1
Radiation Dosimetry
⌂ Units of Absorbed Dose:
Radiation damage ∝ Concentration of absorbed energy in tissue
1. Gray: The basic unit of radiation dose is expressed in terms of
absorbed energy per unit mass of tissue. The SI unit of absorbed
dose is gray (Gy). One gray is expressed in terms of one joule per
kilogram of tissue.
1 𝐺𝑦 = 1 𝐽/𝐾𝑔
The gray is universally applicable to all types of ionizing radiation
dosimetry, such as irradiation due to external fields of γ-rays,
neutrons, β-rays as well as that due to internally deposited
radioisotopes.
2. Rad: Radiation dose was measured by a unit called the rad
(Radiation Absorbed Dose) before the universal acceptance
(adoption) of the SI units.
One rad is defined as an absorbed dose of 100 ergs/g.
𝑒𝑟𝑔𝑠
1 𝑟𝑎𝑑 = 100
= 102 𝑒𝑟𝑔𝑠/𝑔
𝑔
1𝐽
107 𝑒𝑟𝑔𝑠
𝑒𝑟𝑔𝑠
1 𝐺𝑦 =
=
= 102 × (102
) = 100 𝑟𝑎𝑑𝑠.
1 𝐾𝑔
1000 𝑔
𝑔
⌂ Units of Exposure:
1. Exposure unit: The X-ray fields to which an organism may be
exposed are frequently specified in exposure units.
2
The exposure unit (or X-unit) is defined as that quantity X- or
gamma radiation that produces ions (in air) carrying 1 coulomb of
charge (of other sign) per kg air.
𝐶
1 𝑋 𝑢𝑛𝑖𝑡 = 1
𝑎𝑖𝑟
𝑘𝑔
The exposure unit is based on ionization of air. At quantum
energies, less than several KeV and more than several MeV
(~𝐾𝑒𝑉 > 𝐸, 𝐸 > ~𝑀𝑒𝑉), becomes difficult to fulfill the
requirements for measuring the exposure unit. So, X-unit is limited
to X-rays or gamma rays whose quantum energies do not exceed 3
MeV.
We know that the average energy required to produce a single ion
pair in air is 34 eV, and the charge on a single ion is
1.6×10−19 𝐶𝑜𝑢𝑙𝑜𝑚𝑏.
𝐶
1 𝑋 𝑢𝑛𝑖𝑡 = 1
𝑎𝑖𝑟
𝑘𝑔
𝐶
1 𝑖𝑜𝑛
𝑒𝑉
𝐽
−19
= 1 𝑎𝑖𝑟×
×34
×1.6×10
𝑘𝑔
1.6×10−19 𝐶
𝑖𝑜𝑛
𝑒𝑉
𝐺𝑦
×1
= 34 𝐺𝑦 (𝑖𝑛 𝑎𝑖𝑟)
𝐽/𝑘𝑔
The exposure unit is an integrated measure of exposure of the time
over which the exposure occurs.
2. Rontgen: Before the SI unit of exposure was adopted, the unit of Xray exposure was called the rontgen (R).
The rontgen was defined as that quantity of X- or gamma radiation
that produces ions carrying one statcoulomb of charge of either
sign per cubic centimeter of air at 0o C and 760 mm (Hg).
1 𝑅 = 1 𝑠𝐶/𝑐𝑚3
3
Since 1 ion carries a charge of 4.8×10−10 sC, and the mass of 1 cm3
of standard air is 0.001293 g, an exposure of 1R corresponds to an
absorption of 87.7 ergs per gram of air (0.877 rad).
𝑠𝐶
1 𝑐𝑚3 𝑎𝑖𝑟
1 𝑅 = 1 3 𝑎𝑖𝑟×
𝑔
𝑐𝑚
1.29×10−3
𝑎𝑖𝑟
𝑐𝑚3
1 𝑖𝑜𝑛
𝑒𝑉
𝑒𝑟𝑔
𝑒𝑟𝑔
−12
×
×34
×1.6×10
=
87.65
4.8×10−10 𝑠𝐶
𝑖𝑜𝑛
𝑒𝑉
𝑔
𝑒𝑟𝑔
1 𝑟𝑎𝑑
= 87.65
×
𝑒𝑟𝑔 = 0.877 𝑟𝑎𝑑 (𝑡𝑜 𝑎𝑖𝑟)
𝑔
100
𝑔
⌂ Relationship between the exposure unit and the rontgen:
𝐶
𝑗
𝑎𝑖𝑟 = 34 𝐺𝑦 = 34
𝑘𝑔
𝑘𝑔
𝑗 107 𝑒𝑟𝑔𝑠
1 𝑘𝑔
1
= 34
×
×
×
= 3881 𝑅
𝑒𝑟𝑔𝑠/𝑔
𝑘𝑔
𝐽
1000 𝑔
87.7
𝑅
1 𝑋 𝑢𝑛𝑖𝑡 = 1
◊ Problem 1: Medical X-ray shielding design is based on maximum
weekly exposures of 100 mR for controlled areas and 10 mR for
uncontrolled areas. What are the corresponding exposures expressed in
SI units?
