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page -2-
2.1
x
10-6 curies per second, it follows that the amount of radium emanation present
in 1 liter of expired air,--two good breaths-will be-- 6 K 2.1 x 10-6 X 40 x 10-6 =
504 X 10-12 curies, where 6 stands for the time it takes for two breaths. (Hence
per microgram of radium in the person under test the emanation expired is 12.6 x 10-12
curies.
Next calculate the discharge that this amount of emanation will produce in
a Lind electroscope whose calibration constant is 10 x 10-9 an curies emanation.
Evidently the net drift is given by:
504 X 10-12
- 50.4 x 10-3 d.p.sec. or
-
10 X 10-9
=
-
0.0504 div. per sec.
-
When a Lind chamber of only 500 00 capacity is used the net drift will be 0.0252 or
for mingragram
0.00063 d.p.socy. assuming that the calibration constant of the smaller chamber is
the same. Using a chamber of 4 liters capacity with same constant, 40 microgram of
radium (40 X 10-6 curies) produces a net drift of 0.2016 d.p.sec.
In the test Dr. Flinn and I made on Mrs. Dumschoff in Naugatuck, yesterday,
April 23, 1928, we found by gamma ray instrument a net drift of 0.00752 d.p.sec. (or
7.5
x
10-3
d.p.sec.) which represents nearly 42 micrograms of radium in her body.
Three liters of expired air from Mrs. D. in a Lind Chamber, constant esti-
mated to be 10 x 10-9, produced a net drift at maximum of 0.021 d.p.sec.
Hence amount of Rn. present = 10 X 10-9 X . 021 = 2.1 X 10-10 curies.
Emanation produced by 40 micrograms of Ra in 18 seconds, the time taken
to expire 3 liters of air, is given by -- 18 x 40 X 10-6 X 2.1 X 1512 x 10-12curies
The quantity of radium found by expt. in 3 liters of expired air in this case was
210 X 10-12 curies. Hence the percentage of emanation which escapes in breath is:
210 x 10-12
= 210 = 13.9%
-12
This value x 10 seems normal, 1512 but probably still thigh
1512
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"ocrText": "page -2-\n2.1\nx\n10-6 curies per second, it follows that the amount of radium emanation present\nin 1 liter of expired air,--two good breaths-will be-- 6 K 2.1 x 10-6 X 40 x 10-6 =\n504 X 10-12 curies, where 6 stands for the time it takes for two breaths. (Hence\nper microgram of radium in the person under test the emanation expired is 12.6 x 10-12\ncuries.\nNext calculate the discharge that this amount of emanation will produce in\na Lind electroscope whose calibration constant is 10 x 10-9 an curies emanation.\nEvidently the net drift is given by:\n504 X 10-12\n- 50.4 x 10-3 d.p.sec. or\n-\n10 X 10-9\n=\n-\n0.0504 div. per sec.\n-\nWhen a Lind chamber of only 500 00 capacity is used the net drift will be 0.0252 or\nfor mingragram\n0.00063 d.p.socy. assuming that the calibration constant of the smaller chamber is\nthe same. Using a chamber of 4 liters capacity with same constant, 40 microgram of\nradium (40 X 10-6 curies) produces a net drift of 0.2016 d.p.sec.\nIn the test Dr. Flinn and I made on Mrs. Dumschoff in Naugatuck, yesterday,\nApril 23, 1928, we found by gamma ray instrument a net drift of 0.00752 d.p.sec. (or\n7.5\nx\n10-3\nd.p.sec.) which represents nearly 42 micrograms of radium in her body.\nThree liters of expired air from Mrs. D. in a Lind Chamber, constant esti-\nmated to be 10 x 10-9, produced a net drift at maximum of 0.021 d.p.sec.\nHence amount of Rn. present = 10 X 10-9 X . 021 = 2.1 X 10-10 curies.\nEmanation produced by 40 micrograms of Ra in 18 seconds, the time taken\nto expire 3 liters of air, is given by -- 18 x 40 X 10-6 X 2.1 X 1512 x 10-12curies\nThe quantity of radium found by expt. in 3 liters of expired air in this case was\n210 X 10-12 curies. Hence the percentage of emanation which escapes in breath is:\n210 x 10-12\n= 210 = 13.9%\n-12\nThis value x 10 seems normal, 1512 but probably still thigh\n1512"
}