the Weights of Atoms. 295 



of the S/s of col. 4, § 69. With this, equation (20) gives 



Here, in order that the comparison may be between the whole 

 light of the san and the light from an equal apparent area of 

 the sky, we must take 



n = 7r/219-4 2 * = 1/15320, 



being the apparent area of the sun's disc as seen from the 

 earth. As to H, it is what is commonly called the " height 

 of the homogeneous atmosphere w and, whether at the top of 

 Etna or at sea-level, is 



K4,) 



7*988 . 10 5 ( 1 + — \ centimetres ; 



where t denotes the temperature at the place above which II 

 is reckoned. Taking this temperature as 15° C, we find 



i7=8'44 . 10 5 centimetres. 

 Thus (22) becomes 



t ^ {iy ~ D) y ] =\\l-e) .'759 .10- 9 . . (23). 



§ 71. Let us now denote by /and 1—f the proportions of 

 (23) due respectively to the ultimate molecules of air and to 

 dust. We have 



n ^ D '~ V) y =\*f[l-e) .*759 . 10" 9 . (24); 



where n denotes the number of the ultimate molecules in a 

 cubic centimetre of the air at the top of Etna ; and T(I)' ' —D)\D 

 relates to any one of these molecules ; any difference which 

 there may be between oxygen and nitrogen being neglected. 

 Now assuming that the refractivity of the atmosphere is 

 practically due to the ultimate molecules, and that no appre- 

 ciable part of it is due to the dust in the air, we have by § 56 (7), 



•0002 = n r ^~ 1 ^ ..... (25), 

 the first number being approximately enough the refractivity 



* The sun's distance from the earth is 219'4 times his radius. 



