Dosimetry
Crister Ceberg
Medical Radiation Physics
Lund University
Sweden
Part 1 – Introduction
After this lecture you will be able to:
• Define the scientific field of dosimetry
• Describe the scope of dosimetric quantities
• Outline the content of the following lectures on dosimetry
The field of dosimetry
• Ionizing radiation may produce biological effects in living
matter
• Dosimetric quantities are introduced to provide a physical
measure to correlate with biological effects
• Dosimetry is concerned with the definition, calculation and
measurement of dosimetric quantities
The scope of dosimetric quantities
Stochastic
biological effect
Dosimetric quantity
Deterministic
biological effect
Dosimetric quantity
The scope of dosimetric quantities
• Incident ionizing radiation
– Charged particles (e.g. electrons or protons) or
uncharged particles (e.g. photons or neutrons)
– Produce ionizations in a medium (directly or indirectly)
– Practical threshold value Eionizing>Ethreshold≈10 eV
Uncharged
particle
Charged
particle
The scope of dosimetric quantities
• Conversion of energy (blue blobs)
– Energy is transferred to secondary particles
Uncharged
particle
Charged
particle
The scope of dosimetric quantities
• Conversion of energy (blue blobs)
– Energy is transferred to secondary particles
• Deposition of energy (yellow blobs)
– Not re-emitted by ionizing particles
Uncharged
particle
Charged
particle
Content of the following lectures
• Definitions
– Dosimetric quantities
– ICRU 85
• Calculations
– The product of radiometric quantities and interaction coefficients
– Radiation equilibrium
• Measurements
– Detectors and cavity theory
– Perturbation factors
Summary
• Dosimetry is concerned with the definition, calculation and
measurement of dosimetric quantities
• Dosimetric quantities describe how the energy of ionizing
radiation is converted to secondary particles and deposited in
matter
• In the following lectures we will define dosimetric quantities
and discuss the fundamentals of radiation equilibrium and
cavity theory.
Part 2 – conversion of energy
After this lecture you will be able to:
• Define the dosimetric quantities given in ICRU 85, especially
those relating to the conversion of energy
• Perform simple calculations of conversion quantities
Dosimetric quantities and units
• Conversion of energy (blue blobs)
– Kerma
– Exposition
– Cema
Uncharged
particle
Charged
particle
Energy transfer
Scattered
photon, e1
Incoming
photon, ein
Secondary
electron, e2
Example: Compton scattering
Mean energy transferred: 𝑓𝑖
Energy transfer
Mean energy transferred: 𝑓𝑖
All interaction types: 𝑓 =
𝑖 𝑓𝑖 𝜎𝑖
𝑖 𝜎𝑖
Energy transfer
V
Mean energy transferred: 𝑓𝑖
All interaction types: 𝑓 =
𝑖 𝑓𝑖 𝜎𝑖
𝑖 𝜎𝑖
𝜇
𝐸𝑡𝑟 = Φ𝜇𝑉𝜀𝑖𝑛 𝑓 = Ψ 𝑚𝑓
𝜌
Energy transfer
V
Mean energy transferred: 𝑓𝑖
All interaction types: 𝑓 =
𝑖 𝑓𝑖 𝜎𝑖
𝑖 𝜎𝑖
𝜇
𝐸𝑡𝑟 = Φ𝜇𝑉𝜀𝑖𝑛 𝑓 = Ψ 𝑚𝑓
𝜌
𝐸𝑡𝑟,𝑛𝑒𝑡 = 𝐸𝑡𝑟 1 − 𝑔
Kerma
• The kerma, K, for ionizing uncharged particles, is the quotient
of dEtr by dm, where dEtr is the mean sum of the initial kinetic
energies of all the charged particles liberated in a mass dm of
a material by the uncharged particles incident on dm (ICRU
85, 2011)
𝑑𝐸𝑡𝑟
𝐾=
𝑑𝑚
• Unit: J kg-1
• Special name: Gy (gray)
Kerma
𝐸𝑡𝑟
𝜇
= Ψ 𝑚𝑓
𝜌
𝜇𝑡𝑟 𝜇
= 𝑓
𝜌
𝜌
𝑑𝐸𝑡𝑟
𝜇𝑡𝑟
𝐾=
=Ψ
𝑑𝑚
𝜌
𝐾𝑐𝑜𝑙
𝜇𝑒𝑛
=𝐾 1−𝑔 =Ψ
𝜌
Kerma
𝐾=
𝜇𝑡𝑟
Ψ𝐸
𝑑𝐸
𝜌
Ψ𝐸 for a 10 MV beam
From: ESM Ali and DWO Rogers,
Phys. Med. Biol. 57 (2012)
𝐾𝑐𝑜𝑙 =
𝜇𝑒𝑛
Ψ𝐸
𝑑𝐸
𝜌
𝜇𝑒𝑛
for water
𝜌
From : www.nist.gov
Kerma
𝐾 = Φ𝑘Φ
𝑘Φ = 𝐾 Φ
𝑘Φ in ICRU−muscle tissue
From: MB Chadwick et al.,
Med. Phys. 26 (1999)
Kerma rate
• The kerma rate, 𝐾, is the quotient of dK by dt, where dK is the
increment of kerma in the time interval dt (ICRU 85, 2011)
𝑑𝐾
𝐾=
𝑑𝑡
• Unit: J kg-1 s-1
• Special name: Gy s-1 (gray per second)
Do you remember?
