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)