Photon Interactions in Matter

Ho Kyung Kimhokyung@pusan.ac.kr

Pusan National University

Radiation DosimetryAttix 7

References

F. H. Attix, Introduction to Radiological Physics and Radiation Dosimetry, John Wiley and Sons, Inc., 1986

2

1. Compton effect (𝜎𝜎) †

2. Photoelectric effect (𝜏𝜏) †

3. Pair production (𝜅𝜅) †

4. Rayleigh (coherent) scattering• Elastic small-angle scattering with no energy loss

5. Photonuclear interaction• 𝐸𝐸𝛾𝛾 = ℎ𝜈𝜈 > a few MeV

• (𝛾𝛾,𝑛𝑛) reaction

† Resulting in the transfer of energy to electrons (this chapter), which then impart that energy to matter (next chapter) in many small Coulomb-force interactions along their tracks

• 𝐸𝐸𝛾𝛾(= ℎ𝜈𝜈) of the interaction photon

• 𝑍𝑍 of the interaction medium

3

4Attix Fig. 7.1

COMPTON EFFECT

Interaction with electrons being unbound and stationary (assumptions)• Zero-binding free electrons

Kinematics• Relationship between energies and angles

Cross section• Probability that an interaction will occur

5Attix Fig. 7.2

Kinematics

Energy conservation• 𝑇𝑇 = ℎ𝜈𝜈 − ℎ𝜈𝜈′

Momentum conservation• 𝑥𝑥-direction: ℎ𝜈𝜈 = ℎ𝜈𝜈′ cos𝜑𝜑 + 𝑝𝑝𝑝𝑝 cos𝜃𝜃• 𝑦𝑦-direction: ℎ𝜈𝜈′ sin𝜑𝜑 = 𝑝𝑝𝑝𝑝 sin 𝜃𝜃

• Law of invariance: 𝑝𝑝𝑝𝑝 = 𝑇𝑇(𝑇𝑇 + 2𝑚𝑚0𝑝𝑝2)– 𝑚𝑚 = 𝑚𝑚0

1− 𝑣𝑣/𝑐𝑐 2

– 𝑇𝑇 = 𝑚𝑚𝑝𝑝2 − 2𝑚𝑚0𝑝𝑝2

– 𝑝𝑝 = 𝑚𝑚𝑚𝑚

Solutions to the kinematics of Compton interactions:

• ℎ𝜈𝜈′ = ℎ𝜈𝜈

1+ ℎ𝜈𝜈𝑚𝑚0𝑐𝑐2

(1−cos 𝜑𝜑)

• 𝑇𝑇 = ℎ𝜈𝜈 − ℎ𝜈𝜈′

• cot𝜃𝜃 = 1 + ℎ𝜈𝜈𝑚𝑚0𝑐𝑐2

tan𝜑𝜑2

6

7Attix Fig. 7.3

For ℎ𝜈𝜈 < 0.01 MeV, ℎ𝜈𝜈′ = ℎ𝜈𝜈 regardless of 𝜑𝜑 No KE transfer to 𝑒𝑒− Elastic Thomson scattering

Straight-ahead scattering, 𝜑𝜑 = 0

Side scatteringBack scattering

KE of recoiling 𝑒𝑒−

8Attix Fig. 7.4

For ℎ𝜈𝜈 << 1, cot𝜃𝜃 ≅ tan 𝜑𝜑2

or 𝜃𝜃 ≅ 𝜋𝜋2− 𝜑𝜑

2

For ℎ𝜈𝜈 >> 1, 𝜃𝜃 = 𝜑𝜑 at 2.59°→ e.g., All 𝑒𝑒− scattered at 𝜃𝜃 btwn 2.59° & 90°are likewise related to the photons scattered forward btwn 0° & 2.59°

Thomson scattering

The earliest theoretical description of photon scattering with a "free" electron Elastic scattering Only valid to ℎ𝜈𝜈 ≤ 0.01 MeV

• Same as the Compton scattering when ℎ𝜈𝜈 ≤ 0 because of no relativistics

Differential Thomson scattering x-sec. per electron per unit solid angle for a photon scattered at 𝜑𝜑:

d 𝑒𝑒𝜎𝜎0dΩ𝜑𝜑

=𝑟𝑟02

2(1 + cos2 𝜑𝜑)

• cm2 sr-1 per electron

• 𝑟𝑟0 = 𝑒𝑒2

𝑚𝑚0𝑐𝑐2= 2.818 × 10−13 cm, called the "classical electron radius"

• Front-back symmetrical angular distribution• Cylindrical symmetry around the beam axis ⇒ dΩ𝜑𝜑 = 2𝜋𝜋 sin𝜑𝜑 d𝜑𝜑

9

Total Thomson scattering x-sec. per electron:

𝑒𝑒𝜎𝜎0 = �𝜑𝜑=0

𝜋𝜋 d 𝑒𝑒𝜎𝜎0dΩ𝜑𝜑

dΩ𝜑𝜑 = 𝜋𝜋𝑟𝑟02 �𝜑𝜑=0

𝜋𝜋(1 + cos2 𝜑𝜑) sin𝜑𝜑 d𝜑𝜑 =

8𝜋𝜋𝑟𝑟02

3= 6.65 × 10−25

• Note: dΩ𝜑𝜑 = 2𝜋𝜋 sin𝜑𝜑 d𝜑𝜑, the annular element of solid angle

• In units of (cm2 per electron)• Independent of ℎ𝜈𝜈

X-sec.• Can be thought of as an effective target area• Probability that a Thomson scattering event occurs when a single photon passes through a layer

containing on electron per cm2

• Fraction of a large number of incident photons that scatter in passing through the layer (e.g., ~665 events for 1027 photons)

• So long as the fraction of photons interacting in a layer of matter by all processes combinedremains less than 0.05, the fraction may be assumed to be proportional to absorber thickness (i.e., linear approximation); otherwise the exponential law must be used

10

Klein-Nishina x-sec.