A water tank is exposed to 1.25 MeV gamma-radiation.
At a certain point, the fluence rate is 3.5·109 cm-2s-1.
What is the value of Kcol at this point after one hour?
Please write down your answer before moving on to the next
page.
Were you right?
A water tank is exposed to 1.25 MeV gamma-radiation.
At a certain point, the fluence rate is 3.5·109 cm-2s-1.
What is the value of Kcol at this point after one hour?
Solution:
Known data:
Ψ = Φ ℎ𝜈 𝑡 = 3.5 ∙
109 𝑐𝑚−2 𝑠 −1
∙ 1.25𝑀𝑒𝑉 ∙ 1.6 ∙
10−13
𝐽
∙ 3600𝑠 = 2.520 𝐽𝑐𝑚−2
𝑀𝑒𝑉
From www.nist.gov:
𝜇𝑒𝑛
𝑔
= 2.965 ∙ 10−2 𝑐𝑚2 𝑔−1 ∙ 1000
= 29.65𝑐𝑚2 𝑘𝑔−1
𝜌
𝑘𝑔
Thus:
𝐾𝑐𝑜𝑙 = Ψ
𝜇𝑒𝑛
𝜌
= 2.520 𝐽𝑐𝑚−2 ∙ 29.65𝑐𝑚2 𝑘𝑔−1 ⇒ 𝐾𝑐𝑜𝑙 = 74.7 𝐺𝑦
Do you remember?
A water tank is exposed to a 14.5 MeV neutron beam.
At a certain point, the kerma is 74.7 Gy.
What is the neutron fluence at this point?
Please write down your answer before moving on to the next
page.
Were you right?
A water tank is exposed to a 14.5 MeV neutron beam.
At a certain point, the kerma is 74.7 Gy.
What is the neutron fluence at this point?
Solution:
Known data:
𝐾 = 74.7𝐺𝑦
From Caswell et al., Rad. Res. 83:217, 1980:
𝑘Φ = 0.709 ∙ 10−10 𝐺𝑦𝑐𝑚2
Thus:
Φ=
𝐾
74.7 𝐺𝑦
12 𝑐𝑚−2
=
⇒
Φ
=
1.05
∙
10
𝑘Φ 0.709 ∙ 10−10 𝐺𝑦𝑐𝑚2
Exposure
• The exposure, X, is the quotient of dq by dm, where dq is the
absolute value of the mean total charge of the ions of one
sign produced when all the electrons and positrons liberated
or created by photons incident on a mass dm of dry air are
completely stopped in dry air (ICRU 85, 2011)
𝑑𝑞
𝑋=
𝑑𝑚
• Unit: C kg-1
Exposure vs. kerma
Mean energy per ion pair: 𝑊
𝑑𝑞 𝑑𝐸𝑡𝑟
1
𝜇𝑡𝑟
𝑋=
=
1−𝑔
=Ψ
1−𝑔
𝑊
𝑑𝑚
𝑑𝑚
𝜌
𝑒
𝜇𝑒𝑛 1
1
X=Ψ
= 𝐾𝑎𝑖𝑟,𝑐𝑜𝑙
𝑊
𝜌 𝑊 𝑒
𝑒
𝑊 = 33.97𝑒𝑉
−1
𝑊
=
33.97
J𝐶
𝑒
1
𝑊
𝑒
Exposure rate
• The exposure rate, 𝑋, is the quotient of dX by dt, where dX is
the increment of exposure in the time interval dt (ICRU 85,
2011)
𝑑𝑋
𝑋=
𝑑𝑡
• Unit: C kg-1 s-1
Cema
• The cema, C, for ionizing charged particles, is the quotient of
dEel by dm, where dEel is the mean energy lost in electronic
interactions in a mass dm of a material by the charged
particles, except secondary electrons, incident on dm (ICRU
85, 2011)
𝑑𝐸𝑒𝑙
𝐶=
𝑑𝑚
• Unit: J kg-1
• Special name: Gy (gray)
Cema
Scattered
electron, e1
Incoming
electron, ein
Secondary
electron, e2
Example: coulomb interaction
𝑑𝐸𝑒𝑙
𝑆𝑒𝑙
C=
=Φ
𝑑𝑚
𝜌
Cema
𝐶=
𝑆𝑒𝑙
Φ𝐸
𝑑𝐸 =
𝜌
Φ𝐸 for linac electron beams
From: S Righi et al.,
JACMP 14 (2013)
𝐿∞
Φ𝐸
𝑑𝐸
𝜌
𝑆𝑒𝑙
for water
𝜌
From : www.nist.gov
Restricted cema
Scattered
electron, e1
Incoming
electron, ein
Secondary
electron, e2
Example: coulomb interaction
𝐶∆ =
𝐿∆
Φ′𝐸 𝑑𝐸
𝜌
e2>D: included in F’E
e2<D: included in LD
Cema rate
• The cema rate, 𝐶, is the quotient of dC by dt, where dC is the
increment of cema in the time interval dt (ICRU 85, 2011)
𝑑𝐶
𝐶=
𝑑𝑡
• Unit: J kg-1 s-1
• Special name: Gy s-1 (gray per second)
Summary
• Conversion of energy takes place when energy of primary
particles is transferred to secondary particles
• Three important dosimetric quantities defined in the ICRU
report 85 for conversion of energy
– Kerma
– Exposure
– Cema
