> 
574 DR J. HALM ON j 
belonging to the group «=0. On the lowest horizontal line the wave-lengths of the 
first Hydrogen-series (BALMER’S formula) have been drawn on a certain scale (in the { 
original drawing 10 t.m.=1 mm.), beginning with m=3 (A,=6563°07) on the left to 
m= 0 (Ao = 364610) on the right. The points 3, 4,5, . . . thus obtained were then 
connected with a point O arbitrarily fixed in the plane of the paper. If now we draw, 
not necessarily but for the sake of convenience, on the same scale, the wave-lengths of, 
say, the 1st subsidiary series of Lithium on another straight line, for instance the edge 
of a ruler, we can bring this line on our diagram into such a position that its intersections 
with the rays O,, O, . . . O» mark precisely the positions of all the lines of the 
series. The line is indicated in the drawing by the transversal Li. In the same way 
the transversals He, Tl, Zn and In represent the 2nd principal series of Helium 
and the Ist subsidiary series of Thallium, Zine and Indium, which, as we have seen 
before, belong to this particular group u=0. Obviously the “tails” of all these series 
are situated upon the ray Ow, while the “heads” lie upon O, which corresponds to 
m=0(. In line-series these “ heads” are of no special interest, because they correspond 
to wave-lengths outside the range of observation. Nevertheless, they will be considered 
here on account of their great importance in relation to our investigation of the band- 
spectra in subsequent pages. The construction of fig. 2 was based on wave-lengths in 
order to show that the RypBere-TuHreLe formula may also be employed in the form 
given by equation (4). For theoretical purposes, however, an investigation founded on 
wave-frequencies must doubtless be preferred. This has been already pointed out 
by Rypperc, Kayser, ScHusrEeR and others. ‘We shall therefore now revert to our 
formulee referring to wave-frequencies. 
Since the position of O is quite arbitrary, we may conveniently choose it so that 
the two boundary rays Oy and Ov» form a right angle. In this case the cotangent of 
the angle a between the rays Ov and Ov» (fig. 3) is equal to (m+ u)*, if we make 
AOQ=1. Let us further call 6 the angle formed by the ray Ov and the transversal 
which represents our series. Since cc» =v. —v and < Occ =90° —a—, we have from 
triangle Occ : 
sin a sina 
Voo Te Ob GEG) a cha (OEEB) s (11a) 
On the other hand, from triangle Occ, we find 
a cosa | cos a (118) 
Cos (a +B) cos (a+ 8) 
Comparing the first of these formule with equation (2) we see at once that 
Oc _ cos B _ 
The RypBERG-THIELE equation may therefore be written in the form 
| yi 
sin a 
eee. (12) 
V—=Viai= 
