to V {r + d), where 

 d is also a constant. In the paper referred to I have shown that 

 d = 1*33 cm. Thus the ionisation produced by the a particle in 

 traversing the last r cm. of its course may be put equal to 



l(V{r + d) - ^^r)'^ 



where I is a constant and <^=1'33. 



Hence the whole ionisation due to such particles as have 

 ranges between r and r + Sr is equal to 



-\ 1 \{V (r + d) - Vr) Sr 



This must be integrated between the limits oi R — D and 0. 



The final result is 



4si 2 A 2 ^ 



= -(R - D + df- -d^ -RVd-Dx^{R + d-D) 



Nln^ 3 " 3 



^2 ^^R { ^^{R + d) + V{R + d-D)\ 



+ 2Dx/d-^ log. 



V {R+d) ^/^ 1^ ^ ^^_^^^ ^ ^/ ^^1 



The value of I, the current when the material is un- 

 covered, is obtained by putting D — 0. This gives 



4s/ 2 A 2 ^ 



=-(R-^d)^ --d'' - RVd 



Nln, 3' ^3 



The value of ijl is therefore no longer a function of DIR 

 merely, as in the simpler formula found for the case when the 

 variation of ionisation with speed is neglected. Consequently 

 the curves for various values of R are not all of the same form. 

 It appears on trial, however, and it might reasonably have 

 been expected, that the form of the curve is altered but little, 

 even when R is altered considerably. Curve A was plotted 

 for the case 7? = 3, and serves very well in the cases of uran- 

 ium and thorium. It lies very close, as can be found on trial, 

 to the curve obtained from the simpler formula of the case 

 when the variation of ionisation with velocity is neglected. 



* Even if there be errors in the theory which leads to this formula, 

 the present argument is not injured, for the formula correctly represents 

 the actual facts. 



