DISCHARGE OF NEGATIVE ELECTRICITY FROM HOT PLATINUM. 
251 
The following table is calculated from the results given in the previous paper :— 
V - 
A. 
Ap 073 . 
millims. 
0-0013 
10° 
7800 
0-112 
5 x 10 4 
10100 
133 
2 x 10 2 
3600 
Thus, while p is increased 10 5 times, A p 0 " 13 only varies by a factor of 3. Only the 
order of magnitude of A can be obtained from the experimental results, so that it 
appears that the observed variation of A with p is consistent with the equation 
Ap 0 ’ 73 = constant. 
— (Q —P) 
The equations A = Kp~ c and Q = P —2a logp give A = Ke 2a 
Putting P = 79000, a = 2400, K = 9000, and c — + 0'73, and calculating Q from 
the values found for A, the following results are obtained :— 
Q = 6580 log A+19100. 
Gas. 
Pressure. 
A. 
Q 
(calculated). 
Q 
(found). 
Air 
millims. 
1 • 14 x 10 8 
142000 
145000 
Air 
— 
7 x 10 7 
138000 
131000 
h 2 
0-0013 
10 6 
110000 
110000 
h 2 
0-112 
5 x 10 4 
90300 
90000 
h 2 
133 
2 x 10 2 
54000 
56000 
It appears, therefore, that this relation is satisfied by the values of A and Q for a 
wire in air as well as by those for a new wire in hydrogen. 
If we put p = 0 in the equations Q = P —2a logy> and A = K p~ c , we get Q = oo 
and A = co. These equations therefore require modifying to enable them to represent 
all the values of A and Q. If we suppose A = A 0 /(l +ap c ), where a is a constant, 
then, when ap c is large compared with unity, this formula will agree with A = Kp -C , 
and when p = 0, it gives A = A 0 . 
We have K = A 0 /(l + a), which gives a = l - 27 x 10 4 . When p = O’OOl, ap c = 100, 
so that even at this pressure the present formula differs from A = K p~ c only by one 
per cent. 
Since A = A 0 e 2a ", we get Q = Q 0 — 2ac -1 log (1 + ap c ). This formula gives 
Q = Q 0 when p = 0, and for all measurable values of p does not differ appreciably 
from Q = Q 0 — 2a logy» —2ac -1 log a. When p = 1, Q = P, so that this is the same as 
2 k 2 
