1. The document discusses principles of radiation shielding, including the use of time, distance, and shielding materials to reduce radiation exposure. It provides examples showing how to calculate exposure levels based on these principles. 2. Shielding materials like lead, steel, concrete, and tungsten are discussed. Metrics like half-value layer and tenth-value layer are introduced to characterize the shielding ability of different materials against different radiation types. 3. The concept of buildup factor is explained, which accounts for increased radiation levels due to scattering in shielding materials. Proper ordering of shielding layers can help minimize radiation buildup.

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5th BAERA Training Course on Radiation Protection for
Radiation Control Officers (RCOs) of Industrial Practices
Bangladesh Atomic Energy Regulatory Authority
Agargaon, Dhaka
06-09 November 2017
L-16 : Principle of Radiation Shielding

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1. Introduction
 Purpose of radiation shielding is to reduce the radiation exposure
level to persons in the vicinity of radiation sources to a certain
predetermined level
 There are several different types of radiation, such as:
- alpha particle,
- beta particle,
- gamma and x-rays, and
- neutron rays.
 These have different energy and flux levels. accordingly, they are
very hazardous to human health and environment.
 These types of radiation results due to radioactive decay of different
radioactive isotopes.

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 There are different types of common shielding materials
corresponding to each type of radiation.
 As energy levels increase, the complexity of the
required shield also increases
1. Introduction (Contd….)

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2. Control of External Radiation
There are three basic principle for control
of external radiation;
(a) Time
(b) Distance and
(c) Shielding

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(a)Time
 Radiation dose is proportional to the time spent
in the radiation field. Work in a radiation area
should be carried out quickly and efficiently.
 More time spent in a radiation area, receive
more radiation
 Less time spent in a radiation area, receive less
radiation
 Example-1: You are working in a facility where
radiation dose level is found 01 µSv/hr. If you
do stay in this area about 60 minutes per day.
How much radiation you will receive in a year?
 [5 days a week, 50 weeks a year.]

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Solution
 Dose = Dose rate X Time
 Here, Time = 5x60x50/60 hrs
= 250 hrs
 Total Radiation received = 1x250 µSv
= 250 µSv

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(b) Distance
 Ionizing radiation in air travel in straight lines.
In such circumstances the radiation simply
diverges from a radioactive source and the
dose rate decreases as the inverse square of
the distance from the source.
 i,e: I1d1
2 = I2 d2
2
S
d2, I2
d1, I1

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(b) Distance
 Considering distance, there are two different
approaches based on which radiation level can
be measured;
(1) Reference dose level and
(2) Radioactive source activity

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1. Reference dose level
Example-2: A dose rate of 03 mSv/hr is measured
at 04 meter from a gamma emitting source. At
what distance will the dose rate be reduced to
7.5 µSv/hr?
 Solution: I1 = 03 mSv/hr, d1 = 04 m, I2= 7.5
µSv/hr, d2=?
 I1d1
2 = I2 d2
2
 03x16 = 7.5/1000 x d2
2
d = 80 meter (Ans)

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(2) Radioactive source activity
 Here the term Gamma factor/Rhm is
important.
 Gamma factor: Gamma factor is the
absorbed dose rate in mSv/hr at 01 meter
from 1 GBq of the radionuclide.

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(2) Radioactive source activity
The value of Gamma Factor for different sources
Radioisotopes (Gamma Factor)Dose rate
mSv/hr/GBq at 1 meter
Ir-192 0.13
Co-60 0.351
Cs-137 0.081
Yb-169 0.0007

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Radioactive source activity cont’d
Example-3: A 20 Ci Ir-192 is to be used in open place to
perform Industrial radiography. What distance is to be
maintained to produce control area boundary layer
keeping dose level within 7.5 µSv/hr ? [ 01 Ci=37GBq]
 dose rate = gamma factor x activity/(distance)2
 Example-4: A dose rate is of 2 mSv/hr is measured at 15
cm from a Cesium-137 source. What is the source’s
activity?
 Example-5 : A 40 Ci (17/05/2011) Ir-192 was bought
to be used in open place to perform Industrial
radiography. What distance is to be maintained to
produce control area boundary layer keeping dose level
within 7.5 µSv/hr ? [Half life of Ir-192 is 73 days]

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Solution
 Example-3:
 Activity = 20 Ci = 740 GBq
 dose rate = gamma factor x activity/(distance)2
 Distance, d = [0.13x740x1000/7.5]1/2
 d = 113.3 meter (Ans)
 Example-4: dose rate = gamma factor x
activity/(distance)2
 = (0.081x
activity)/0.0225 mSv/hr
 Activity = (2x 0.0225)/0.081
 = 0.556 GBq