Solution: We know,
3881 𝑅 = 1 𝑋 𝑢𝑛𝑖𝑡 = 1 𝐶/𝑘𝑔
∴1𝑅 =
1
𝑋 𝑢𝑛𝑖𝑡
3881
4
For controlled areas,
𝐸𝑥𝑝𝑜𝑠𝑢𝑟𝑒 = 100 𝑚𝑅 = 100×10−3 𝑅 = 0.1 𝑅
∴ 0.1 𝑅 =
0.1
𝐶
𝜇𝐶
𝑥 𝑢𝑛𝑖𝑡 = 25.8×10−6
= 25.8
3881
𝑘𝑔
𝑘𝑔
For uncontrolled areas,
𝐸𝑥𝑝𝑜𝑠𝑢𝑟𝑒 = 10 𝑚𝑅 = 10×10−3 𝑅 = 0.01 𝑅
∴ 0.01 𝑅 =
0.01
𝐶
𝜇𝐶
𝑥 𝑢𝑛𝑖𝑡 = 2.58×10−6
= 2.58
3881
𝑘𝑔
𝑘𝑔
⌂ Exposure Measurement:
(i)
(ii)
By the free air chamber – this is primary lab. Standard (Heavy)
By the air wall chamber- this is portable.
The Air Wall Chamber:
This instrument could be made by compressing the air around the
measuring cavity. In practice, it would be quite difficult to construct an
instrument whose walls are made of compressed air. However, it is
possible to make an instrument with walls of “air equivalent” material.
The material whose X-ray absorption properties are very similar to those
of air is called air equivalent material.
5
Fig1: Condenser type pocket ionization chamber
The air equivalent chamber can be built in the form of an electrical
capacitor as shown in fig. 1. It consists of an outer cylindrical wall (about
4.75 mm thick, made of electrically conducting plastic mixture of plastic
& conducting element). Coaxial with the outer wall, but separated from
it by a very high-quality insulator, is a center wire. The center wire (or
center anode) is positively charged with respect to the wall. When the
chamber is exposed to X- or to gamma radiation, the ionization produced
in the measuring cavity as a result of interactions between photons and
the wall, discharges the condenser, thereby decreasing the potential of
the anode.
𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑎𝑛𝑜𝑑𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 ∝ 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑
𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 ∝ 𝑟𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝑒𝑥𝑝𝑜𝑠𝑢𝑟𝑒
When this air wall chamber is used, care must be taken that the
walls are of the proper thickness for the energy of the radiation being
measured.
If the walls are too thin, an insufficient number of photons will
interact (because other photons will pass through the wall, 𝐼 = 𝐼0 𝑒 −𝜇𝑡 )
to produce primary electrons (which ionize the gas in the chamber).
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If the walls are too thick, the primary radiation (gamma ray) will be
absorbed to a significant degree by the wall, and an attenuated primary
electron flux will result, because primary electrons produced at the outer
surface of the wall have not sufficient energy to pass through the wall
into the cavity.
◊ Problem 2:
Given
chamber volume = 2 cm3,
chamber is filled with air at STP,
capacitance = 5 pF,
voltage across chamber before exposure to radiation = 180 V,
voltage across chamber after exposure to radiation = 160 V, and
exposure time = 0.5 h.
Calculate the radiation exposure and the exposure rate.
Solution:
The exposure is calculated as follows:
𝐶 × ∆𝑉 = ∆𝑄
where
C = 5 pF = 5×10-12 F
∆𝑉 = 180 – 120 = 60 volts
Density of air at STP, 𝜌 = 1.293× 10−6 kg/cm3
V = 2 cm3
∴ ∆𝑄 = 5 × 10−12 F × (180 − 160) V = 1 × 10−10 C.
