dosimetry - medicinsk strålningsfysik, lund · • define the scientific field of dosimetry ......

Dosimetry

Crister Ceberg

Medical Radiation Physics

Lund UniversitySweden

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 livingmatter

• 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

Deterministicbiological effect

Dosimetric quantity

Stochasticbiological effect

Dosimetric quantity


• Incident ionizing radiation

– Charged particles (e.g. electrons or protons) oruncharged particles (e.g. photons or neutrons)

– Produce ionizations in a medium (directly or indirectly)

– Practical threshold value Eionizing>Ethreshold≈10 eV

Unchargedparticle

Chargedparticle


• Conversion of energy (blue blobs)

– Energy is transferred to secondary particles

Unchargedparticle

Chargedparticle



– Energy is transferred to secondary particles

• Deposition of energy (yellow blobs)

– Not re-emitted by ionizing particles

Unchargedparticle

Chargedparticle

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


• 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


– Kerma

– Exposition

– Cema

Unchargedparticle

Chargedparticle

Energy transfer

Incomingphoton, ein

Scatteredphoton, e1

Secondaryelectron, e2

Example: Compton scattering

Mean energy transferred: 𝑓𝑖

Energy transfer


All interaction types: 𝑓 = 𝑖 𝑓𝑖𝜎𝑖 𝑖 𝜎𝑖

Energy transfer



𝐸𝑡𝑟 = Φ𝜇𝑉𝜀𝑖𝑛𝑓 = Ψ𝜇

𝜌𝑚𝑓

V

Energy transfer



𝐸𝑡𝑟 = Φ𝜇𝑉𝜀𝑖𝑛𝑓 = Ψ𝜇

𝜌𝑚𝑓

𝐸𝑡𝑟,𝑛𝑒𝑡 = 𝐸𝑡𝑟 1 − 𝑔

V

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)

• Unit: J kg-1

• Special name: Gy (gray)

𝐾 =𝑑𝐸𝑡𝑟𝑑𝑚

Kerma

𝐸𝑡𝑟 = Ψ𝜇

𝜌𝑚𝑓

𝜇𝑡𝑟𝜌

=𝜇

𝜌𝑓

𝐾 =𝑑𝐸𝑡𝑟𝑑𝑚

= Ψ𝜇𝑡𝑟𝜌

𝐾𝑐𝑜𝑙 = 𝐾 1 − 𝑔 = Ψ𝜇𝑒𝑛𝜌

Kerma

𝐾 = Ψ𝐸

𝜇𝑡𝑟𝜌𝑑𝐸 𝐾𝑐𝑜𝑙 = Ψ𝐸

𝜇𝑒𝑛𝜌

𝑑𝐸

From : www.nist.govFrom: ESM Ali and DWO Rogers,Phys. Med. Biol. 57 (2012)

Ψ𝐸 for a 10 MV beam𝜇𝑒𝑛𝜌

for water

Kerma

𝐾 = Φ𝑘Φ 𝑘Φ= 𝐾 Φ

From: MB Chadwick et al.,Med. Phys. 26 (1999)

𝑘Φ in ICRU−muscle tissue

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)

• Unit: J kg-1 s-1

• Special name: Gy s-1 (gray per second)

𝐾 =𝑑𝐾

𝑑𝑡

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 ∙ 109𝑐𝑚−2𝑠−1 ∙ 1.25𝑀𝑒𝑉 ∙ 1.6 ∙ 10−13𝐽

𝑀𝑒𝑉∙ 3600𝑠 = 2.520 𝐽𝑐𝑚−2

From www.nist.gov:𝜇𝑒𝑛𝜌

= 2.965 ∙ 10−2𝑐𝑚2𝑔−1 ∙ 1000𝑔

𝑘𝑔= 29.65𝑐𝑚2𝑘𝑔−1

Thus:

𝐾𝑐𝑜𝑙 = Ψ𝜇𝑒𝑛

𝜌= 2.520 𝐽𝑐𝑚−2 ∙ 29.65𝑐𝑚2𝑘𝑔−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?


