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

268 CHAPTER 6
or
−8 Gy μGy
s
8.19 × 10 × 106 × 3.6 × 103 = 295 μGy/h(29.5 mrads/h).
s Gy h
In the example above, the neutron beam was monoenergetic and thus only one
neutron energy was considered. If a beam contains neutrons of several energies,
then the calculation must be carried out separately for each energy group.
Thermal Neutrons
For thermal neutrons, two reactions are considered, namely, the 14 N(n, p) 14 C reaction
and the 1 H(n, γ ) 2 H reaction. For the former reaction, the dose rate may be
calculated from the equation
˙Dn,p =
φNNσNQ × 1.6 × 10 −13
1 J/kg
Gy
J
MeV
, (6.108)
where
φ = thermal flux, neutrons/cm2 /s,
NN = number of nitrogen atoms per kg tissue, 1.49 × 1024 ,
σ N = absorption cross section for nitrogen, 1.75 × 10−24 cm2 , and
Q = energy released by the reaction = 0.63 MeV.
The latter reaction, 1 H(n, γ ) 2 H is equivalent to having a uniformly distributed
gamma-emitting isotope throughout the body and results in an auto-integral gammaray
dose. The specific activity of this distributed gamma emitter, the number of
reactions per second per gram, is governed by the neutron flux and is given by
Eq. (6.109)
A = φNHσH
where
′′ Bq ′′
kg
, (6.109)
φ = thermal flux, neutrons/cm 2 /s,
NH = number of hydrogen atoms per kg tissue = 5.98 × 10 25 , and
σH = absorption cross section for hydrogen = 0.33 × 10 −24 cm 2 .
W Example 6.22
What is the absorbed dose rate to a 70-kg person from a whole body exposure to a
mean thermal flux of 10,000 neutrons/cm 2 /s?