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    • Abstract: GAMMA RADIATION SHIELDINGLinear Attenuation Coefficient, μFor some, the word attenuation is somewhat of a misnomer; a better word might be collision or interaction.The linear attenuation coefficient (μ) for photons is analogous to the macroscopic cross section (Σ) for

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Linear Attenuation Coefficient, μ
For some, the word attenuation is somewhat of a misnomer; a better word might be collision or interaction.
The linear attenuation coefficient (μ) for photons is analogous to the macroscopic cross section (Σ) for
neutrons, and hence, the μ for a mixture is
μ mix = μ1 + μ 2 + L (1)
The mass attenuation coefficient is μ/ρ where ρ is the material density. For a mixture (compound) based on
constituent weight fractions, ωi
(μ / ρ )mix = ∑ ω i ( μ / ρ ) i (2)
The mean free path, which is the average distance that a photon moves between interactions, is mfp = 1/μ.
Photon mass attenuation (μ/ρ) and mass-energy attenuation (μen/ρ) coefficients for silicon and gallium
arsenide [Source: NIST].
Monoenergetic Photon Attenuation
The intensity (I) of photons that penetrate a target to a distance x without a collision is
I ( x) = I 0 e − μ x = I 0 e − ( μ / ρ ) ρ x (3)
This decreasing exponential absorption characteristic is shown below. It is important to note that I(x) is not
necessarily all of the photons present, since some may have scattered to lower energy via Compton
scattering or annihilation radiation may have been produced.
Gamma intensity
EEE460-Handout K.E. Holbert
Half and Tenth Thickness
The half value layer (or half thickness) is the thickness of any particular material necessary to reduce the
intensity of an X-ray or gamma-ray beam to one-half its original value.
I ( x1 2 ) 1 −μ x ln(2)
= =e 12 ⇒ x1 2 ≡ (4)
I ( 0) 2 μ
In similar fashion, one can define a tenth thickness as the depth required to reduce the photon intensity by a
factor of ten.
Point Source
For suitable geometry, the point source approximation is oftentimes employed for isotropic emission from a
source, which emits S particles per unit time. The expression below, which assumes no attenuation (e.g., in
a vacuum), nonetheless is useful in describing a source in air:
S A(t ) (no. particles / disint )
φ (r ) = 2
= (5)
4π r 4π r 2
Example: A nuclear weapon is exploded at an altitude of 1 km. Assuming 1018 fissions in the explosion
and that each fission yields 2.5 prompt neutrons, estimate the neutron fluence at 2 km from ground zero.
1 km
2 km
Solution: First, the linear distance from the explosion point to the location of interest is computed:
r = (1000 m) 2 + (2000 m) 2 = 2236 m
Treating the explosion as a point source and neglecting the attenuating effect of air.
EEE460-Handout K.E. Holbert
(1018 fissions)(2.5 neutrons/fission ) neutrons
Φn = 2
= 3.98 × 1010
4π (2236 m) m2
Inverse Square Law
For point sources of gamma and X radiation, the photon flux (or intensity) is inversely proportional to the
& &
squared distance from the source. Since the exposure rate ( X ) and dose rate ( D ) are directly proportional
to the flux (φ), then the ratio of intensities at distances R1 and R2 from the point source are
2 & &
I 1 φ ( R1 ) R2 X ( R1 ) D( R1 )
= = 2 = = (6)
I 2 φ ( R2 ) R1 & &
X ( R2 ) D ( R2 )
This leads to a well-known radiation shielding rule that doubling the distance from the source decreases the
radiation by a factor of four.
Buildup Factor, B
Attenuation of gamma intensity through a material is more realistically described by:
φ ( x ) = φ ( 0) B e − μ x B ≥1 (7)
where φ(x) is the intensity at x, φ(0) is the intensity before attenuation, x is the thickness of material, and B is
the buildup factor to account for buildup by scattering, or secondary gamma emission on absorption of the
original gamma ray.
intensity of primary and secondary radiation φ b
B= = (8)
intensity of primary radiation only φu
where φu is the uncollided flux and φb is the buildup flux. The buildup factor, B, accounts for the amount of
forward scattering by the shield; B is a function of material and γ-ray energy as well as geometry. B is
generally determined from tables, or empirical formulas.
Definition of Common Shielding Flux Terminology
Example fluxes for a point source of γ-rays is given to the right of each term
S po int
Unshielded flux : φ 0 φ0 ( R) =
4π R 2
S po int − μ R
Uncollided flux : φu = φ 0 e − μ x φu ( R ) = e (9)
4π R 2
S po int
Buildup flux : φb = φ 0 B e − μ x = φu B φb ( R ) = 2
B( μ R) e −μ R
4π R
where μR = number of mean free paths (mfp).
First, determine the unshielded flux 5 cm from a 100-mCi point source that emits a 0.5 MeV gamma ray for
each decay. Second, if a 10-cm diameter, spherical lead shield encapsulates the point source, determine the
uncollided gamma flux on the surface of the shield. For lead, the linear attenuation coefficient at 0.5 MeV is
1.64 cm–1.
The unshielded flux at a radius of 5 cm from the point source is
EEE460-Handout K.E. Holbert
(100 × 10 −3 Ci)(3.7 × 1010 Bq/Ci)(1 γ/decay) γ
φ 0 (5 cm) = 2
= 11.78 × 10 6
4 π (5 cm) cm 2 sec
The uncollided flux on the surface of the Pb shield is
(100 × 10 −3 Ci)(3.7 × 1010 Bq/Ci)(1 γ/decay) −(1.64 cm −1 )(5 cm ) γ
φu (5 cm) = 2
e = 3235
4 π (5 cm) cm 2 sec
Energy Absorption Coefficient
Since photon attenuation does not mean that all the photon energy is absorbed (e.g., consider Compton
scattering in which only a fraction of the photon energy is liberated to an electron), it is necessary to
introduce another quantity—the energy absorption coefficient, μenergy. In comparing the photon attenuation
versus absorption coefficient
μ atten ≥ μ energy (10)
To contrast the attenuation and absorption coefficients, one should use μatten in the calculation of
probabilities of removing radiation from a beam, and use μenergy in the calculation of radiation dose (i.e.,
energy deposition). A note of caution when looking these values up, as some references use µa to denote
attenuation while others use µa to denote absorption, so one should be careful in extracting such data.
Alpha Concrete
EEE460-Handout K.E. Holbert

Use: 0.1027