Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

516 CHAPTER 10
Solution
His total dose includes the dose during approach to the source, viewing the source,
and then leaving. The dose during the approach, from Eq. (10.6), is
H (approach) = Ɣ × A × w R
v × d0
H (viewing) = Ɣ × A × t × w R
d 2 0
= 54.2 × 10 −3 rem
H (leaving) = Ɣ × A × w R
v × d0
R · m2
1.3 × 10 Ci × 1 rem/R
=
Ci · h
1m· s−1 × 3600 s/h × 1m = 3.6 × 10−3 rem.
1.3
=
R · m2
Ci · h
15 s
× 10 Ci × × 1 rem/R
3600 s/h
(1m) 2
R · m2
1.3 × 10 Ci × 1 rem/R
=
Ci · h
= 1.8 × 10
2m/s × 3600 s/h × 1m
−3 rem
Dose commitment = H = 10 −3 (3.6 + 54.2 + 1.8) = 60 × 10 −3 rem
= 60 mrems (600 μSv).
Line Source
In the case of a line source of radiation, such as a pipe carrying contaminated
liquid waste, the variation of dose rate with distance is somewhat more complex
mathematically than in the case of the point source. If the linear concentration of
activity in the line is Cl MBq or curies per unit length of a gamma emitter whose
source strength is Ɣ, then the dose rate at point p (Fig. 10-1), at a distance h from
the infinitesimal length dl is given by
d ˙Dp = Ɣ × Cl × dl
l 2 + h2 , (10.7)
and for the dose rate due to the activity in the total length of pipe, we have
˙Dp = Ɣ × Cl
˙Dp =
Ɣ × Cl
h
l1
0
dl
l 2 + Ɣ × Cl
+ h2 l2
−1
l1 l2
tan + tan−1
h h
0
dl
l 2 + h 2
˙Dp = Ɣ × Cl × θ
. (10.8)
h