Mass of air in the chamber, m = 𝜌𝑉 =(1.293× 10−6 kg/cm3)( 2 cm3)
∆𝑄
10−10 𝐶
𝐶
= 3.867×10−5
𝑚
𝑘𝑔
𝐶 3881 𝑅
= 3.867×10−5
×
= 0.15 𝑅
𝐶
𝑘𝑔
𝑘𝑔
0.15 𝑅
150 𝑚𝑅
∴ 𝐸𝑥𝑝𝑜𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑒 =
=
= 300 𝑚𝑅/ℎ𝑟
𝐸𝑥𝑝𝑜𝑠𝑢𝑟𝑒 =
=
1.293×10−6×2 𝑘𝑔
0.5 ℎ𝑟
0.5 ℎ𝑟
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⌂ Exposure-Dose relationship:
Absorption of radiation is approximately proportional to the
electronic density of the absorber, medium. Air wall chamber measures
the energy absorption in air. In most instances, we are interested in the
energy absorbed in tissue.
The electronic density of muscle tissue (consisting of H, O, N, C) is
3.28×1023 electrons/g.
Absorption of radiation (in tissue) ∝ 3.28×1023 electrons/g ………… (1)
Absorption of radiation (in air) ∝ 3.01×1023 electrons/g ………………. (2)
Absorption of radiation (in tissue) 3.28×1023
=
Absorption of radiation (in air)
3.01×1023
or, Absorption of radiation (in tissue)
3.28
=
× Absorption of radiation (in air)
3.01
=
3.28
3.01
×1
𝐶
𝑘𝑔
=
3.28
3.01
× 34
J
kg
= 37
𝐽
𝑘𝑔
Corresponding to an exposure of 1 C/kg air.
1
𝐶
𝑘𝑔
𝑎𝑖𝑟 = 1 𝑋 𝑢𝑛𝑖𝑡 = 37 𝐽/𝑘𝑔(𝑖𝑛 𝑡𝑖𝑠𝑠𝑢𝑒).
⌂ Problem 3: Consider a gamma-ray beam of quantum energy 0.3
MeV. If the photon flux is 1000 quanta/cm2/s and the air temperature is
20◦C, what is the exposure rate at a point in this beam and what is the
absorbed dose rate for soft tissue at this point? Given, µa = 3.46 × 10-5
cm-1 for 0.3 MeV photon & 𝜌a = 1.293 × 10-6 kg/cm3 at 0o C (or at 273 K),
µm = 0.0312 cm-1, 𝜌m = 0.001 kg/cm3.
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Solution:
we know,
1 𝑒𝑉 = 1.6×10−19 𝐽
1 𝑀𝑒𝑉 = 1.6×10−13 𝐽
𝐶
1 𝑋 𝑢𝑛𝑖𝑡 = 1 = 34 𝐽/𝑘𝑔
𝑔
𝑞𝑢𝑎𝑛𝑡𝑎
𝜑 = 1000
/𝑠
𝑐𝑚2
µa = 3.46 × 10-5 cm-1
𝜌a = 1.293 × 10-6 kg/cm3
273
𝜌a(R.T) = 1.293 × 10-6 × at 293 K
279
-1
µm = 0.0312 cm
𝜌m = 0.001 kg/cm3
The exposure rate expressed in C/kg is given by
𝑋̇ =
𝑑𝑋
𝑑𝑡
=
𝜑
𝑝ℎ𝑜𝑡𝑜𝑛𝑠
𝑀𝑒𝑉
𝐽
×𝐸𝑝ℎ𝑜𝑡𝑜𝑛×1.6×10−13𝑀𝑒𝑉×𝜇𝑎 𝑐𝑚−1
2
𝑐𝑚 /𝑠
𝑘𝑔
𝐽/𝑘𝑔
𝜌𝑎 3×34
𝐶/𝑘𝑔
𝑐𝑚
………………………. (1)
1000×0.3×1.6×10−13 ×3.46×10−5
𝐶
=
= 4×10−11
/𝑠
273
𝑘𝑔
−6
1.29×10 ×
×34
293
The absorbed dose rate in Gy/s is given by
𝐷̇ =
𝑑𝐷
𝑑𝑡
=
𝜑
𝑝ℎ𝑜𝑡𝑜𝑛𝑠
𝑀𝑒𝑉
𝐽
×𝐸𝑝ℎ𝑜𝑡𝑜𝑛×1.6×10−13𝑀𝑒𝑉×𝜇𝑚 𝑐𝑚−1
𝑐𝑚2 /𝑠
𝑘𝑔
𝐽/𝑘𝑔
𝜌𝑚 3 ×1 𝐺𝑦
𝑐𝑚
……………………….. (2)
1000×0.3×1.6×10−13 ×0.0312
=
= 1.5×10−9 𝐺𝑦/𝑠
0.001×1
Dividing eqn (2) by (1)
𝐷̇
𝜑×𝐸×1.6×10−13 𝜇𝑚 /𝜌𝑚
=
𝜑×𝐸×1.6×10−13 𝜇𝑎 /𝜌𝑎
𝑋̇
𝜇𝑚 /𝜌𝑚
𝐷̇ = 34×
×𝑋̇ 𝐺𝑦/𝑠
𝜇𝑎 /𝜌𝑎
This is the relationship between exposure rate and dose rate.
