41 
1919-20.] The Absorption of X-Rays. 
Since in an absorption band g is nearly equal to p, the third term in the 
bracket is the largest, and we can neglect the others. The expression for 
k thus becomes 
K — 
\.Ne 2 p 
Tcvm(p 2 - g 2 )(p - g) 
(1 - cos (p-g)t) 
( 11 ) 
As r increases the bracket oscillates between 0 and 2, but the denominator 
steadily increases. Thus for a large value of r we have k = 0. Half the 
time the light wave does work on the electron, and half the time the 
electron does work on the wave. Consequently with an undisturbed 
motion we get no absorption at all. 
To account for absorption we must assume impacts, stoppages, and 
interruptions of the motion. If these disturbances were absolutely at 
random, they would give as much energy to the electron as they would 
take from it ; hence they must vary according to some definite law. The 
energy lost by the electron goes presumbly into fluorescent and corpuscular 
radiation and the heat motion of the atom, but how it gets there we do not 
know. The law according to which the impacts and stoppages take place 
is also unknown to us, but the simplest assumption to make is that the 
electron stops dead at fixed intervals r, no matter what the phase of the 
impressed force. 
The above discussion assumes that the electric intensity is a maximum 
when the electron starts to vibrate. If it vibrates for intervals of t seconds 
duration, the phase of the electric intensity must have all possible values 
when the vibration starts. Consequently the above discussion requires 
amplification. 
Retain (9), but assume now that x and dx/dt are equal to zero when 
t = t 0 . Then the solution becomes 
/(*) = * = 
Ae 
m(p 2 - g 2 ) 
- Ae 
cos gt - cos gt 0 cos pit - ^ 0 ) + — sin gt 0 sin p(t - t 0 ) 
(cos gt - R cos {p(t - t 0 ) + <£}), 
m(p 2 — g 2 ) 
where R 2 = cos 2 gt 0 +(g/p) 2 sin 2 gt 0 and tan <p — (g/p) tan gt 0 . Differentiate 
this, multiply by cos gt, and integrate, retaining only the term multiplied 
by 1 Kp—g)' Then we obtain instead of (10) 
AeR p 
sin (\{p -g)T + cf>- gt 0 ) sin J(p - g)r. 
m(p 2 -g 2 )(p-g) 
Since g is nearly equal to p we can write <p=gt 0 and R = l. Hence we 
obtain 
2ANe 2 p 
rcvm(p 2 - g 2 )(p - g) 
sin 2 ^(p - g)r . 
( 12 ) 
