ACADs (08-006) Covered
3.3.3.9
3.3.3.10
3.3.3.11.1
3.3.3.11.2
4.11.9.1
4.11.9.2
4.11.9.3
4.11.10
3.3.3.11.3
3.3.3.11.4
Keywords
Gamma, shielding, attenuation, scatter, buildup, half value, tenth value.
Description
Supporting Material
3.3.3.12
4.11.8
Gamma Shielding
Calculations
Radiation Protection III
NUCP2331
SHIELDING
• Gamma Radiation
– Gamma rays do not give up energy
continuously as charged particles do
– Gamma ray intensity decreases due to
absorption or scattering
– Efficiency of the shield depends upon the
probability of interaction and thickness of
material
– Attenuation-Process by which radiation is
reduced in intensity when passing through
some material
Attenuation Coefficients
• l (cm-1) Linear Absorption Coefficient is
defined as the probability of interaction per
unit path length of material
• m (cm2/g) Mass Absorption Coefficient
is defined as / where  is the linear
absorption coefficient and  is density
• μl = μm ρ
– μm = mass attenuation coefficient
– ρ = density of material in g/cc
EM Formula
• I = Io e (-μlx)
–
–
–
–
I = intensity with the shield,
Io = intensity without shield,
μl = linear attenuation coefficient,
x = thickness of shield
Example
•
•
•
•
I0 = 1000 mR/hr
Your shield material is Lead
Shield thickness is 2 inches
Radionuclide is Cs-137
• What do you do first?
Example
• I0 = 1000 mR/hr
• Shield is Lead
– thickness 5 cm (2 in), μm = 0.115 c/g, ρ = 11.35g/cc
• What is reading on other side of shield?
• I = 1000 e -(0.115)(11.35)(5)
• I = 1.47 mR/hr
Example
•
•
•
•
I0 = 2000 mR/hr
Your shield material is Iron
Shield thickness is 4 inches
Radionuclide is Cs-137
• What do you do first?
Example
• I0 = 2000 mR/hr
• Shield is Iron
– thickness 10 cm (4 in), μm = 0.0745 cc/g, ρ = 7.8
g/cc
• What is reading on other side of shield?
• I = 2000 e -(0.0745)(7.8)(10)
• I = 5.988 mR/hr
Scatter
• Remember the Compton scatter
• Some of the radiation interacting with the
shield will create some new EM radiation,
some of which may escape the shield
• This will add exposure to the area on the
backside of the shield ,
• So how does on take this into account?
Build up
• So if you have another parameter in your
equation that changes your dose after
your shield, your shield will need to be
thicker to take into account eh greater
dose
• So that means you have 2 unknowns in
your equations, not good
• Need to estimate the shield thickness
calculate the B (get close) and use that
Build up Factors
• I = BIo e (-μlx)
• B= build up factor
• based on μx
• Dependant on energy and density of
material
• Caused by Compton scattering
• How will your answers change?
Example
•
•
•
•
I0 = 1000 mR/hr
Your shield material is Lead
Shield thickness is 2 inches
Radionuclide is Cs-137
• What do you do first?
• Build up factor?
Example
• I0 = 1000 mR/hr
• B =? μx= 6.5 energy is .662 MeV
• Shield is Lead
– thickness 5 cm (2 in), μm = 0.115 c/g, ρ = 11.35g/cc
• What is reading on other side of shield?
• I = 2.3 1000 e -(0.115)(11.35)(5)
• I = 3.38 mR/hr
Example
•
•
•
•
I0 = 2000 mR/hr
Your shield material is Iron
Shield thickness is 4 inches
Radionuclide is Cs-137
• What do you do first?
Example
• I0 = 1000 mR/hr
• B μx= 6 energy is .662 MeV
• Shield is Iron
– thickness 10 cm (4 in), μm = 0.0745 cc/g, ρ = 7.8
g/cc
• What is reading on other side of shield?
• I = 9.25 2000 e -(0.0745)(7.8)(10)
• I = 55.3 mR/hr
Half Value Layers
• Half-Value Layer
That thickness of material which reduces
the radiation intensity to one-half of its
initial value
• Tenth-Value Layer
That thickness of material which reduces
the radiation intensity to one-tenth of its
initial value
Both take into account build up
How many
• How many half value layers would it take
to reduce the radiation intensity from 500
mR/hr to less than 5 mR/hr?
Total thickness
• Number of HVL X the thickness of each
HVL
• What is total thickness of material need to
reduce 500 mR/hr to less than 5 mR/hr?
• Of
– Iron
– Concrete
– Lead
Problem
• How many TVL does it take to decrease a
reading of 350 mR/hr to under 5 mR/hr?
• How many HVL are there in a TVT?
Medical X ray
• Several factors need to be taken into
account when designing shielding for an
X-ray machine
• Not really for the people getting x rayed
but for the people outside the X-ray room
• The dose to those people are limited, as
opposed to the dose to the patient
• What energy the machined operates, how
many patients will be x rayed, and where
the X-ray room are all important to the
Medical X-ray
• 3 sources of radiation
– Primary radiation (from the machine)
– Full beam from the machine interacting with wall or floor
– Secondary radiation
• Scattered radiation ( from patient)
• Leakage radiation ( from machine)
• Dose rate at some distance (primary)
• H (unshielded) =WUT/ d2
Medical X-ray
• Shielded/Unshielded(Dose/mA minute) =
PD2/WUT
– P- dose level in area needs to be r/wk
– D-Distance from the machine (m)
– W- workload (mA min/wk)
– U-use factor how often the primary beam is
faced on that barrier
– T-occupancy percentage time area is in use
Xray room design
•
Secondary barrier
D primary
D secondary
Primary
barrier
Questions?