Here, µm/𝜌m is the mass absorption co efficient of the medium & µa/𝜌a
is the mass absorption co efficient of the air.
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Thus, the radiation dose absorbed from any given exposure is
determined by the ratio of the mass absorption coefficient of the
medium to that of air.
⌂ Absorbed Dose Measurement: Bragg-Gray Principle:
According to the Bragg-Gray principle, the amount of ionization
produced in a small gas-filled cavity surrounded by a solid absorbing
medium (whose radiation absorption is similar to that of tissue) is
proportional to the energy absorbed by the solid.
i.e. ionization in the gas ∝ energy absorbed by the solid
If the cavity is surrounded by a solid of proper thickness, then the
energy absorbed per unit mass of walls, dEm/dMm, is related to the
energy absorbed per unit mass of gas in the cavity dEg/dMg, by
𝑑𝐸𝑚
𝑑𝑀𝑚
=
𝑆𝑚
𝑆𝑔
×
𝑑𝐸𝑔
𝑑𝑀𝑔
……………………….. (1)
Where, Sm ⇒ mass stopping power of the wall material
Sg ⇒ mass stopping power of the gas.
The mass rate of energy loss of a radiation while passing through a
material is called mass stopping power (S) and is defined by the
equation, 𝑆 =
𝑑𝐸/𝑑𝑥
(unit=
𝜌
𝑀𝑒𝑉/𝑐𝑚
𝑔/𝑐𝑚3
)
Where, 𝜌 is the density of the material or medium and x is the depth of
penetration of the radiation through the medium. The unit of S is
MeV.g-1.cm2.
1 𝑑𝐸
1
𝑀𝑒𝑉
[ . 𝑚→
= 𝑐𝑚2]
−1
2.
𝑆𝑚 𝑑𝑀𝑚
𝑀𝑒𝑉.𝑔
.𝑐𝑚
𝑔
Now, dEg/dMg, the energy absorbed per unit mass of the gas, is
expressed
𝑑𝐸𝑔
= 𝑤×𝐽 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑓𝑜𝑟𝑚 𝑖𝑜𝑛 𝑝𝑎𝑖𝑟𝑠
𝑑𝑀𝑔
𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑔𝑎𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑎𝑣𝑖𝑡𝑦 …………………. (2)
Where w → mean energy dissipated in the production of an ion pair in
the gas,
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J → number of ion pairs per unit mass of gas.
[Ionization in the gas ∝
𝑑𝐸𝑔
𝑑𝑀𝑔
(=
𝑆𝑔 𝑑𝐸𝑚
𝑆𝑚 𝑑𝑀𝑚
𝑑𝐸𝑚
Or, ionization in the gas ∝
𝑑𝑀𝑚
)
]
From eqn (1) and (2), we get
𝑑𝐸𝑚
𝑆
= 𝑚 ×𝑤×𝐽 = 𝜌𝑚 ×𝑤×𝐽 ………………………. (3)
𝑑𝑀𝑚
𝑆𝑔
Where, 𝜌𝑚 is the ratio of the mass stopping power of the solid relative
to that of the gas.
𝑆
[𝜌𝑚 = 𝑚 = 1.137 𝑓𝑜𝑟 1.25 𝑀𝑒𝑉 𝑓𝑜𝑟 𝑔𝑎𝑚𝑚𝑎 𝑟𝑎𝑦, 𝑖. 𝑒. 𝑆𝑚 > 𝑆𝑔 ]
𝑆𝑔
◊ Problem 4: Calculate the absorbed dose rate from the following data
on a tissue-equivalent chamber with walls of equilibrium thickness
embedded within a phantom and exposed to 60Co gamma rays for 10
minutes. The volume of the air cavity in the chamber is 1 cm3, the
capacitance is 5 μF, and the gamma-ray exposure results in a decrease
of 72 V across the chamber.
