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Now referring back on page 2, to the value 6.3 x 10-4 d.p.sec. discharge per
microcurie of the emanation in the electroscope, based on the idea that all generated
is expired, let us see what the actual amount, which we will call 1/8 of the total,
will produce in the way of discharge. Evidently the net drift will not be more
than 1/8 of 6.3 x 10-4 = 8 X 10-5 d.p.sec. This is the value in the 1/2 liter
chamber. For 10 micrograms the actual net drift would be proportional, and numeri-
cally is 8 x 10-4 d.p.sec. = 0.0008 d.p.sec., in other words we may say that in
practice 10 microcuries produce a net drift of 0.0008 d.p.sec. Now note that in the
gamma ray instrument used for these tests 10 micrograms Ra in a person produce a net
drift of 0.0018 d.p.sec. which is about twice as great an effecto Looking at the
matter in this way then we see the gamma ray methoix is actually fully as sensitive
as the emanation method.
AMOTHNR COMPARISON OF the SENSITIVENESS OF THE GAMMA RAY METHOD AND THE
EXPIRED AIR METHOD OF TESTING FOR RADIUM IN A LIVING RADIOACAL
TIVE
PERSON.
For the Lind instrument assume that we can detect positively a net drift of
0.002 depesec. What quantity of emanation is then present in the electroscope?
The answer to that is: 0.002 x 10 X 10-9 = 2 x 10-11 euries = 20 x 10-12 curies.
Suppose we had used the chamber 1/2 1. capacity. What quanti ty of radium
will in 3 sec. produce this emanation? 1 microgram Ra. produces in 3 sec.
1 x 10-6 X 2.1 x 10-6 x 3 = 6.3 X 10-12 curies. To get the detectable quan-
-12
tity, 20 X 10-12 curies, will take as many microcuries as 6.3 X 10 is contai ned
into 20 X 10-12 =
20 x 10-12
= 3.1 microcuries.
6.3 x 10-12
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"ocrText": "page -3-\nNow referring back on page 2, to the value 6.3 x 10-4 d.p.sec. discharge per\nmicrocurie of the emanation in the electroscope, based on the idea that all generated\nis expired, let us see what the actual amount, which we will call 1/8 of the total,\nwill produce in the way of discharge. Evidently the net drift will not be more\nthan 1/8 of 6.3 x 10-4 = 8 X 10-5 d.p.sec. This is the value in the 1/2 liter\nchamber. For 10 micrograms the actual net drift would be proportional, and numeri-\ncally is 8 x 10-4 d.p.sec. = 0.0008 d.p.sec., in other words we may say that in\npractice 10 microcuries produce a net drift of 0.0008 d.p.sec. Now note that in the\ngamma ray instrument used for these tests 10 micrograms Ra in a person produce a net\ndrift of 0.0018 d.p.sec. which is about twice as great an effecto Looking at the\nmatter in this way then we see the gamma ray methoix is actually fully as sensitive\nas the emanation method.\nAMOTHNR COMPARISON OF the SENSITIVENESS OF THE GAMMA RAY METHOD AND THE\nEXPIRED AIR METHOD OF TESTING FOR RADIUM IN A LIVING RADIOACAL\nTIVE\nPERSON.\nFor the Lind instrument assume that we can detect positively a net drift of\n0.002 depesec. What quantity of emanation is then present in the electroscope?\nThe answer to that is: 0.002 x 10 X 10-9 = 2 x 10-11 euries = 20 x 10-12 curies.\nSuppose we had used the chamber 1/2 1. capacity. What quanti ty of radium\nwill in 3 sec. produce this emanation? 1 microgram Ra. produces in 3 sec.\n1 x 10-6 X 2.1 x 10-6 x 3 = 6.3 X 10-12 curies. To get the detectable quan-\n-12\ntity, 20 X 10-12 curies, will take as many microcuries as 6.3 X 10 is contai ned\ninto 20 X 10-12 =\n20 x 10-12\n= 3.1 microcuries.\n6.3 x 10-12"
}