the Natural Ionization in a Closed Vessel. 221 



electroscope, and that they are absorbed according to an 

 exponential law with the distance traversed. 



Let dl be the ionization produced in a layer of thickness 

 dx at a distance x from the side of the box, and let I dx be 

 that produced in the layer adjacent to the side. 



dl = l e~ x dx, 



and the whole ionization due to the secondary rays, if the 

 length of the box is a, is 



Assuming the rays to be of the /3 ray type, X will be pro- 

 portional to the density of the gas, i. e. to the pressure, and 

 so will I , since I dx is the ionization in a thin layer of 

 constant thickness dx. 



Therefore the total ionization due to the secondary rays at 

 different pressures is given by 



B(i-<r AaF ) 5 



where B is a constant, p the pressure, and X the coefficient 

 of absorption at atmospheric pressure. 



Since the 7 rays are not appreciably absorbed in their 

 passage across the vessel, the ionization due to them will be 

 given by a term of the form Ap, where A is a constant, 

 having the value of the ionization due to the 7 rays alone at 

 atmospheric pressure. 



We have then, if I is the whole ionization in the 

 electroscope, I = Ap + B (1 - e~ kpa ). 



Now, Eve * gives the value of X for aluminium for the 

 secondary rays from aluminium as 24. Assuming the density 

 law and taking the density of aluminium as 2'G, and that of 

 air *00129, we obtain for our value of X *0119. 



It will be noticed that for high pressures the equation 

 becomes I = Ap + B. 



Therefore the intercept of this straight line with the 

 ionization axis gives the value of B, and the tangent of the 

 angle made by this straight line with the pressure-axis gives 

 the value of A. 



Producing the final portion of the curve backwards, we get 

 then 



A = -030, B = l-78. 



The theoretical curve is drawn in the figure and the 



* Eve, Phil. Mag\ Dec. 1904. 