Solution:
The charge collected by the chamber is
∆𝑄 = 𝐶 × ∆𝑉 = (5×10−12 𝐹 )(72𝑉 ) = 3.6×10−10 𝐶
The number of electrons collected (= no. of ion pairs formed in the
cavity) by the chamber is
3.6×10−10 𝐶
1.6×10−19 𝐶/𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛
= 2.25×109 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠
The number of ion pairs per unit mass of the gas is
2.25×109 𝑖𝑜𝑛 𝑝𝑎𝑖𝑟𝑠
𝐽=
(1.293×10−6 𝑘𝑔/𝑐𝑚3 )(1 𝑐𝑚3 )
Cavity volume Vc = 1 cm3
Density of air, 𝜌 = 1.293×10−6 𝑘𝑔/𝑐𝑚3
Mean energy dissipated in the production of ion pair in air is
𝑤 = 34 𝑒𝑉 = 34×1.6×10−19 𝐽
11
So,
𝑑𝐸𝑚
𝑑𝑀𝑚
𝐽
= 𝜌𝑚 ×𝑤×𝐽 = 1.137× (34×1.6×10−19 )
𝑖𝑝
𝑖𝑝
×(
2.25×109 𝑘𝑔
𝑘𝑔
1.293×10−6 3
𝑐𝑚
1
).
𝐽
(
(1 𝑘𝑔 )
The dose rate is
𝑑 𝑑𝐸
0.0108 𝐺𝑦
𝐷̇ = ( 𝑚 ) =
= 0.00108
𝑑𝑡 𝑑𝑀𝑚
10 𝑚𝑖𝑛
= 0.0108 𝐺𝑦
𝐺𝑦
𝐺𝑦
𝑚𝑖𝑛
)
= 1.08 𝑚𝐺𝑦/𝑚𝑖𝑛.
Kerma: The initial kinetic energy of the primary ionizing particles (e.g.
photoelectrons, Compton electrons etc) produced by the interaction of
the incident indirectly ionizing radiation (such as X-rays, gamma rays, fast
neutrons) per unit mass of the interacting medium is called the kerma.
Kerma in SI system is measured in J/kg or Gy.
The kerma decreases continuously with increasing depth in an
absorbing medium because of the continuous decrease in the flux of the
radiation. The absorbed dose, however, increases with increasing depth
(as the density of the primary ionizing particles and the secondary
particles that they produce increases) until a maximum is reached, after
which the absorbed dose decreases with the increase in depth (fig 1).
The maximum dose occurs at a depth approximately equal to the
maximum range of primary ionizing particles.
12
Fig 1: Relationship between kerma and absorbed dose for photon
radiation
⌂ Source Srength: Specific Gamma-ray Emission
The radiation intensity from any given gamma ray source is a
measure of source strength. It is also called specific gamma ray emission,
Γ.
The gamma radiation exposure rate from a point source of unit
activity (1- MBq point source) at unit distance (1m) is called the specific
gamma ray emission. Its unit in SI system is C/Kg/MBq/hr.
The source strength may be calculated if the decay scheme of the
isotope is known. The decay scheme of 131I gives the following
information (column 1 & 2)
Along with corresponding absorption coefficients:
Photon energy, Photons per disintegration
Energy absorption
f(MeV/photon)
(photons/tps)
coefficient, µ for air(m-1)
0.723
0.016
3.8 × 10-3
0.673
0.069
3.9 × 10-3
0.503
0.003
3.8 × 10-3
13
0.326
0.177
0.365
0.284
0.080
0.164
3.8 × 10-3
3.4 × 10-3
3.8 × 10-3
3.7 × 10-3
3.2 × 10-3
3.3 × 10-3
0.002
0.002
0.853
0.051
0.051
0.006
Gamma radiation exposure level (exposure rate) for each quantum
energy, calculated by the following equation:
𝑋̇ =
𝑓
𝑝ℎ𝑜𝑡𝑜𝑛𝑠
𝑀𝑒𝑉
𝐽
106 𝑡𝑝𝑠
𝑠 𝜇
×𝐸
×1.6×10−13
×1×
×3.6×103 ×
𝑡
𝑝ℎ𝑜𝑡𝑜𝑛
𝑀𝑒𝑉
𝑀𝐵𝑞
ℎ 𝑚
𝑘𝑔
𝐽/𝑘𝑔
4𝜋(1 𝑚)2 ×𝜌 3 ×34
𝐶/𝑘𝑔
𝑚
Where,
𝑋̇ = exposure rate (C/kg/MBq/hr)
f = fraction of transformations that result in a photon of the quantum
energy under consideration
E = quantum energy (MeV)
𝜇 = linear energy absorption coefficient (m-1)
𝜌 = density of air (kg/m3)
Using 𝜌 = 1.293 kg/m3, the source strength, Γ is calculated by the
expression
Γ = ∑𝑖 𝑋̇𝑖
𝐶/𝑘𝑔
Γ = 1.043×10−6 ∑𝑖 𝑓𝑖 ×𝐸𝑖 ×𝜇𝑖
……………. (2)
𝑀𝐵𝑞.𝐻𝑟
Where
fi → photons per transformation of the ith photon
Ei → energy of the ith photon (MeV)
𝜇𝑖 → linear absorption coefficient in air of the ith photon
For quantum energies from 0.060 MeV to 2 MeV, µ varies little with
energy, and µ is about 3.5 × 10-3 m-1.