• The restricted cema is applicable when the energy transport
with high-energy delta particles cannot be disregarded
Part 3 – deposition of energy
After this lecture you will be able to:
• Define the dosimetric quantities given in ICRU 85, especially
those relating to the deposition of energy
• Perform simple calculations of deposition quantities
Dosimetric quantities and units
• Deposition of energy (yellow blobs)
– Energy deposit
– Energy imparted
– Absorbed dose
Uncharged
particle
Charged
particle
Energy deposit
• The energy deposit, 𝜀𝑖 , is the energy deposited in a single
interaction, i,
𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄
where 𝜀𝑖𝑛 is the energy of the incident ionizing particle
(excluding rest energy), 𝜀𝑜𝑢𝑡 is the sum of the energies of all
charged and uncharged ionizing particles leaving the
interaction (excluding rest energy), and Q is the change in rest
energies of the nucleus and of all elementary particles
involved in the interaction (ICRU 85, 2011)
• Unit: J
Energy deposit
Scattered
electron, e1
Incoming
electron, ein
Secondary
electron, e2
𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄
Example: coulomb interaction, Q=0
Energy deposit
hn
Scattered
electron, e1
Incoming
electron, ein
Secondary
electron, e2
EA
𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄
𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀1 + 𝜀2 + ℎ𝜈 + 𝐸𝐴
Example: coulomb interaction, Q=0
Energy deposit
• Energy deposit may appear as
– Visible light
– Chemical bindning energy
– Heat
• Stochastic quantity
– Subject to random fluctuations
– Associated with a probability distribution
Energy imparted
• The energy imparted, 𝜀, to the matter in a given volume is the
sum of all energy deposits in the volume
𝜀=
𝜀𝑖
𝑖
where the summation is performed over all energy deposits,
𝜀𝑖 , in that volume (ICRU 85, 2011)
• Unit: J
Energy imparted
𝜀=
𝜀𝑖
𝑖
𝜀=
𝜀𝑖𝑛 −
𝑖
𝜀𝑜𝑢𝑡 +
𝑖
𝑄
𝑖
Energy imparted
f(e)
e
e
Energy imparted
Rin
Rout
𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄
Energy imparted
𝑑𝐴
Ψ
𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄
𝜀=−
𝜀=−
Ψ𝑑 𝐴 + Σ𝑄
𝑑𝑖𝑣 Ψ𝑑𝑉 + Σ𝑄
Energy imparted
𝑑𝐴
Ψ
𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄
𝜀=−
𝜀=−
Ψ𝑑 𝐴 + Σ𝑄
𝑑𝑖𝑣 Ψ𝑑𝑉 + Σ𝑄
𝑑𝜀
1
1 𝑑(Σ𝑄)
= − 𝑑𝑖𝑣Ψ +
𝑑𝑚
𝜌
𝜌 𝑑𝑉
Do you remember?
A 6 MeV photon interacts by compton scattering in a given volume.
The secondary electron spend half its kinetic energy in electronic
collisions and one fourth in radiation losses. The scattered photon
carries 4 MeV when it escapes the volume. What are the values of
energy transfer, net energy transfer, and energy imparted?
Please write down your answer before moving on to the next page.
Were you right?
A 6 MeV photon interacts by compton scattering in a given volume.
The secondary electron spend half its kinetic energy in electronic
collisions and one fourth in radiation losses. The scattered photon
carries 4 MeV when it escapes the volume. What are the values of
energy transfer, net energy transfer, and energy imparted?
𝜀𝑜𝑢𝑡,1 = 4 𝑀𝑒𝑉
Solution:
𝜀𝑖𝑛 = 6 𝑀𝑒𝑉
𝐸𝑡𝑟 = 2 𝑀𝑒𝑉
Were you right?
