368 
THE OUTSIDE OF A STAR 
equal to e hv l RT and neglect stimulated emission. Then the emission per cu. 
frequency traversing the cubic centimetre per second is 
cl ( v , T) Av — 87rhv 3 Ave~ hv l RT /c 2 . 
Hence by division, the absorption coefficient per cu. cm. is 
and the coefficient per gram is 
For A = 3950 A, AA = 1 i, we find Av = 1-92.10 11 , and hence 
k' = 9-0.10 8 . 
The mass M of a column of 1 sq. cm. section extending from the base 
to the top of the chromosphere is 
The same result can also be found (without previously calculating k') by 
equating the weight of the column to the difference of radiation pressure 
at its extremities due to radiation in the range A A. 
The density is of order 10~ 17 gm. per cu. cm. and the free path must 
be very great. Milne, treating the particles as neutral, finds a free path of 
6000 km.; in that case the gas pressure can scarcely be said to “support ” 
the matter of the chromosphere, but it replaces the falling particles by 
projecting fresh ones into the region. Probably, however, when account 
is taken of the charges of the particles the free path is not so formidable. 
The equilibrium of a calcium atom at the top of the chromosphere is 
unstable. Although the value of g and the intensity of the radiation both 
fall off according to the inverse-square law as the distance increases, the 
radiation force diminishes less rapidly than gravity. This is because the 
absorbing power of a given mass of material is proportional to the number 
of unexcited atoms present; the proportion of excited atoms is always small, 
but there is a slight loss of efficiency because of them. With increasing 
distance from the star the efficiency increases towards unity, so that 
radiation force gains slightly in comparison with gravity. Presumably 
there is some escape of atoms from the top of the chromosphere but it 
is doubtful whether they can travel far. Ultimately the calcium atom will 
cm. per sec. is 
(254-81), 
and this will also be equal to the absorption. The radiation of H and K 
(254-9). 
Taking the base to be given by (254-63) 
V =4(1- r)/3r = 12, 
so that 
M = 1-33.10~ 8 gm./cm. 2