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 ∙ 10−10𝐺𝑦𝑐𝑚2

Thus:

Φ =𝐾

𝑘Φ=

74.7 𝐺𝑦

0.709 ∙ 10−10𝐺𝑦𝑐𝑚2⇒ Φ = 1.05 ∙ 1012𝑐𝑚−2

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)

• Unit: C kg-1

𝑋 =𝑑𝑞

𝑑𝑚

Exposure vs. kerma

Mean energy per ion pair: 𝑊

𝑋 =𝑑𝑞

𝑑𝑚=𝑑𝐸𝑡𝑟𝑑𝑚

1 − 𝑔1

𝑊𝑒

= Ψ𝜇𝑡𝑟𝜌

1 − 𝑔1

𝑊𝑒

X = Ψ𝜇𝑒𝑛𝜌

1

𝑊𝑒

= 𝐾𝑎𝑖𝑟,𝑐𝑜𝑙1

𝑊𝑒

𝑊 = 33.97𝑒𝑉

𝑊𝑒 = 33.97 J𝐶−1

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)

• Unit: C kg-1 s-1

𝑋 =𝑑𝑋

𝑑𝑡

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)

• Unit: J kg-1


𝐶 =𝑑𝐸𝑒𝑙𝑑𝑚

Cema

C=𝑑𝐸𝑒𝑙𝑑𝑚

=Φ𝑆𝑒𝑙𝜌

Example: coulomb interaction

Incomingelectron, ein

Scatteredelectron, e1


Cema

𝐶 = Φ𝐸

𝑆𝑒𝑙𝜌𝑑𝐸 = Φ𝐸

𝐿∞𝜌𝑑𝐸

From : www.nist.govFrom: S Righi et al.,JACMP 14 (2013)

Φ𝐸 for linac electron beams𝑆𝑒𝑙𝜌

for water

Restricted cema

𝐶∆ = Φ′𝐸𝐿∆𝜌𝑑𝐸

e2>D: included in F’E

e2<D: included in LD




Example: coulomb interaction

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)



𝐶 =𝑑𝐶

𝑑𝑡

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


• 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

Unchargedparticle

Chargedparticle

Energy deposit

• The energy deposit, 𝜀𝑖, is the energy deposited in a single interaction, i,

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

𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄

Energy deposit




Example: coulomb interaction, Q=0

𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄

Energy deposit




hn

EA

𝜀𝑖 = 𝜀𝑖𝑛 − 𝜀𝑜𝑢𝑡 + 𝑄𝜀𝑖 = 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

where the summation is performed over all energy deposits, 𝜀𝑖, in that volume (ICRU 85, 2011)

• Unit: J

𝜀 =

𝑖

𝜀𝑖

Energy imparted

𝜀 =

𝑖

𝜀𝑖

𝜀 =

𝑖

𝜀𝑖𝑛 −

𝑖

𝜀𝑜𝑢𝑡 +

𝑖

𝑄

Energy imparted

e

f(e)

e

Energy imparted

𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄Rin Rout

Energy imparted

Ψ𝑑 𝐴

𝜀 = 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡 + Σ𝑄

𝜀 = − Ψ𝑑 𝐴 + Σ𝑄

𝜀 = − 𝑑𝑖𝑣Ψ𝑑𝑉 + Σ𝑄

Energy imparted

Ψ𝑑 𝐴


𝜀 = − Ψ𝑑 𝐴 + Σ𝑄

𝜀 = − 𝑑𝑖𝑣Ψ𝑑𝑉 + Σ𝑄

𝑑 𝜀

𝑑𝑚= −

1

𝜌𝑑𝑖𝑣Ψ +

1

𝜌

𝑑(Σ𝑄)

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 electroniccollisions and one fourth in radiation losses. The scattered photoncarries 4 MeV when it escapes the volume. What are the values ofenergy transfer, net energy transfer, and energy imparted?


