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

EXTERNAL RADIATION SAFETY 527
layer (1 HVL) to account for buildup. The HVL of lead for 0.661-MeV gamma
rays is
HVL =
ln 2
λ
0.693
= = 0.56 cm.
1.24 cm−1 The estimated shield thickness therefore is 3.84 plus 0.56, or 4.40 cm, which corresponds
to 1.24 × 4.40 = 5.5 relaxation lengths. From Figure 10-9, we find (by
interpolation) the dose-buildup factor for 0.661-MeV gamma rays to be 2.12 for a
lead shield of this thickness. Substituting these values for B and t into Eq. (10.17),
we have
−1.24 × 4.4
I = 2890 × 2.12 × e
= 26 μSv/h (2.6 mrems/h).
This calculated reduction in gamma-ray dose rate is just slightly more than the
desired value of 2.5 mR/h. The thickness of 4.4, as calculated above, is therefore just
about correct. In this example, we found the correct thickness after one attempt in
a trial-and-error method. If the calculated reduction in radiation dose rate had not
turned out to be so close to the design value with the estimated shield thickness, we
would have continued, by trial and error, to estimate thicknesses until the one that
results in the desired reduction of dose rate would have been obtained.
W Example 10.5
Design a spherical, lead storage container that will attenuate the exposure rate from
1Ciof 24 Na to 10 mR/h at a distance of 1 m from the source.
Solution
In each disintegration of a 24 Na atom, two gamma rays are emitted in cascade:
one of 2.75 MeV and one of 1.37 MeV. The dose rate, at a distance of 1 m, due to
each of these photons is, from Eq. (6.18):
Ɣ2.75 = 0.5 × 2.75 = 1.38
Ɣ1.37 = 0.5 × 1.37 = 0.69
R · m2
C · h ,
R · m2
C · h .
To reduce the exposure rate from the high-energy photon to 10 mR/h for the
condition of good geometry, we use the interpolated value of 0.485 cm −1 (Table 5-3)
for the attenuation coefficient μ.
I
=
I0
10 −0.485t
= e
1380
t = 10.16 cm.