So eqn (2) becomes
𝐶/𝑘𝑔
Γ = 3.65×10−9 ∑𝑖 𝑓𝑖 ×𝐸𝑖
……………… (3)
𝑀𝐵𝑞.𝐻𝑟
14
⌂ Specific effective energy (SEE) for internally deposited radioisotopes
producing corpuscular radiation:
For an infinitely large medium containing a uniformly distributed
radioisotope, the concentration of absorbed energy must be equal to the
concentration of energy emitted by the isotope.
The energy absorbed per unit mass per transformation is called the
specific effective energy (SEE).
The health physics purposes “infinitely large” means the tissue
mass whose dimensions exceed the range of the radiation from the
distributed isotope. For ∝ and β radiation, this condition is easily met in
practice. The SEE for ∝ and β radiation is simply the average energy of
the radiation divided (per transformation) by the mass of the tissue in
which it is distributed:
𝐸 (𝛼 𝑜𝑟 𝛽) 𝑀𝑒𝑉/𝑡𝑝𝑠
(
)
𝑆𝐸𝐸 𝛼 𝑜𝑟 𝛽 =
𝑚
𝑘𝑔
◊ Problem 5: Calculate the daily dose rate to a tissue (an organ) that
weighs 18 g and has 6660 Bq of 35S uniformly distributed throughout the
organ. Sulfur is a pure beta emitter whose maximum-energy beta
particle is 0.1674 MeV and whose average energy is 0.0488 MeV.
Solution:
Given,
q = 6660 Bq
𝐸
0.0488 𝑀𝑒𝑉
𝑆𝐸𝐸 = =
𝑚
0.018 𝑘𝑔
The β-ray dose rate from q Bq uniformly distributed in m kg of tissue is
𝐷̇(𝛽 ) =
=
𝑀𝑒𝑉/𝑡𝑝𝑠
)×(1.6×10−13
𝑘𝑔
𝐽/𝑘𝑔
1 𝐺𝑦
6660×(0.0488)×(1.6×10−13)×(8.64×104 )
(𝑞 𝐵𝑞)×(1 𝑡𝑝𝑠/𝐵𝑞)×(𝑆𝐸𝐸
0.018×1
𝐽/𝑀𝑒𝑉)×(8.64×104 𝑠/𝑑𝑎𝑦)
= 2.5×10−4 𝐺𝑦/𝑑𝑎𝑦
15
⌂ Effective half-life:
The total dose absorbed during any given time interval after the
deposition of the sulfur in any organ (or in the testis) may be calculated
by integrating the dose rate over the required time interval.
𝑡=5 𝑦𝑒𝑎𝑟𝑠 𝑑𝐷
[e.g. Total dose absorbed = ∫𝑡=0
𝑑𝑡]
𝑑𝑡
Two factors must be considered for the calculation of the total dose
absorbed:
(a)
In situ, radioactive decay of the isotope → decided by
disintegration constant, λR.
(b)
Biological elimination of the isotope → decided by integration
constant, λB(called the biological elimination constant).
In most instances, biological elimination (disposal) follows first-order
kinetics, and is calculated by 𝑒 − 𝜆𝐵 𝑡 with time t.