Were you right?


Solution:

𝜀𝑖𝑛 = 6𝑀𝑒𝑉

𝜀𝑜𝑢𝑡,1 = 4𝑀𝑒𝑉

𝐸𝑡𝑟 = 2𝑀𝑒𝑉

Were you right?


Solution:



0.5 𝑀𝑒𝑉

𝐸𝑡𝑟 = 2𝑀𝑒𝑉𝐸𝑡𝑟,𝑛𝑒𝑡 = 1.5 𝑀𝑒𝑉

Were you right?


Solution:



1 𝑀𝑒𝑉

0.5 𝑀𝑒𝑉𝜀𝑜𝑢𝑡,2 = 0.5 𝑀𝑒𝑉

𝐸𝑡𝑟 = 2𝑀𝑒𝑉𝐸𝑡𝑟,𝑛𝑒𝑡 = 1.5 𝑀𝑒𝑉

𝜀 = 1.0 𝑀𝑒𝑉

Absorbed dose

• The absorbed dose, D, is the quotient of 𝑑 𝜀 by dm, where 𝑑 𝜀is the mean energy imparted by ionizing radiation to matter of mass dm (ICRU 85, 2011)

• Unit: J kg-1


𝐷 =𝑑 𝜀

𝑑𝑚

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)



𝐷 =𝑑𝐷

𝑑𝑡

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


• 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)


Rin Rout 𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡 𝑑𝑖𝑣 Ψ = 0


⇒ 𝐷 =𝑑 𝜀

𝑑𝑚=1

𝜚

𝑑(Σ𝑄)

𝑑𝑉


Rin Rout 𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡 (𝑑𝑖𝑣 Ψ = 0)

Distance greater thanrange of particle


𝑅𝑖𝑛 = 𝑅𝑜𝑢𝑡 (𝑑𝑖𝑣 Ψ = 0)

𝑅𝑖𝑛,𝑖 = 𝑅𝑜𝑢𝑡,𝑖 𝑑𝑖𝑣 Ψ𝑖 = 0

Distance greater thanrange of particle i

Rin Rout

Partial radiation equilibrium

Charged particle equilibrium (CPE): 𝑅𝑖𝑛,𝑐 = 𝑅𝑜𝑢𝑡,𝑐 𝑑𝑖𝑣 Ψ𝑐 = 0

d-particle equilibrium: 𝑅𝑖𝑛,𝛿= 𝑅𝑜𝑢𝑡,𝛿 (𝑑𝑖𝑣 Ψ𝛿 = 0)

Partial d-particle equilibrium: 𝑅𝑖𝑛,𝛿,Δ = 𝑅𝑜𝑢𝑡,𝛿,Δ (𝑑𝑖𝑣 Ψ𝛿,Δ = 0)

Charged particle equilibrium (CPE)

Incoming beam ofuncharged particles



𝑅𝑖𝑛,𝑢

𝑅𝑜𝑢𝑡,𝑢

𝑅𝑖𝑛,𝑐

𝑅𝑜𝑢𝑡,𝑐

𝜀 ≈ 𝐸𝑡𝑟

Distance greater thanrange of charged particles


Distance greater thanrange of charged particles


𝑅𝑖𝑛,𝑢

𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑

𝜀 = 𝐸𝑡𝑟 − 𝑅𝑜𝑢𝑡,𝑢,𝑟𝑎𝑑

𝑅𝑜𝑢𝑡,𝑢,𝑛𝑜𝑛𝑟𝑎𝑑

𝑅𝑜𝑢𝑡,𝑢𝑅𝑖𝑛,𝑐



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)


