208 Mr. E. V. Appleton and Dr. B. van der Pol on 



Consider the case of a filament A B (see fig. 6). Let i 2 be 

 the total current leaving the negative end B and i x the 



• Fig. 6. 



0.3 l'a A { 



Ax> 



x 2 B 



total current entering the positive end A. Then (i 2 — tj) is 

 the emitted electron current. The heat W developed in the 

 filament per second is given by 



w 



= 1 i x 2 r x dx — <f>(i 2 —i^ 



where <jf> is ihe latent heat of evaporation o£ the electrons 

 and r x the resistance of the filament per unit length at the 

 point a). We are here neglecting the kinetic energy the elec- 

 trons still possess after they have left the filament *, If we 

 make the further approximation that r x is independent of the 

 value of x, and also that the current distribution is linear, 

 we have 



iz = if + {^-y 1 -0-3)4, 



where l = ,<u 2 — .v 1 =. length of filament. 



Hence 



and 



W=2> 2 R+2i z>yf^^ 



^i/U + 0-Ai a ifU-<l>i a , 



where R is the total filament resistance. 



Now, if W is the energy supplied to the filament per 



# Richardson, ' Emission of Electricity from Hot Bodies,' p. 164. 