A 6 MeV photon interacts by compton scattering in a given volume.
The secondary electron spend half its kinetic energy in electronic
collisions and one fourth in radiation losses. The scattered photon
carries 4 MeV when it escapes the volume. What are the values of
energy transfer, net energy transfer, and energy imparted?
𝜀𝑜𝑢𝑡,1 = 4 𝑀𝑒𝑉
Solution:
𝜀𝑖𝑛 = 6 𝑀𝑒𝑉
𝐸𝑡𝑟 = 2 𝑀𝑒𝑉
𝐸𝑡𝑟,𝑛𝑒𝑡 = 1.5 𝑀𝑒𝑉
0.5 𝑀𝑒𝑉
Were you right?
A 6 MeV photon interacts by compton scattering in a given volume.
The secondary electron spend half its kinetic energy in electronic
collisions and one fourth in radiation losses. The scattered photon
carries 4 MeV when it escapes the volume. What are the values of
energy transfer, net energy transfer, and energy imparted?
𝜀𝑜𝑢𝑡,1 = 4 𝑀𝑒𝑉
Solution:
𝜀𝑖𝑛 = 6 𝑀𝑒𝑉
1 𝑀𝑒𝑉
0.5 𝑀𝑒𝑉
𝐸𝑡𝑟 = 2 𝑀𝑒𝑉
𝐸𝑡𝑟,𝑛𝑒𝑡 = 1.5 𝑀𝑒𝑉
𝜀 = 1.0 𝑀𝑒𝑉
𝜀𝑜𝑢𝑡,2 = 0.5 𝑀𝑒𝑉
Absorbed dose
• The absorbed dose, D, is the quotient of 𝑑𝜀 by dm, where 𝑑 𝜀
is the mean energy imparted by ionizing radiation to matter of
mass dm (ICRU 85, 2011)
𝑑𝜀
𝐷=
𝑑𝑚
• Unit: J kg-1
• Special name: Gy (gray)
Absorbed dose rate
• The absorbed dose rate, 𝐷, is the quotient of dD by dt, where
dD is the increment of absorbed dose in the time interval dt
(ICRU 85, 2011)
𝑑𝐷
𝐷=
𝑑𝑡
• Unit: J kg-1 s-1
• Special name: Gy s-1 (gray per second)
Summary
• Deposition of energy takes place when energy is locally
absorbed
• Three important dosimetric quantities defined in the ICRU
report 85 for deposition of energy
– Energy deposit
– Energy imparted
– Absorbed dose
• There is a close relation between radiation transport and
dosimetry
Part 4 – radiation equilibrium
After this lecture you will be able to:
• Define different types of radiation equilibrium and discuss its
consequences for absorbed dose calculations
• Perform simple calculations of absorbed dose under different
types of radiation equilibrium
Radiation equilibrium
Rin
Rout
𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡
(𝑑𝑖𝑣 Ψ = 0)
Radiation equilibrium
Rin
Rout
𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡
𝑑𝑖𝑣Ψ = 0
𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄
𝑑𝜀
1 𝑑(Σ𝑄)
⇒ 𝐷=
=
𝑑𝑚 𝜚 𝑑𝑉
Radiation equilibrium
Rin
Distance greater than
range of particle
Rout
𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡
(𝑑𝑖𝑣 Ψ = 0)
Radiation equilibrium
Rin
Rout
𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡
𝑅𝑖𝑛,𝑖 = 𝑅𝑜𝑢𝑡,𝑖
Distance greater than
range of particle i
(𝑑𝑖𝑣 Ψ = 0)
𝑑𝑖𝑣Ψ𝑖 = 0
Partial radiation equilibrium
Charged particle equilibrium (CPE): 𝑅𝑖𝑛,𝑐 = 𝑅𝑜𝑢𝑡,𝑐
𝑑𝑖𝑣Ψ𝑐 = 0
d-particle equilibrium:
