DISCHARGE OF NEGATIVE ELECTRICITY FROM HOT PLATINUM. 271 
The experiments show that Q and A do not vary much with the temperature, but it 
is quite possible that they really vary to some extent, for they cannot be found very 
exactly. Consequently I think it is justifiable to regard the two'formulae for It as 
equivalent over the range of temperature of the experiments, and so to put 
2 log (D/A) = 87rN p () eH 2 /3 () 2 6/Jw 2 . 
Now Nw/J = 47rcr£eN/J is the value of ft when 0 = 0, so it will be approximately 
equal to Q. 
Hence we get w = QJ/N and Q 2 log (D/A) = 4:irN 3 e 3 p o t 2 (3 o 2 0/J 3 . 
If we suppose that the hydrogen in the layer consists of positively charged atoms, 
it will increase cr without altering t. If, then, a value of D can be found which will 
make Q 2 log (D/A) a constant, this may be taken to be the true value of D. Let 
Q 2 log (D/A) = C, and then D and C can be found from two values of A and the 
corresponding values of Q. Taking the values 
Q. 
A. 
145000 
1•14 x10 8 
90000 
5x 10 4 
gives D = l'44x 10 10 and C = 10‘2x 10 10 . Using these values of C and D, we obtain 
the following values of Q 2 log D/A :— 
Q. 
A. 
Q 2 log D/A. 
145000 
1•14 x10 8 
10-2 x 10 10 
131000 
6-9 x10 7 
9-0 x 10 10 
110000 
10 s 
11-6 x 10 10 
90000 
5 x 10 4 
10-2 x 10 1 " 
56000 
2 x 10 2 
5 • 7 x 10 10 
The five values of C agree as well as could be expected. We have, therefore, 
10 11 = Av~N 3 e% 2 t 2 Po d/J 3 . 
According to Ktchardson’s theory we have 
p 0 = = D x 3 x 10 9 (irNm/Jf, 
where m is the mass of one electron. This gives, taking c/m = —-LSxlO 7 in E.M. 
units and Ne = — 2 , 9xl0 lt in E.S. units, p 0 — — 273 xlO 14 electrostatic units per 
cubic centimetre. 
