EXPERIMENT 9—2
217
where the x—ray tube may be either a gas—lled tube operated by
an induction coil or a hot—cathode tube operated at low voltage,
say 50 to 80 kilovolts.
The lead plates  and B each should be
3 to 4 millimeters thick and should be pierced With holes of
approximately 1 millimeter diameter. They should be separated
from each other by approximately 10 centimeters. Adjustment
for maximum intensity of the rays which pass through  and B
may be made With the aid of a uorescent screen.
The crystal C
is supported on a brass or lead—alloy shield whose hole is slightly
larger than the Width of the x-ray beam at this point. The rock
salt crystal is prepared by tapping a large piece with the sharp
edge of a knife so as to break it along its natural cleavage planes.
It should be about 0.05 centimeter thick and may be mounted
over the hole With wax or cement or, if small, supported on a thin
piece of glass or celluloid over the hole. The photographie plate
is enclosed in a light tight paper cover and issupported as shown,
4 to 6 centimeters from the crystal. A lead disc D, 2 to 3 millimeters thick and 1 centimeter in diameter may be supported by
ne threads over the central part of the plate in order to avoid the
halation from the intense direct beam.
A uoroscope may
be
used for this adjustment.
The exposure time should be about one-half hour for each
20 milliamperes in the lower voltage tubes and the same With
approximately 10 milliamperes in the higher voltage tubes.
Measure the distance R from the center of the crystal to the
photographie plate.
_
After a properly exposed Laue photograph has been obtained
'
and a positive print made, draw in the axes & and 6. Then, follow
the procedure outlined in section
Lack of symmetry
of the
SpOts may be corrected for by measuring the distance between
diametrically opposite spots
and dividing by 2 to get 7”. As an
example, let R = 5.9 centimeters. For a selected spot, measure—
ments give r = 3.53 centimeters and çb = 33.5°. Then, tan
çb =
0.662, tan 26 = 0.598 (equation 9—11), 26 = 30° 53’ and sm
0.266.
Now, 0.662 is nearly equal to %, so that, from equation
9—12, we choose % = 2 and [= 3.
If, then, /z = 1, we nd that
the right hand side of equation 9—13 is equal to 0.267, in essential
agreement With 0.266.
In similar manner, identify all of the more
intense spots, recording their indices near them on the photograph.
217
where the x—ray tube may be either a gas—lled tube operated by
an induction coil or a hot—cathode tube operated at low voltage,
say 50 to 80 kilovolts.
The lead plates  and B each should be
3 to 4 millimeters thick and should be pierced With holes of
approximately 1 millimeter diameter. They should be separated
from each other by approximately 10 centimeters. Adjustment
for maximum intensity of the rays which pass through  and B
may be made With the aid of a uorescent screen.
The crystal C
is supported on a brass or lead—alloy shield whose hole is slightly
larger than the Width of the x-ray beam at this point. The rock
salt crystal is prepared by tapping a large piece with the sharp
edge of a knife so as to break it along its natural cleavage planes.
It should be about 0.05 centimeter thick and may be mounted
over the hole With wax or cement or, if small, supported on a thin
piece of glass or celluloid over the hole. The photographie plate
is enclosed in a light tight paper cover and issupported as shown,
4 to 6 centimeters from the crystal. A lead disc D, 2 to 3 millimeters thick and 1 centimeter in diameter may be supported by
ne threads over the central part of the plate in order to avoid the
halation from the intense direct beam.
A uoroscope may
be
used for this adjustment.
The exposure time should be about one-half hour for each
20 milliamperes in the lower voltage tubes and the same With
approximately 10 milliamperes in the higher voltage tubes.
Measure the distance R from the center of the crystal to the
photographie plate.
_
After a properly exposed Laue photograph has been obtained
'
and a positive print made, draw in the axes & and 6. Then, follow
the procedure outlined in section
Lack of symmetry
of the
SpOts may be corrected for by measuring the distance between
diametrically opposite spots
and dividing by 2 to get 7”. As an
example, let R = 5.9 centimeters. For a selected spot, measure—
ments give r = 3.53 centimeters and çb = 33.5°. Then, tan
çb =
0.662, tan 26 = 0.598 (equation 9—11), 26 = 30° 53’ and sm
0.266.
Now, 0.662 is nearly equal to %, so that, from equation
9—12, we choose % = 2 and [= 3.
If, then, /z = 1, we nd that
the right hand side of equation 9—13 is equal to 0.267, in essential
agreement With 0.266.
In similar manner, identify all of the more
intense spots, recording their indices near them on the photograph.