Post on 21-Dec-2015
IAEA 2
To discuss shielding for photon beams and the increase in photon transmission through a shield resulting from buildup
Objective
IAEA 3
HVL and TVL
• The amount of shielding required to reduce the incident radiation levels by ½ is called the “half-value layer” or HVL
• The HVL is dependent on the energy of the photon and the type of material.
• Similarly, the amount of shielding required to reduce the incident radiation levels by 1/10 is called the “tenth-value layer” or TVL.
IAEA 4
HVL and TVL
HVL (cm) TVL (cm)
Isotope Photon E (MeV)
Concrete Steel Lead Concrete Steel Lead
137Cs 0.66 4.8 1.6 0.65 15.7 5.3 2.1
60Co 1.17, 1.33
6.2 2.1 1.2 20.6 6.9 4
198Au 0.41 4.1 0.33 13.5 1.1
192Ir 0.13 to 1.06
4.3 1.3 0.6 14.7 4.3 2
226Ra 0.047 to 2.4
6.9 2.2 1.66 23.4 7.4 5.5
IAEA 5
HVL and TVL
The half value layer (HVL) and tenth value layer (TVL) are mathematically related as follows:
HVL = ln(2)
TVL = ln(10)
=TVLHVL
ln(10)
ln(2)
= ln(10)ln(2)
=2.3030.693
= 3.323
TVL = 3.323 x HVL
and
IAEA 6
Shielding
Shielding is intended to reduce the radiation level at a specific location
The amount of shielding (thickness) depends on:
the energy of the radiation the shielding material the distance from the source
IAEA 7
Inverse Square Law
If the radiation emanates from a point source, the radiation follows what is commonly known as the “Inverse Square Law” or ISL.
Most real sources which are considered to be “point” sources are actually not “point” sources. Most sources such as a 60Co teletherapy source have finite dimensions (a few cm in each direction). These sources appear to behave like point sources at some distance away but as one gets closer to the source, the physical dimensions of the source result in a breakdown of the ISL.
IAEA 8
Inverse Square Law
If the source of the radiation is not a point but is a line, a flat surface (plane) or a finite volume, the ISL does not apply.
However, if one gets far enough away from a finite line, plane or volume, they appear to be a point and the ISL applies with some acceptable error.
For the remaining discussion let’s assume we have a point source. The ISL predicts that the intensity of the radiation will decrease as distance from the source increases even without any shielding.
IAEA 9
These photons should not strike the individual. But due to scatter, they do, so the calculated value is too low. It needs to be increased by the buildup factor.
Scatter
IAEA 10
Photon Attenuation and Absorption
• Absorption refers to the total number of photons absorbed by the material (dark blue arrows)
Attenuation refers to total number of photons removed from incident beam (absorbed + scattered) (dark blue and light blue arrows)
IAEA 11
Photon Attenuation
Ix = Io e-x = Io ewhere:
Ix = photon intensity after traversing x cm of some material
Io = initial or incident photon intensity
x = thickness of material (cm)
= linear attenuation coefficient (cm-1)
= density (g/cm3)
/ = mass attenuation coefficient (cm2/g)
(x)-
IAEA 13
I = Io B e(-x)
primary photons + scattered photonsprimary photonsB =
If there are no scattered photons, then B = 1
If there are scattered photons, then B > 1
Buildup
IAEA 14
What amount of lead shielding is needed to reduce the dose rate beyond the shield from 1 mSv/hr to 0.02 mSv/hr for a 1 MeV photon beam?
Sample Buildup
IAEA 16
Sample Buildup
The mass attenuation coefficient (/) for a1 MeV photon incident on a lead shield is:
(/) = 0.0708 cm2/g
The density of lead = 11.35 g/cm3
The linear attenuation coefficient is:
(/) x = 0.0708 cm2/g x 11.35 g/cm3 = 0.804 cm-1
IAEA 17
I = Io B e(-x)
I = 1 mSv/hr
Io = 0.02 mSv/hr
B = 1 (assumed)
= 0.804 cm-1
Solve for “x”
0.02 mSv/hr = (1 mSv/hr) (1) e(-x)
Sample Buildup
IAEA 18
Sample Buildup
ln(0.02) = ln[e(-x)]-3.91 = -x
Although it is not required:x = -3.91/-0.804 = 4.86 cmThis would be the calculated value of the lead shield if scatter was not considered.
0.02 mSv/hr
1 mSv/hr= e(-x)
IAEA 20
Sample Buildup
x = 3.91 Let’s interpolate:when x = 4, B = 2.26when x = 2, B = 1.69
1.27 = 0.09/(2.26 - x)(2.26 - x) = 0.09/1.27 = 0.0712.26 – 0.071 = x = 2.19
(4-2)
(2.26-1.69)
(4-3.91)(2.26 - x)
=
IAEA 21
Solve for x using B = 2.2
I = I0 B e(-x)
0.02 mSv/hr = (1 mSv/hr) (2.2) e(-x)
ln(0.02/2.2) = ln[e(-x)]
-4.71 = - x
so x = -4.71/-0.804 = 5.86 cm
The thickness of the shield increased from 4.86 cm to 5.86 cm (20%) due to buildup
Sample Buildup
IAEA 22
Skyshine
Photons can scatter or “bounce” off atoms in materials such as the ceiling of a room or even air molecules !
IAEA 23
Where to Get More Information
Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009)
International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)