270 PROFESSOR H. A. WILSON ON THE EFFECT OF HYDROGEN ON THE 
Integrating this gives 
F F 0 -a+(F 0 + a)^ v , 
— = ==-hpr--(- r , where a = — incr. 
a F 0 —« — (F 0 + a) e lp 
The equation 8np 0 /3 + F 0 2 = a 2 gives F 0 — a — — 8np 0 fi/(F 0 + a) = p 0 /3/<x very nearly, 
because F 0 and a must be nearly equal. Hence 
— F/inCT = (1 
8lT(T~ 
^in<rx/p\ If l+ 
H \ 8 tto"' 
>4tt ox/Pj 
This gives 
c? 
Y = — | F clx = —(3 log (1 + 
477 cr 2 
Pofi 
477cr 2 
-OiTrcrt/fZ 
+ 
Po(3 
1677 cr 2 
_ g + 4v r<rt/p 
/3 log 1 + + 
Po /3 I6770" 
In this expression 
PoP e w/p 
16 77 o -2 
is quite negligible for platinum, and p„/3/l6n(r 2 is small compared with 477 a 2 /p 0 (3, hence 
V = -p log + e- 4 "®) - p log (l + ^4) . 
\ poP / \ p»P I 
This is almost the same as 
Y = -0 log e~ in<Tt/p j 1 - ^- 2 1 
or 
V = inert + p 0 fi 2 finer 2 . 
Let incite = w, and then 
Ye = w + inp 0 eH 2 /3 2 w~ 2 . 
Now Ye is the work which an electron must do before it can escape, so that it N 
denotes the number of electrons in one gram-molecule, R = NYe/J, where J is the 
mechanical equivalent of heat. Let /3 = /3 0 6, and we get 
j, _ N™ , 477 N p»eH 2 p 0 2 6 2 
K " T + ' 
If 6 increases one degree, R therefore increases by 877 N p o e s t 2 fi o 2 0/Jw 2 . 
In the last section we got R = Q + 20 log (L)/A), according to which R increases 
uniformly with 6. But the experiments only extend over a small range of temperature, 
so that they are not sufficient to distinguish between the equations R = Q + kd and 
R = Q + k'0 2 , where k and k! are constants. The equation R = Q + 20 log (D/A) 
depends on the assumption that Q, D, and A are all independent of the temperature. 
