Voiv. 6, 1920 
PHYSICS: E. H. HALL 
615 
of the free electrons produces. The term {dP^ dT) has a correspond- 
ing meaning with regard to the associated electrons. 
The relation between Pj and Pa being expressed by equation (2), we 
must suppose at least one of the ratios {dPf dT) and {dP^ -^ dT) to 
have a finite value. I shall assume the value of {dPa dT) to be negligi- 
ble, and therefore we shall have, from equation (2), 
%f = ^.{s - 2.5). (5) 
di e 
Since 
p = 11 RT, (6) 
and, according to a previous assumption, 
n = zT\ , (7) 
where z and q are constants, we have 
^g, = + + g). (8) 
Substituting from (5) and (S) in (4) and taking {dPa dT) as zero, we 
get 
^/7(^ + .-l-5)+f_)^-...g. (9) 
According to my conception of thermal conduction, we have only to 
multiply either side of this equation by X, the amount of energy required 
to free {\ e) electrons within the metal, to get the value of B, the thermal 
conductivity of the metal. That is, we can write 
dP 
B = — X^rt— . ergs/cm., sec, deg. C. (10) 
We must now deal with the individual factors in the second member 
of this equation. From (9), remembering that k = ka -\- kf, we get 
= - - T-is + q - l-o), (11) 
di e k 
and X is (1 e) times the X' of equation (1). 
Hence we have 
Of course, since k is known, ka is known if kf'is known. Substantially, 
my method of procedure with each metal is to find by trial values of Xo 
s, {kf k), and q, that will in combination account for Bridgman's value 
of a, the Thomson effect, and then from the many combinations that will 
do this to find the smaller number that will account also for the known 
value of B. Ultimately I must undertake to find from the combinations 
that meet these two tests the ones that will account for or be consistent 
with the Peltier effect, but I shall not do that in this paper. 
