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

W = workload, Gy/wk,
T = occupancy factor, and
F = actual cross-sectional area of the beam at 1 meter, cm 2 .
EXTERNAL RADIATION SAFETY 563
The barrier transmission factor for leakage radiation is given in NCRP 147 as
BL = 1000 × P × d2 L
.
W × T
The factor of 1000 accounts for the fact that leakage radiation at a distance of 1 m
from the target may not exceed 0.1% of the workload.
The scattered radiation barrier transmission is found by substituting the appropriate
values into Eq. (10.33). The value of α for 90 ◦ scatter is found in Table 10-8.
Bps = P
αWT × d2 sca × d2 400
sec ×
F
−6 Sv
20 × 10
Bps =
wk
4.26 × 10−4 × 800 Gy
× (1 m)
× 1
wk 2 × (6.3m) 2 400 cm2
×
225 cm2 = 4.1 × 10−3 .
The number of TVLs to attain this degree of attenuation is
n = log 1
Bps
1
= log
= 2.4.
−3
4.1 × 10
The required maximum transmission of leakage radiation is calculated by substituting
the appropriate values in Eq. (10.33):
BL = 1000 × P × d2 L
W × T
=
−6 Gy
1000 × 20 × 10 × (6.3m)2
wk
800 Gy
= 9.9 × 10
wk
−4 .
The number to TVLs required to reach 9.9 × 10 −4 is
n = log 1
BL
1
= log
= 3. −4
9.9 × 10
According to the recommended design practice, when the two calculated barrier
thicknesses are close together, as in this case, we add 1 HVL to the thicker barrier.
If the two barrier thicknesses differ by a TVL or more, we simply use the thicker
barrier. In this example, the barrier thickness for the leakage radiation is not
much greater than that for the scattered radiation. We therefore add 1 HVL to the
thicker barrier. In Table 10-9, we find TLV1 and the TLVe for 6-MV X-rays to be 34and
29-cm concrete respectively. Since 3.3 HVL = 1 TVL, the required barrier
thickness is
tsecondary = 34 cm + (3 − 1) × 29 cm +
29 cm
3.3
= 101 cm.