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© Shielding
 There are two approaches;
 (1) Thumb rules approaches
 (2) Theoritical approaches

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(1) Thumb rules approaches
The radiation shielding capacity of materials can be
measured in terms of
 Half-Value Layer (HVL) and
 Tenth-Value Layer (TVL)
(a) Half Value Layer (HVL): The HVL is defined as the
thickness of shielding material required to reduce the
radiation intensity to one-half of its original value
(b) Tenth Value Layer (TVL): The TVL is defined as the
thickness of shielding material required to reduce the
radiation intensity to one-tenth of its original value

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HVL and TVL
A table of HVL and TVL values for radioisotopes and common
shielding materials
Shielding Material and Thickness (cm) Source
Ir-192 Co-60
Concrete TVL 14.74 22.86
HVL 4.82 6.85
Steel TVL 2.90 7.36
HVL 0.87 2.20
Lead TVL 1.62 4.11
HVL 0.48 1.24
Tungsten TVL 1.09 2.62
HVL 0.33 0.79
Uranium TVL 0.93 2.29
HVL 0.28 0.69

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Radiation Lead Concrete
HVL TVL HVL TVL
(a) 100 kVp X-rays 0.026 0.087 1.65 5.42
(b) 200 kVp X-rays 0.043 0.142 2.59 8.55

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Example-6
A container containing Co-60 gamma radiation
source inside it. The radiation dose rate at the
surface of this container is 20 µSv/hr.
(a) If additional 2.2 cm thickness steel sheet is
attached to the container for shielding, what will be
the surface dose ?
(b) What thickness of additional steel sheet is required
to reduce the surface dose below 02 µSv/hr ?

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(d) MULTILAYERED SHIELDS
 The buildup factor is defined as the ratio of the actual gamma flux to
that which would be calculated using a simple exponential attenuation
with the linear absorption coefficient.
 The gamma ray buildup factor represents a necessary correction factor in the
design calculations of the reactors shielding, medical physics field and
nuclear laboratories, where the shielding is an important principle to
protection from nuclear radiation.
Importance of Buildup Factor:
 It will be observed that the numerical values of Bm can be very large, which
shows the importance of the buildup of scattered radiation in shielding
calculation.
 This means that if buildup were omitted from a calculation of the
effectiveness of such a shield, the resulting exposure rate would be in error .
The buildup factor, therefore, cannot be ignored.

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 The buildup flux at P is depend upon which of the two layers comes first. Thus
while a monoenergetic flux incident on the first region and producing a continuous
spectrum incident on the second, gives one value of Фb at P, AND If the media are
interchanged, Фb would not be the same. And, indeed, the two values of Фb in
general, are different.
 Consider, for example, 0.5-MeV gamma-rays incident on layers composed of lead
and water. It will be seen that the buildup factor of water (35.9) is much larger than
that of lead (2.08) for the same thickness (10 Mean free path).
 It follows that there is a greater buildup of low-energy scattered radiation in water
than in lead.
 Because in water, the low-energy radiation propagates with little photoelectric
absorption. In lead, by contrast, owing to its high value of Z, photoelectric
absorption is important at low energies and a buildup of low-energy radiation Is not
possible.
(d) Multilayered Shields

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 If, therefore, the water were placed before the lead, the
buildup radiation from the water would be absorbed as it
passed through the lead, and the overall buildup through
the shield would be small.
 On the other hand, if the lead were placed before the
water, the subsequent buildup of radiation in the water
would reach the observer and the overall buildup would
be larger.
 Unfortunately, there is no simple way to obtain the exact
value of the buildup factor for a layered shield in terms of
the buildup factors of each layer.
(d) Multilayered Shields

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Example-7
(a) You are carrying an industrial radiographic source into
exposure projector through the vehicle from storage pit
to working place. During carrying the source, surface
radiation level of that projector was about 01 mSv/hr.
What arrangement to be made such that the vehicle
driver can receive the dose within 02 µSv/hr?
(b) A dose rate is of 2 mSv/hr is measured at 02 cm from
a Cobalt-60 source container surface that one is used
for nucleonic gauge purpose. During maintenance work
this source is to be kept into safe and secured storage
room. What arrangement to be made such that the
general workers around there can do their routine work
within Background Radiation level?

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