𝐷 =𝑑 𝜀

𝑑𝑚=𝑑𝐸𝑡𝑟𝑑𝑚

1 − 𝑔 = 𝐾 1 − 𝑔

𝐷 = 𝐾𝑐𝑜𝑙


DepthDepthRange ofcharged particle

Collision kerma, Kcol

Absorbed dose, D

Without attenuation With attenuation

Range ofcharged particle


Depth

Collision kerma, Kcol

Absorbed dose, D

With attenuation

d

𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 − 𝑥

𝐾𝑐𝑜𝑙 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 − 𝑥 𝑒−𝜇 𝑥

𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 𝑒𝜇 𝑥

𝐷 𝑑 = 𝐾𝑐𝑜𝑙 𝑑 (1 + 𝜇 𝑥)

𝑥

Charged particle interactions

𝜀 =

𝑖

𝜀𝑖

𝜀 = 𝑁1

𝑁

𝑖

𝜀𝑖 = 𝑁 𝜀𝑖

𝜀 = 𝑁 𝜀𝑖


𝐷 =𝑑 𝜀

𝑑𝑚=𝑑 𝑁

𝑑𝑚 𝜀𝑖

𝑑 𝑁

𝑑𝑚= Φ

𝑁𝐴𝑀𝜎

𝐷 = Φ𝑁𝐴𝑀𝜎 𝜀𝑖


𝐷 =𝑑 𝜀

𝑑𝑚=𝑑 𝑁

𝑑𝑚 𝜀𝑖

𝑑 𝑁

𝑑𝑚= Φ

𝑁𝐴𝑀𝜎

𝐷 = Φ𝑁𝐴𝑀𝜎 𝜀𝑖

𝐷 = Φ𝑆𝑒𝑙𝜌𝑘

Delta-particle equilibrium

Distance greater thanrange of delta particles

𝑅𝑖𝑛,𝑐


𝑅𝑖𝑛,𝛿

𝑅𝑜𝑢𝑡,𝛿

𝑘 = 1

Charged particlefluence

Delta-particle equilibrium

𝐷 = Φ𝑆𝑒𝑙𝜌𝑘 = Φ

𝑆𝑒𝑙𝜌

𝐷 = 𝐶

k=1

Partial delta-particle equilibrium

Distance greater thanrange of slow delta particles

Charged particlefluence

𝑅𝑖𝑛,𝑐


𝑅𝑖𝑛,𝛿,∆

𝑅𝑜𝑢𝑡,𝛿,∆

𝑘 < 1

Partial delta-particle equilibrium

𝐷 = Φ𝑆𝑒𝑙𝜌𝑘 = Φ′

𝐿∆𝜌

𝐷 = 𝐶∆

k<1

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?


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 ∙ 1011𝑐𝑚−2𝑠−1 ∙ 1.25𝑀𝑒𝑉 ∙ 1.6 ∙ 10−13𝐽

𝑀𝑒𝑉= 4.20 ∙ 10−2𝐽𝑐𝑚−2

From www.nist.gov:𝜇𝑒𝑛𝜌

= 2.965 ∙ 10−2𝑐𝑚2𝑔−1 ∙ 1000𝑔

𝑘𝑔= 29.65𝑐𝑚2𝑘𝑔−1

Given that CPE exists, we get:

𝐷 = 𝐾𝑐𝑜𝑙 = Ψ𝜇𝑒𝑛

𝜌= 4.20 ∙ 10−2𝐽𝑐𝑚−2 ∙ 29.65𝑐𝑚2𝑘𝑔−1 ⇒ 𝐷 = 1.25 𝐺𝑦

Do you remember?

A thin disc of lithium fluoride (LiF) is irradiated by an electronfluence of 4.1·109 cm-2 with energy 6 MeV. Assuming delta-particle equilibrium, what is the absorbed dose in the disc?


Were you right?

A thin disc of lithium fluoride (LiF) is irradiated by an electronfluence of 4.1·109 cm-2 with energy 6 MeV. Assuming delta-particle equilibrium, what is the absorbed dose in the disc?

