246 
THE COEFFICIENT OF OPACITY 
Conservation of angular momentum gives 
moV = m'o'V'... 
Conservation of energy gives 
m'c 2 = me 2 + Ze 2 jo r 
(170-11). 
(170-12), 
since Ze 2 !o' is the loss of potential energy. The law of change of mass with 
where b = e 2 /mc 2 . Preliminary trials show that o' is very small compared 
with a so that o' 2 can be neglected. Hence 
Formally this makes o' negative, but allowing a reasonable margin 
for the errors of the data the actual conclusion is that 2 o' ¡Zb is, say, less 
than 1, or o' is less than 3.10 -12 cm. Thus it suggests itself that the true 
target is the nucleus which has a radius of the order 10~ 12 cm., or perhaps 
a fairly full hit on the nucleus may be necessary corresponding to o' = 0. 
The precise value of o' (if of nuclear dimensions) makes little difference 
to cr, so we shall take o' = 0. Then by (170-3) 
The total distance travelled by all the free electrons per gram per second is 
Hence dividing by A, the number of captures is 
Multiplying (170-5) by the average energy of a quantum 2-1BT we obtain 
the total emission, which is equal to the absorption kacT 4 per gm. per sec. 
velocity gives m 2 ^ _ p 2 / c2 ) = m 'a (i _ p'a/ c a) (170-13). 
Eliminating m' and V between these three equations we obtain 
cr 2 = O -' 2 + Zb (Zb + 2cr') c 2 /F 2 (170-2), 
(170-2), 
(170-3). 
Inserting numerical values this gives 
Zbc 
a= ~V~ 
(170-4). 
Hence by (151-91) 
1 1 Ah V 2 
7 tNo 2 7 T p Z 2 b 2 C 2 
V 
(1 +/)' 
Also V must be replaced by its harmonic mean value ¿77 F 0 - Ittu 0 TK The 
(170-6)