RATE OF DIFFUSION OF IODINE IN KI 493 



For this special case, wlien t = 0, ii = 0, from x = to a; = r/2; u = 

 from .r = ?■ '2 to x = r, hence, 



a„ = 2ug/r \ cos nvx/r ■ dx — 2uJn-K ■ sin mr/2 



Hence: Oi = 2mo, a^ = 0, as = 2u(,/3, a^ = 0, as = 2uo/5 ■ • 



(Note that we can not obtain a^ from the above fomiula because it takes 

 the indeterminate form 0/0. To determine a take our equation before 

 integrating and put < = oo . f (x) or u now clearly becomes ^uq, and all 

 other terms on the right except the first or "zero" term vanish; hence, 

 do = 2 "o- 



We can now write the complete solution for our case, which is: 



(IV) I f (x) ■ dx = \ u ■ dx = rwo/4 + Swo/t • rir ■ sin ir/2 - e r^ + 



Jx=0 Jx=0 



- 9r-kt 



+ 2H,o/37r ■ r/3ir ■ sin 37r/2 ■ e r^ + ■ • • 



Note that all of the even terms fall out because they contain the expres- 

 sion sin rmr, where ?w is an integer. Also as the series stands we see that 

 the sines are alternately + 1 and — 1, hence we can write, 



(V) I }{x) ■ dx = 2uor I 1/8 + l/x^ • e^ - l/g^r^ ■ e^^ + 



-25ir2ii -i9-n-m 



1/25^2 . g-^r- _ i/49,r2 . g~w- . . . 



Numerical considerations. For our case, 



r = 50 approximately 

 k = 1.2 



The third term in (V) becomes therefore: 



-Qtt^X 1.5X3 



X = 1/89 ■ e "50— = 1/89 ■ e"* 



Put e+^ = N, then will log^ iV = 8; and hence. 



Log N = .4343 X 8 = 3.47 (approximately) 



Finding the number corresponding to this logarithm from tables, we 

 have; 



N = 3000 (approximately) 