Solution:

Known data: Φ = 4.1 ∙ 109𝑐𝑚−2

From www.nist.gov:𝑆𝑒𝑙𝜌

= 1.547 𝑀𝑒𝑉𝑐𝑚2𝑔−1 ∙ 1000𝑔

𝑘𝑔∙ 1.6 ∙ 10−13

𝐽

𝑀𝑒𝑉= 2.48 ∙ 10−10 𝐽𝑐𝑚2 𝑘𝑔−1

Given that delta-particle equilibrium exists, we get:

D = C = Φ𝑆𝑒𝑙

𝜌= 4.1 ∙ 109𝑐𝑚−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


• 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 ofionizing particles

𝐷𝑚𝑒𝑑

Cavity theory


𝐷𝑑𝑒𝑡 = 𝑐𝑀

Cavity theory


𝐷𝑚𝑒𝑑 = 𝐷𝑑𝑒𝑡𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡

Ideal case – large detector

Charged particle equilibrium (CPE): 𝐷 = 𝐾𝑐 = Ψ𝜇𝑒𝑛

𝜌

𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

Ψ𝑚𝑒𝑑𝜇𝑒𝑛𝜌

𝑚𝑒𝑑

Ψ𝑑𝑒𝑡𝜇𝑒𝑛𝜌

𝑑𝑒𝑡


Ideal case – large detector

Charged particle equilibrium (CPE): 𝐷 = 𝐾𝑐 = Ψ𝜇𝑒𝑛

𝜌

𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

Ψ𝑚𝑒𝑑𝜇𝑒𝑛𝜌

𝑚𝑒𝑑

Ψ𝑑𝑒𝑡𝜇𝑒𝑛𝜌

𝑑𝑒𝑡

=

𝜇𝑒𝑛𝜌

𝑚𝑒𝑑

𝜇𝑒𝑛𝜌

𝑑𝑒𝑡

≡𝜇𝑒𝑛𝜌

𝑑𝑒𝑡

𝑚𝑒𝑑

Samefluence


Ideal case – small detector


Delta-particle equilibrium: 𝐷 = 𝐶 = Φ𝑆𝑒𝑙

𝜌

𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

Φ𝑚𝑒𝑑𝑆𝑒𝑙𝜌

𝑚𝑒𝑑

Φ𝑑𝑒𝑡𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

Ideal case – small detector


Delta-particle equilibrium: 𝐷 = 𝐶 = Φ𝑆𝑒𝑙

𝜌

𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

Φ𝑚𝑒𝑑𝑆𝑒𝑙𝜌

𝑚𝑒𝑑

Φ𝑑𝑒𝑡𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

=

𝑆𝑒𝑙𝜌

𝑚𝑒𝑑

𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

≡𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

𝑚𝑒𝑑

Samefluence

Electron fluence – large detector

Attenuation of primary beam is neglected

DepthFtotal

Fmed

Fdet

Medium Detector Medium

Electron fluence – small detector

Depth



Ftotal

Fmed

Fdet

Electron fluence – mid-size detector

Depth



Ftotal

Fmed

Fdet

Practical application – ion chamber


Practical application – ion chamber


𝐷𝑑𝑒𝑡 =𝑀

𝜌𝑉 𝑊𝑒

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 chargedparticles. Assuming the Bragg-Gray condition is fulfilled, what is the absorbed dose in the adjacent polystyrene wall?


Were you right?


Solution

Known data:

𝑀 = 11.55 ∙ 10−9 𝐶𝑉 = 0.1 𝑐𝑚3

𝜌 = 1.293 ∙ 10−6𝑘𝑔𝑐𝑚3

𝑊𝑒 = 33.97 J𝐶−1

⇒ 𝐷𝑑𝑒𝑡 =𝑀

𝜌𝑉 𝑊𝑒 =3.03 𝐺𝑦

Were you right?


