D O C U M E N T 1 2 5 S E P T E M B E R 1 9 2 3 1 2 1
The measurements ranged over 4 scale graduations, always from 20 to 16.
Thereupon the secondary radiator was removed and Z1 determined again; one
obtained Z1 = 0,001. The secondary radiator was put back again and observed an-
ew; the values obtained were very similar to the first ones. Hence, the average ob-
tained from all the trials for the ratio
3.5 at 90° = ϑ.
5) Now the chamber was set at ϑ = 15°. (I unfortunately couldn’t take a smaller
angle because then the primary ray directly hits the edge of the chamber, which I
wanted to avoid. I am now having the chamber constructed a little narrower and
better shielded so that the angle can be diminished.) Then the experiment was re-
peated exactly as above, alternately without and with the foil. The sought-after ratio
resulted in:
= 5.3 at ϑ = 15°.
This would qualitatively correspond to the expected redshift of the Rh light at
the position ϑ = 90°. It amounts to
the absorption edge of the Mo lies at 618, 0 XE =
0,618Å.[9]
According to Compton’s
formula,[10]
the shift is:
at 15°
at 90°
thus, much more can go through at 90° than at 15°.
There still are two things that aren’t at all nice:
I. That hard leakage radiation is present besides the Rh radiation. This is in there
in larger quantities in the case of ϑ = 15° than in the case of ϑ = 90° (which is
proved by Z2 Z1) and, naturally, easily goes through the Mo sheet, as (calculated
from the tension inside the tube) it has a hardness of approximately λ ~ 0.3–0.4. It
has the consequence that the “intensity with foil” comes out too large in the case of
ϑ = 15°, hence the found ratio 5.3 is probably much too small. This doesn’t disturb
the qualitative result, of course, but is unpleasant on the quantitative side.
II. That the values Z2 and Z1 differ so much in size. I hope to eliminate this by
having the chamber heavily shielded and having the experiment performed in this
way:
A.) At 90°: Scattering without secondary radiator = Z1
intensity without foil
intensity with foil
------------------------------------------------- - =
intensity without foil
intensity with foil
------------------------------------------------- - =
λRhKα λ0 612,1 XE 0,6121 Å = = =
λϑ λ0 0 0484
2---
ϑ
2
sin , + 0 6121 0.00083 + , 0 61293 , = = =
λϑ λ0 0 0484
245
sin , + 0 6121 0.0242 + , 0 6363 , = = =