The quantity of radioisotope within an organ at any time t after
deposition of a quantity Q0 is given by
𝑄(𝑡) = (𝑄0 𝑒 −𝜆𝑅𝑡 )(𝑒 − 𝜆𝐵 𝑡 ) ……………….. (1)
Where, 𝜆𝑅 → radioactive decay constant
𝜆𝐵 → biological elimination constant
Or, 𝑄(𝑡) = (𝑄0 𝑒 −(𝜆𝑅+𝜆𝐵)𝑡 ) ……………………. (2)
Or, 𝑄(𝑡) = (𝑄0 𝑒 −𝜆𝐸𝑡 ) …………………………. (3)
Where, 𝜆𝐸 = (𝜆𝑅 + 𝜆𝐵 ) is called the effective elimination constant.
The effective half-life is defined by
0.693
𝑇𝐸 =
Or,
1
𝑇𝐸
=
Or, 𝑇𝐸 =
𝜆𝐸
=
0.693
𝑇𝑅 𝑇𝐵
𝑇𝑅 +𝑇𝐵
𝜆𝐸
𝜆𝑅
0.693
+
𝜆𝐵
0.693
=
……………… (4)
Where, 𝑇𝐵 = biological half-life
𝑇𝑅 = half-life due to radioactive decay
1
𝑇𝑅
+
1
𝑇𝐵
16
◊ Problem 6: Calculate the effective half-life and effective half-life
elimination of pure beta emitter 35S uniformly distributed throughout an
organ. Given, Radioactive half-life, TR = 87.1 days & Biological half-life, T
= 623 days for beta emission & biological elimination, respectively.
Solution:
WE know,
𝑇 𝑇
87.1×623
𝑇𝐸 = 𝑅 𝐵 =
= 76.4 𝑑𝑎𝑦𝑠
𝑇𝑅 +𝑇𝐵
87.1+623
The effective elimination constant is
0.693
0.693
𝜆𝐸 =
=
= 0.009 𝑑𝑎𝑦 −1
𝑇𝐸
76.4
⌂ Find an expression for the total dose in case of multi-compartment
scheme in terms of effective clearance rates.
Total Dose: Dose Commitment:
The dose, dD, during an infinitesimally small period of time dt,
between t and (t+dt) after an initial dose rate𝐷̇0 is
𝑑𝐷 = (𝐷0̇ 𝑒 −𝜆𝐸𝑡 )dt ……………………. (1)
The total dose during a time interval 𝜏 after deposition of the isotope is
𝛕
𝐷 = 𝐷̇0 ∫0 𝑒 −𝜆𝐸𝑡 dt …………………. (2)
𝑒
𝐷 = 𝐷̇0 [
=
𝐷̇0
𝜆𝐸
−𝜆𝐸 𝑡
−𝜆𝐸
]𝛕0
[1 − 𝑒 −𝜆𝐸τ ] ……………………. (3)
If τ = ∞ (i.e. when the isotope is completely gone)
𝐷̇
We have, 𝐷 = 0 = 𝐷0̇ ×𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑖𝑓𝑒, 𝑇𝑎
(since, 𝑇𝑎 =
𝜆𝐸
1
𝜆𝐸
)
For practical purposes, an “infinitely long time” corresponds to about
6(six) effective half-life (TE).
Infinite time ≈ 6 𝑇𝐸
The dose due to total decay is equal to the product of the initial
dose rate, 𝐷0̇ , and the average life of the radioisotope within the organ,
1/𝜆𝐸 , and this total dose is called the dose commitment of that organ.
17
The dose commitment can also be defined as the absorbed dose from a
given particle or from a given exposure.
Eqn (3) represents the total dose during a time interval τ after
deposition of radioisotope in a single compartment. Generally, if there is
more than one compartment, the body burden (quantity of
radioisotope) at any time τ after deposition of Q0 amount of the isotope
is given by
𝑄(τ) = 𝑄10 𝑒 −𝜆1𝐸τ + 𝑄20 𝑒 −𝜆2𝐸τ + ⋯ … … … … . . + 𝑄𝑛0 𝑒 −𝜆𝑛𝐸τ…………. (4)
Where, 𝑄10 , 𝑄20 , … … … . ., 𝑄𝑛0 = amount deposited in compartments 1,
2, ……….., n, 𝑄0 = 𝑄10 + 𝑄20 + ⋯ … … … + 𝑄𝑛0 and 𝜆1𝐸 , 𝜆2𝐸 , ….., 𝜆𝑛𝐸 are
effective clearance rates for compartments 1, 2, ………, n.
