Calculations:
i. ‘Using highest count with no background or probe contamination deducted:
t = 3747010.) = o,
500 (168)
Oot M/F
1
I, = 17}? = 0,31(122.6) = 38 mr/ar @ H+ 1
t= 55
Dose to Infinity:
t2 =
o (Infinity)
D = $8,
_,[55°-2-o] =
38
== 28-[0.449]
== 85 mr
Dose to Monitoring:
tz = 300
== -38__f550.2_
poy bs? 3 3900-2]
28.[0.449 ~ 0,319) = 25 or
Dose to End First Decontamination:
to = 70
38 [5° . 2 _ 799-2]
.
d=- 2
28[0.449 - 0.427] = 4.2 ur
2.
Subtracting background of 2400 cpm and probe contamination of 11,235 cpm:
I=—
37,470 - 2400 = 11,235 = 23,835 cpm
23,835(0.7)
500 (168) _ 0.20 mr/hr
I, = 0.20(122.6) = 24.5 mr/hr @ H+ 1
Dose to End First Decontamination:
D= as? [0.449 ~ 0.427] = 2.7 mr
“2
t5 = 70
DOE ARCHES
FE