Solution

From www.nist.gov:𝑓 = 0.467

𝐸𝑒 =1

2ℎ𝜈 𝑓 =

1

21.25 𝑀𝑒𝑉 0.467 = 292 𝑘𝑒𝑉

𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

= 2.11 𝑀𝑒𝑉𝑐𝑚2𝑔−1

𝑆𝑒𝑙𝜌

𝑚𝑒𝑑

= 2.34 𝑀𝑒𝑉𝑐𝑚2𝑔−1

Were you right?


SolutionGiven that the Bragg−Gray condition is fulfilled:

𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

𝑆𝑒𝑙

𝜌 𝑑𝑒𝑡

𝑚𝑒𝑑⇒ 𝐷𝑚𝑒𝑑 = 3.03 𝐺𝑦

2.34

2.11⇒ 𝐷𝑚𝑒𝑑 = 3.36 𝐺𝑦

Ion chamber dosimetry

𝐷 = 𝐶∆ = Φ′𝐸𝐿∆𝜌𝑑𝐸

𝐷𝑑𝑒𝑡 =

Δ

𝐸𝑚𝑎𝑥

Φ′𝐸,𝑑𝑒𝑡𝐿Δ𝜌

𝑑𝑒𝑡

𝑑𝐸 + Φ′𝐸,𝑑𝑒𝑡(Δ)Δ

𝑆𝑒𝑙𝜌

𝑑𝑒𝑡


𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=

Δ𝐸𝑚𝑎𝑥Φ′𝐸,𝑚𝑒𝑑

𝐿Δ𝜌 𝑚𝑒𝑑

𝑑𝐸 +Φ′𝐸,𝑚𝑒𝑑(Δ)Δ

𝑆𝑒𝑙𝜌 𝑚𝑒𝑑

Δ𝐸𝑚𝑎𝑥Φ′𝐸,𝑑𝑒𝑡

𝐿Δ𝜌

𝑑𝑒𝑡𝑑𝐸 +Φ′

𝐸,𝑑𝑒𝑡(Δ)Δ𝑆𝑒𝑙𝜌

𝑑𝑒𝑡


𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡=



𝑑𝐸 +Φ′𝐸,𝑚𝑒𝑑(Δ)Δ



𝐿Δ𝜌

𝑑𝑒𝑡𝑑𝐸 +Φ′

𝐸,𝑑𝑒𝑡(Δ)Δ𝑆𝑒𝑙𝜌

𝑑𝑒𝑡

𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 =



𝑑𝐸 + Φ′𝐸,𝑚𝑒𝑑(Δ)Δ



𝐿Δ𝜌

𝑑𝑒𝑡𝑑𝐸 + Φ′

𝐸,𝑚𝑒𝑑(Δ)Δ𝑆𝑒𝑙𝜌

𝑑𝑒𝑡


𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡= 𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝

𝑆𝑚𝑒𝑑,𝑑𝑒𝑡 =


𝐿Δ𝜌

𝑚𝑒𝑑𝑑𝐸 + Φ′


𝑚𝑒𝑑


𝐿Δ𝜌 𝑑𝑒𝑡

𝑑𝐸 + Φ′𝐸,𝑚𝑒𝑑(Δ)Δ

𝑆𝑒𝑙𝜌 𝑑𝑒𝑡

𝑝 =


𝐿Δ𝜌

𝑑𝑒𝑡𝑑𝐸 + Φ′


𝑑𝑒𝑡


𝐿Δ𝜌 𝑑𝑒𝑡

𝑑𝐸 + Φ′𝐸,𝑑𝑒𝑡(Δ)Δ

𝑆𝑒𝑙𝜌 𝑑𝑒𝑡


𝐷𝑚𝑒𝑑

𝐷𝑑𝑒𝑡= 𝑠𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝

𝐷𝑑𝑒𝑡 =𝑀

𝜌𝑉 𝑊𝑒

𝐷𝑚𝑒𝑑 =𝑀

𝜌𝑉 𝑊𝑒 𝑠𝑚𝑒𝑑,𝑑𝑒𝑡 𝑝

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

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