Since the ability in each compartment contributes to the dose to
that organ or tissue, eq (3) becomes for the multi-compartment case,
𝐷 (𝑡 ) =
𝐷̇10
𝜆1𝐸
(1 − 𝑒 −𝜆1𝐸τ ) +
𝐷̇20
(1 − 𝑒 −𝜆2𝐸τ ) + ⋯ . +
𝜆2𝐸
𝐷̇𝑛0
𝜆𝑛𝐸
(1 − 𝑒 −𝜆𝑛𝐸τ )
…………. (5)
When the radionuclide has completely benn eliminated, eqn (5) reduces
to
𝐷 (𝑡 ) =
𝐷̇10
𝜆1𝐸
+
𝐷̇20
+ ⋯.+
𝜆2𝐸
𝐷̇𝑛0
𝜆𝑛𝐸
………………… (6)
◊ Problem 7: Calculate the total absorbed dose during the first 5 days
after deposition of the radio sulfur in an organ, and also calculate the
dose commitment.
Given, initial dose rate to the organ, 𝐷̇0 = 2.5×10−4 𝐺𝑦/𝑑𝑎𝑦 & effective
elimination constant, 𝜆𝐸 = 0.009 𝑑𝑎𝑦 −1 .
Solution:
Here,
𝐷𝑡𝑜𝑡 =
𝐷̇0
𝜆𝐸
=
[1 − 𝑒 −𝜆𝐸τ ]
𝐺𝑦
2.5×10−4 𝑑𝑎𝑦
−0.009 𝑑𝑎𝑦
−1 (1 − 𝑒
0.009 𝑑𝑎𝑦
−3
= 1.2×10
Again,
𝐷𝑐𝑜𝑚𝑚𝑖𝑡 =
𝐷̇0
𝜆𝐸
=
−1 ×5 day
𝐺𝑦 = 0.0012 𝐺𝑦
𝐺𝑦
2.5×10−4 𝑑𝑎𝑦
0.009 𝑑𝑎𝑦 −1
= 0.028 𝐺𝑦.
)
18
⌂ Show that the dose rate at the center of a spherical tissue of radius
𝟒𝝅
R for internally deposited γ-emitting radioisotope is 𝑫 = 𝑪𝛤 (𝟏 −
𝝁
𝒆
−𝝁𝑹)
).
Dose rate for gamma-emitters:
Gamma ray can travel great distances within tissue, and even can
leave the tissue without interacting. So, we cannot simply calculate the
absorbed dose by assuming the organ to be infinitely large.
Figure:1
For a uniformly distributed gamma emitting isotope within tissue,
the dose rate at any point p due to the isotope in the infinitesimal volume
dV located at a distance r from the point p is
𝑑𝐷 = 𝐶𝛤
Where,
𝑒 −𝜇𝑟
𝑟2
𝑑𝑉 …………………. (1)
𝐶 → concentration of the isotope
𝛤 → specific gamma-ray emission/source strength &
𝜇 → linear energy absorption coefficient
The dose rate at point p due to all isotope in the tissue
𝐷 = ∫ 𝑑𝐷 = ∫ 𝐶𝛤
𝑒 −𝜇𝑟
𝑟2
𝑑𝑉 =
𝑉 𝑒 −𝜇𝑟
𝐶𝛤 ∫0 2 𝑑𝑉
𝑟
………..(2)
19
Fig 02: Geometry for evaluating eqn (2) for the center of a sphere
In case of a sphere (fig-02), the dose rate at the center is
𝐷=
𝑟=𝑅 𝜃=𝜋/2 𝜑=𝜋 𝑒 −𝜇𝑟
4{𝐶𝛤 ∫𝑟=0 ∫𝜃=0 ∫𝜑=0 2 (𝑟𝑑𝜃)(𝑟𝑐𝑜𝑠𝜃𝑑𝜑)(𝑑𝑟)}
𝑟
𝑟=𝑅
𝜃=𝜋/2
= 4𝐶𝛤 ∫𝑟=0 𝑒 −𝜇𝑟 𝑑𝑟 ∫𝜃=0
= 4𝐶𝛤(
𝑒 −𝜇𝑅
−𝜇
−
𝑒 −𝜇×0
−𝜇
∴𝑫=
𝜑=𝜋
𝑐𝑜𝑠𝜃𝑑𝜃 ∫𝜑=0 𝑑𝜑
𝜋
)(sin − sin 0𝑜 )(𝜋 − 0)
2
𝟒𝝅𝑪𝜞
𝝁
(𝟏 − 𝒆−𝝁𝑹 ) ……………. (3)
[showed]