𝑅𝑖𝑛,𝛿 = 𝑅𝑜𝑢𝑡,𝛿
Partial d-particle equilibrium:
𝑅𝑖𝑛,𝛿,Δ = 𝑅𝑜𝑢𝑡,𝛿,Δ (𝑑𝑖𝑣Ψ𝛿,Δ = 0)
(𝑑𝑖𝑣Ψ𝛿 = 0)
Charged particle equilibrium (CPE)
Incoming beam of
uncharged particles
Charged particle equilibrium (CPE)
Incoming beam of
uncharged particles
𝑅𝑜𝑢𝑡,𝑢
𝑅𝑖𝑛,𝑐
𝑅𝑖𝑛,𝑢
𝑅𝑜𝑢𝑡,𝑐
Distance greater than
range of charged particles
𝜀 ≈ 𝐸𝑡𝑟
Charged particle equilibrium (CPE)
Incoming beam of
uncharged particles
𝑅𝑜𝑢𝑡,𝑢,𝑛𝑜𝑛𝑟𝑎𝑑
𝑅𝑖𝑛,𝑐
𝑅𝑖𝑛,𝑢
Distance greater than
range of charged particles
𝑅𝑜𝑢𝑡,𝑢
𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑
𝑅𝑜𝑢𝑡,𝑐
𝜀 = 𝐸𝑡𝑟 − 𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑
Charged particle equilibrium (CPE)
𝜀 = 𝑅𝑖𝑛,𝑢 + 𝑅𝑖𝑛,𝑐 − 𝑅𝑜𝑢𝑡,𝑢 − 𝑅𝑜𝑢𝑡,𝑐 + Σ𝑄
𝑅𝑖𝑛,𝑐 = 𝑅𝑜𝑢𝑡,𝑐
𝜀 = 𝑅𝑖𝑛,𝑢 − 𝑅𝑜𝑢𝑡,𝑢 + Σ𝑄
𝜀 = 𝑅𝑖𝑛,𝑢 − 𝑅𝑜𝑢𝑡,𝑢,𝑛𝑜𝑛𝑟𝑎𝑑 − 𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑 + Σ𝑄
𝜀 = 𝐸𝑡𝑟 − 𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑
𝜀 = 𝐸𝑡𝑟 (1 − 𝑔)
Charged particle equilibrium (CPE)
𝑑𝜀
𝑑𝐸𝑡𝑟
𝐷=
=
1−𝑔 = 𝐾 1−𝑔
𝑑𝑚
𝑑𝑚
𝐷 = 𝐾𝑐𝑜𝑙
Charged particle equilibrium (CPE)
Without attenuation
Range of
charged particle
With attenuation
Depth
Range of
charged particle
Collision kerma, Kcol
Absorbed dose, D
Depth
Charged particle equilibrium (CPE)
With attenuation
𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 − 𝑥
𝐾𝑐𝑜𝑙 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 − 𝑥 𝑒 −𝜇𝑥
𝑥
𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 𝑒 𝜇𝑥
𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 (1 + 𝜇𝑥)
d
Collision kerma, Kcol
Absorbed dose, D
Depth
Charged particle interactions
𝜀=
𝜀𝑖
𝑖
1
𝜀=𝑁
𝑁
𝜀 = 𝑁𝜀𝑖
𝜀𝑖 = 𝑁𝜀𝑖
𝑖
Charged particle interactions
𝑑𝜀
𝑑𝑁
𝐷=
=
𝜀𝑖
𝑑𝑚 𝑑𝑚
𝑑𝑁
𝑁𝐴
=Φ 𝜎
𝑑𝑚
𝑀
𝑁𝐴
𝐷 = Φ 𝜎𝜀𝑖
𝑀
Charged particle interactions
𝑑𝜀
𝑑𝑁
𝐷=
=
𝜀𝑖
𝑑𝑚 𝑑𝑚
𝑑𝑁
𝑁𝐴
=Φ 𝜎
𝑑𝑚
𝑀
𝑁𝐴
𝐷 = Φ 𝜎𝜀𝑖
𝑀
𝑆𝑒𝑙
𝐷=Φ
𝑘
𝜌
Delta-particle equilibrium
Charged particle
fluence
𝑅𝑜𝑢𝑡,𝑐
𝑅𝑖𝑛,𝛿
𝑅𝑖𝑛,𝑐
𝑅𝑜𝑢𝑡,𝛿
Distance greater than
range of delta particles
𝑘=1
Delta-particle equilibrium
k=1
𝑆𝑒𝑙
𝑆𝑒𝑙
𝐷=Φ
𝑘=Φ
𝜌
𝜌
𝐷=𝐶
Partial delta-particle equilibrium
Charged particle
fluence
𝑅𝑜𝑢𝑡,𝑐
𝑅𝑖𝑛,𝛿,∆
𝑅𝑖𝑛,𝑐
Distance greater than
range of slow delta particles
𝑘<1
𝑅𝑜𝑢𝑡,𝛿,∆
Partial delta-particle equilibrium
k<1
𝑆𝑒𝑙
𝐿∆
𝐷=Φ
𝑘 = Φ′
𝜌
𝜌
𝐷 = 𝐶∆
Do you remember?
A water tank is exposed to 1.25 MeV gamma-radiation.
At a certain point, the fluence is 2.1·1011 cm-2.
Assuming CPE, what is the absorbed dose at this point?
Please write down your answer before moving on to the next
page.
Were you right?
A water tank is exposed to 1.25 MeV gamma-radiation.
At a certain point, the fluence is 2.1·1011 cm-2.
Assuming CPE, what is the absorbed dose at this point?
