272 
PROFESSOR H. A. WILSON ON THE EFFECT OF HYDROGEN ON THE 
We have also /3 0 = 2‘8 x 10 7 , so that at 0 = 1600 we get 
(4-2 x 10 7 ) 3 x 10 11 = 4t rt 2 (2'9 x 10 14 ) 3 (2’8 x 1CT 7 ) 2 1600 x 273 x 10 14 , 
which gives t = 2’6 x 1CT 8 centim. 
The thickness of the double layer on platinum polarised by depositing hydrogen on 
it from dilute sulphuric acid has been calculated from the polarisation capacity by 
several observers who have found t about 2xl0 -8 centim. We should, of course, 
expect the thickness to be about of the order of the diameter of a molecule, which is 
believed to be about 2x 10 -8 centim. I think the fact that the theory proposed here 
leads to a value of t in agreement with that which might have been expected on other 
grounds shows that the theory is substantially correct. 
Since platinum is an octovalent element we may take the number of free electrons 
associated with each atom as 8. The charge p 0 carried by the free electrons in 
one cubic centimetre of platinum ought therefore to be 8sNe/M, where s denotes the 
density and M the atomic weight of the platinum. This gives 
p 0 = —8x21'5x2'9 x 10 14 -+ 195 = -2‘6x 10 14 , 
which agrees very well with the value —27 x 10 14 obtained from the negative leak. 
The agreement between these two values of p 0 shows that all the 8 electrons associated 
with each atom of platinum are included in n 0 , the number per cubic centimetre on 
which the negative leak depends. 
Substituting the values found for D, A and Q in the formula R = Q+-2d log (D/A), 
we get the following values of R :— 
Gas. 
Pressure. 
R. 
millims. 
Air 
— 
145000+ 9-680 
H, 
0-0013 
110000 + 19-150 
h 2 
0-112 
90000 + 25-140 
Ho 
133-0 
56000 + 36-180 
At d = 2000 these values are 164000, 148300, 140280 and 128360. The values of 
R at other pressures can be calculated, if desired, by means of the formula given in the 
last section. 
A difficulty in the electron theory of the conductivity of metals has been pointed 
out by J. J. Thomson in his book ‘ The Corpuscular Theory of Matter.’ If the 
energy required to raise the temperature of the electrons one degree is calculated, it 
comes out greater than the specific heat of the metal. We have 
P = -/3p 0 =-R,pod = %mn 0 U 2 , 
where U" is the mean value of the square of the velocity of agitation of the electrons. 
