507 



R = l:27S,i resp. = 72,44 to 71,67 (the value — :/? is = 80,915). 



Let us now calculate the value of bjc- From bk: 0^=: 2y follows 

 with 6„ = v, = 1 : Do : 



2y 2y X ^ 



D, D,X 22412' 



when bjc is calculated in "normal" units, and per Gr. atomic 



weight. We must, therefore, know D^. From the relation for the 



ideal straight diameter (while namely the vapour density can be 



7' B, 



neglected) D ^ D. — 2yZ)/. — follows with Dk = : 



^ ^ " ' ' Tt 2(1 +y) 



T 



D^DAl - 



Unfortunately, however, the liquid density for Lithium is unknown 

 But Z)=: 0,5935 holds for solid Lithium at 15° C. For liquid Lithium 

 D is therefore slightly smaller than this value, perhaps 27o smaller. 

 Thus we have: 



/ 288 \ 



< 5935 = D, { 1-0,589 X = 0,930 D, 



\ ° V ^ 2410y ' 



/ 288 \ 

 ot=:DA 1—0,598 X = 0,936 D, 



"V 27ooy 



so that D^ becomes <^ 0,638 or <^ 0,634 (according as Tk = 2410° or 

 2700° abs.) Hence 



^ 2,865 X 6,94 19,88 



hk y— = — — = 139 . 10-5 



"^0,638x22412 14310 



2,975 X 6,94 20,64 



or hjc > — — — — = 145 . 10-5 



-^0,634X22412 14210 



The value of bk can therefore be at most 2 "/o greater, i.e. from 

 142 to 148.10-5. Now bk = 55 for F, = 70 for 0, = 85 for jV, 

 = 100.10-5 for Q . yyg might, therefore, expect for B the value 

 115, for Be 130, and for Li the value 145.10-5. If this last value 

 is correct, Tk would have to lie between 2400° and 2700° for Lithium, 

 e. g. it would be about 2550°. 



Now the value ak follows from 



TkXbk 

 ak = , 



8 X 



in which ff =: (see above). This gives : 



27 H 



33* 