Solution:
Known data:
Ψ = Φ ℎ𝜈 = 2.1 ∙
1011 𝑐𝑚−2 𝑠 −1
∙ 1.25𝑀𝑒𝑉 ∙ 1.6 ∙
10−13
𝐽
= 4.20 ∙ 10−2 𝐽𝑐𝑚−2
𝑀𝑒𝑉
From www.nist.gov:
𝜇𝑒𝑛
𝑔
= 2.965 ∙ 10−2 𝑐𝑚2 𝑔−1 ∙ 1000
= 29.65𝑐𝑚2 𝑘𝑔−1
𝜌
𝑘𝑔
Given that CPE exists, we get:
𝐷 = 𝐾𝑐𝑜𝑙 = Ψ
𝜇𝑒𝑛
𝜌
= 4.20 ∙ 10−2 𝐽𝑐𝑚−2 ∙ 29.65𝑐𝑚2 𝑘𝑔−1 ⇒ 𝐷 = 1.25 𝐺𝑦
Do you remember?
A thin disc of lithium fluoride (LiF) is irradiated by an electron
fluence of 4.1·109 cm-2 with energy 6 MeV. Assuming deltaparticle equilibrium, what is the absorbed dose in the disc?
Please write down your answer before moving on to the next
page.
Were you right?
A thin disc of lithium fluoride (LiF) is irradiated by an electron
fluence of 4.1·109 cm-2 with energy 6 MeV. Assuming deltaparticle equilibrium, what is the absorbed dose in the disc?
Solution:
Known data:
Φ = 4.1 ∙ 109 𝑐𝑚−2
From www.nist.gov:
𝑆𝑒𝑙
𝑔
𝐽
= 1.547 𝑀𝑒𝑉𝑐𝑚2 𝑔−1 ∙ 1000
∙ 1.6 ∙ 10−13
= 2.48 ∙ 10−10 𝐽𝑐𝑚2 𝑘𝑔−1
𝜌
𝑘𝑔
𝑀𝑒𝑉
Given that delta-particle equilibrium exists, we get:
D=C=Φ
𝑆𝑒𝑙
𝜌
= 4.1 ∙ 109 𝑐𝑚−2 ∙ 2.48 ∙ 10−10 𝐽𝑐𝑚2 𝑘𝑔−1 ⇒ 𝐷 = 1.02 𝐺𝑦
Summary
• Radiation equilibrium can only be approximate in a medium
exposed to external irradiation
• Different types of partial radiation equilibrium have important
consequences for absorbed dose calculations
– Charged particle equilbrium (CPE)
– Delta-particle equilibrium
– Partial delta-particle equilibrium
• Unless CPE exists, absorbed dose calculations need to account
for charged particle interactions
Part 5 – Cavity theory
After this lecture you will be able to:
• Discuss absorbed dose measurements in the context of cavity
theory
• Describe ideal cases of large and small cavities, and discuss
the case of mid-size cavities in terms of charged particle
fluence
• Perform simple calculations based on the Bragg-Gray theory
• Describe the particular case of a gas-filled ion chamber, and
explain the concept of perturbation factor
Cavity theory
Incoming beam of
ionizing particles
𝐷𝑚𝑒𝑑
Cavity theory
Incoming beam of
ionizing particles
𝐷𝑑𝑒𝑡 = 𝑐𝑀
Cavity theory
Incoming beam of
ionizing particles
𝐷𝑚𝑒𝑑 = 𝐷𝑑𝑒𝑡
𝐷𝑚𝑒𝑑
𝐷𝑑𝑒𝑡
Ideal case – large detector
Charged particle equilibrium (CPE): 𝐷 = 𝐾𝑐 = Ψ
𝜇
Ψ𝑚𝑒𝑑 𝜌𝑒𝑛
𝐷𝑚𝑒𝑑
𝑚𝑒𝑑
=
𝜇
𝐷𝑑𝑒𝑡
Ψ𝑑𝑒𝑡 𝑒𝑛
𝜌 𝑑𝑒𝑡
Incoming beam of
ionizing particles
𝜇𝑒𝑛
𝜌
Ideal case – large detector
Charged particle equilibrium (CPE): 𝐷 = 𝐾𝑐 = Ψ
𝜇
Ψ𝑚𝑒𝑑 𝜌𝑒𝑛
𝐷𝑚𝑒𝑑
𝑚𝑒𝑑
=
𝜇
𝐷𝑑𝑒𝑡
Ψ𝑑𝑒𝑡 𝑒𝑛
𝜌 𝑑𝑒𝑡
Incoming beam of
ionizing particles
𝜇𝑒𝑛
𝜌
Ideal case – large detector
Charged particle equilibrium (CPE): 𝐷 = 𝐾𝑐 = Ψ
Same
𝜇
𝜇
Ψ𝑚𝑒𝑑 𝜌𝑒𝑛 fluence 𝜌𝑒𝑛
𝐷𝑚𝑒𝑑
𝜇𝑒𝑛
𝑚𝑒𝑑
𝑚𝑒𝑑
=
=
≡
𝜇
𝜇𝑒𝑛
𝜌
𝐷𝑑𝑒𝑡
Ψ𝑑𝑒𝑡 𝑒𝑛
𝜌 𝑑𝑒𝑡
𝜌 𝑑𝑒𝑡
Incoming beam of
ionizing particles
𝜇𝑒𝑛
𝜌
𝑚𝑒𝑑
𝑑𝑒𝑡
Ideal case – small detector
Delta-particle equilibrium: 𝐷 = 𝐶 = Φ
𝑆
Φ𝑚𝑒𝑑 𝜌𝑒𝑙
𝐷𝑚𝑒𝑑
𝑚𝑒𝑑
=
𝑆
𝐷𝑑𝑒𝑡
Φ𝑑𝑒𝑡 𝑒𝑙
𝜌 𝑑𝑒𝑡
Incoming beam of
ionizing particles
𝑆𝑒𝑙
𝜌
Ideal case – small detector
Delta-particle equilibrium: 𝐷
Same
𝑆
Φ𝑚𝑒𝑑 𝜌𝑒𝑙 fluence
𝐷𝑚𝑒𝑑
𝑚𝑒𝑑
=
=
𝑆
𝐷𝑑𝑒𝑡
Φ𝑑𝑒𝑡 𝑒𝑙
𝜌 𝑑𝑒𝑡
Incoming beam of
ionizing particles
𝑆𝑒𝑙
𝜌
=𝐶=Φ
𝑆𝑒𝑙
𝜌 𝑚𝑒𝑑
𝑆𝑒𝑙
≡
𝑆𝑒𝑙
𝜌
𝜌 𝑑𝑒𝑡
𝑚𝑒𝑑
𝑑𝑒𝑡
Electron fluence – large detector
Medium
Ftotal
Fmed
Fdet
Detector
Medium
Depth
Attenuation of primary beam is neglected
Electron fluence – small detector
Medium
Ftotal
Fmed
Fdet
Detector
Medium
Depth
Attenuation of primary beam is neglected
Electron fluence – mid-size detector
Medium
Ftotal
Fmed
Fdet
Detector
Medium
Depth
Attenuation of primary beam is neglected
Practical application – ion chamber
Incoming beam of
ionizing particles
Practical application – ion chamber
Incoming beam of
ionizing particles
𝐷𝑑𝑒𝑡
𝑀 𝑊
=
𝑒
𝜌𝑉
Do you remember?
An air-filled ion chamber is exposed to 1.25 MeV gamma radiation and
the collected charge is 11.55 nC. The gas volume is 0.1 cm3, and the
polystyrene wall is thicker than the range of the secondary charged
particles. Assuming the Bragg-Gray condition is fulfilled, what is the
absorbed dose in the adjacent polystyrene wall?
Please write down your answer before moving on to the next page.
Were you right?
An air-filled ion chamber is exposed to 1.25 MeV gamma radiation and
the collected charge is 11.55 nC. The gas volume is 0.1 cm3, and the
polystyrene wall is thicker than the range of the secondary charged
particles. Assuming the Bragg-Gray condition is fulfilled, what is the
absorbed dose in the adjacent polystyrene wall?
Solution
Known data:
𝑀 = 11.55 ∙ 10−9 𝐶
𝑉 = 0.1 𝑐𝑚3
𝜌 = 1.293 ∙ 10−6 𝑘𝑔𝑐𝑚3
−1
𝑊
=
33.97
J𝐶
𝑒
⇒ 𝐷𝑑𝑒𝑡
𝑀 𝑊
=
=3.03 𝐺𝑦
𝑒
𝜌𝑉
Were you right?
An air-filled ion chamber is exposed to 1.25 MeV gamma radiation and
the collected charge is 11.55 nC. The gas volume is 0.1 cm3, and the
polystyrene wall is thicker than the range of the secondary charged
particles. Assuming the Bragg-Gray condition is fulfilled, what is the
absorbed dose in the adjacent polystyrene wall?
Solution
From www.nist.gov:
𝑓 = 0.467
1
1
𝐸𝑒 = ℎ𝜈 𝑓 = 1.25 𝑀𝑒𝑉 0.467 = 292 𝑘𝑒𝑉
2
2
𝑆𝑒𝑙
= 2.11 𝑀𝑒𝑉𝑐𝑚2 𝑔−1
𝜌 𝑑𝑒𝑡
𝑆𝑒𝑙
= 2.34 𝑀𝑒𝑉𝑐𝑚2 𝑔−1
𝜌 𝑚𝑒𝑑
Were you right?
An air-filled ion chamber is exposed to 1.25 MeV gamma radiation and
the collected charge is 11.55 nC. The gas volume is 0.1 cm3, and the
polystyrene wall is thicker than the range of the secondary charged
particles. Assuming the Bragg-Gray condition is fulfilled, what is the
absorbed dose in the adjacent polystyrene wall?
Solution
Given that the Bragg−Gray condition is fulfilled:
𝐷𝑚𝑒𝑑
𝐷𝑑𝑒𝑡
=
𝑆𝑒𝑙 𝑚𝑒𝑑
𝜌 𝑑𝑒𝑡
⇒ 𝐷𝑚𝑒𝑑 = 3.03 𝐺𝑦
2.34
2.11
⇒ 𝐷𝑚𝑒𝑑 = 3.36 𝐺𝑦
Ion chamber dosimetry
𝐷 = 𝐶∆ =
𝐿∆
Φ′𝐸 𝑑𝐸
𝜌
𝐸𝑚𝑎𝑥
𝐷𝑑𝑒𝑡 =
Φ′𝐸,𝑑𝑒𝑡
Δ
𝐿Δ
𝜌
𝑑𝐸 +
𝑑𝑒𝑡
Φ′
𝑆𝑒𝑙
𝐸,𝑑𝑒𝑡 (Δ)Δ
𝜌
𝑑𝑒𝑡
Ion chamber dosimetry
𝐷𝑚𝑒𝑑
=
𝐷𝑑𝑒𝑡
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑑𝑒𝑡
Δ
𝑆𝑒𝑙
𝐿Δ
′
𝜌 𝑚𝑒𝑑 𝑑𝐸 + Φ 𝐸,𝑚𝑒𝑑 (Δ)Δ 𝜌 𝑚𝑒𝑑
𝑆𝑒𝑙
𝐿Δ
′
𝑑𝐸 + Φ 𝐸,𝑑𝑒𝑡 (Δ)Δ
𝜌 𝑑𝑒𝑡
𝜌 𝑑𝑒𝑡
Ion chamber dosimetry
𝐷𝑚𝑒𝑑
=
𝐷𝑑𝑒𝑡
𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 =
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑑𝑒𝑡
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝑆𝑒𝑙
𝐿Δ
′
𝜌 𝑚𝑒𝑑 𝑑𝐸 + Φ 𝐸,𝑚𝑒𝑑 (Δ)Δ 𝜌 𝑚𝑒𝑑
𝑆𝑒𝑙
𝐿Δ
′
𝑑𝐸 + Φ 𝐸,𝑑𝑒𝑡 (Δ)Δ
𝜌 𝑑𝑒𝑡
𝜌 𝑑𝑒𝑡
𝐿Δ
′
𝑑𝐸
+
Φ
𝐸,𝑚𝑒𝑑 (Δ)Δ
𝜌 𝑚𝑒𝑑
𝐿Δ
𝑑𝐸 + Φ′ 𝐸,𝑚𝑒𝑑 (Δ)Δ
𝜌 𝑑𝑒𝑡
𝑆𝑒𝑙
𝜌
𝑆𝑒𝑙
𝜌
𝑚𝑒𝑑
𝑑𝑒𝑡
Ion chamber dosimetry
𝐷𝑚𝑒𝑑
= 𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝
𝐷𝑑𝑒𝑡
𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 =
𝑝=
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑚𝑒𝑑
Δ
𝐸𝑚𝑎𝑥
Φ′𝐸,𝑑𝑒𝑡
Δ
𝐿Δ
𝜌
𝐿Δ
𝜌
𝐿Δ
′
𝑑𝐸
+
Φ
𝐸,𝑚𝑒𝑑 (Δ)Δ
𝜌 𝑚𝑒𝑑
𝐿Δ
𝑑𝐸 + Φ′ 𝐸,𝑚𝑒𝑑 (Δ)Δ
𝜌 𝑑𝑒𝑡
𝑆𝑒𝑙
𝑑𝐸 + Φ 𝐸,𝑚𝑒𝑑 (Δ)Δ 𝜌
𝑑𝑒𝑡
𝑑𝑒𝑡
𝑆𝑒𝑙
′
𝑑𝐸 + Φ 𝐸,𝑑𝑒𝑡 (Δ)Δ
𝜌 𝑑𝑒𝑡
𝑑𝑒𝑡
′
𝑆𝑒𝑙
𝜌
𝑆𝑒𝑙
𝜌
𝑚𝑒𝑑
𝑑𝑒𝑡
Ion chamber dosimetry
𝐷𝑚𝑒𝑑
= 𝑠𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝
𝐷𝑑𝑒𝑡
𝐷𝑑𝑒𝑡
𝑀 𝑊
=
𝑒
𝜌𝑉
𝐷𝑚𝑒𝑑
𝑀 𝑊
=
𝑠𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝
𝑒
𝜌𝑉
Summary
• A radiation detector forms a cavity in the medium
• Cavity theory deals with the conversion of absorbed dose in
the detector to absorbed dose in the medium
• While ideal cases are illustrative, the effect on the particle
fluence of a real detector needs to be taken inte account
• This can be accounted for through the concept of
perturbation factors