Relativistic treatment to unbound electrons using the Dirac's relativistic theory Differential KN scattering x-sec. per electron per unit solid angle for a photon scattered at

𝜑𝜑:d 𝑒𝑒𝜎𝜎dΩ𝜑𝜑

=𝑟𝑟02

2ℎ𝜈𝜈′

ℎ𝜈𝜈

2 ℎ𝜈𝜈ℎ𝜈𝜈′

+ℎ𝜈𝜈′

ℎ𝜈𝜈− sin2 𝜑𝜑

• See Attix Fig. 7.5

– Forward scattering at high energies

• ℎ𝜈𝜈′ ≅ ℎ𝜈𝜈 for low energies;d 𝑒𝑒𝜎𝜎dΩ𝜑𝜑

≅ 𝑟𝑟02

22 − sin2 𝜑𝜑 = 𝑟𝑟02

2(1 + cos2 𝜑𝜑):

– Identical to the Thomson scattering!

11

12Attix Fig. 7.5

Same as the Thomson scattering

Unlikely backscattering at high ℎ𝜈𝜈

Total KN scattering x-sec. per electron:

𝑒𝑒𝜎𝜎 = 2𝜋𝜋𝑟𝑟021 + 𝛼𝛼𝛼𝛼2

2(1 + 𝛼𝛼)1 + 2𝛼𝛼

−ln(1 + 2𝛼𝛼)

𝛼𝛼+

ln(1 + 2𝛼𝛼)2𝛼𝛼

−1 + 3𝛼𝛼

(1 + 2𝛼𝛼)2

• Note: 𝑒𝑒𝜎𝜎 = 2𝜋𝜋 ∫𝜑𝜑=0𝜋𝜋 d 𝑒𝑒𝜎𝜎

dΩ𝜑𝜑sin𝜑𝜑 d𝜑𝜑 = 𝜋𝜋𝑟𝑟02 ∫0

𝜋𝜋 ℎ𝜈𝜈′

ℎ𝜈𝜈

2 ℎ𝜈𝜈ℎ𝜈𝜈′

+ ℎ𝜈𝜈′

ℎ𝜈𝜈− sin2 𝜑𝜑 sin𝜑𝜑 d𝜑𝜑

• 𝛼𝛼 = ℎ𝜈𝜈𝑚𝑚0𝑐𝑐2

with ℎ𝜈𝜈 in MeV

• See Attix Fig. 7.6

– 𝑒𝑒𝜎𝜎 ∝ (ℎ𝜈𝜈)−1 for higher photon energies

– Independent of 𝑍𝑍 or 𝑒𝑒𝜎𝜎 ∝ 𝑍𝑍0

• Because the electron binding energy has been assumed to be zero (unbound electrons)

– KN x-sec. per atom: 𝑎𝑎𝜎𝜎 = 𝑍𝑍 𝑒𝑒𝜎𝜎 (cm2/atom) ⟹ 𝑎𝑎𝜎𝜎 ∝ 𝑍𝑍

• Compton mass attenuation coefficient𝜎𝜎𝜌𝜌

=𝑁𝑁𝐴𝐴𝑍𝑍𝐴𝐴 𝑒𝑒𝜎𝜎 ⟹

𝜎𝜎𝜌𝜌∝ 𝑍𝑍0

– 𝑁𝑁𝐴𝐴 = 6.022 × 1023 mole-1

– 𝑁𝑁𝐴𝐴𝑍𝑍𝐴𝐴

= number of electrons per gram of material

13

14Attix Fig. 7.6

𝑒𝑒𝜎𝜎0 = 6.65 × 10−25 cm2/e at ℎ𝜈𝜈 = 0.01 MeV

𝑒𝑒𝜎𝜎 ∝ (ℎ𝜈𝜈)−1

𝑒𝑒𝜎𝜎𝑠𝑠, x-sec for the energy carried by the scattered photon

𝑒𝑒𝜎𝜎 = 𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 + 𝑒𝑒𝜎𝜎𝑠𝑠

𝑒𝑒𝜎𝜎 becomes constant & 𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 diminishes with

decreasing ℎ𝜈𝜈 below 0.5 MeV

Energy-transfer x-sec.

In each Compton scattering; incident (ℎ𝜈𝜈) = scattered (ℎ𝜈𝜈′) + recoiled electron (𝑇𝑇) Related to the "kerma," it is interesting to know �𝑇𝑇/ℎ𝜈𝜈 or �𝑇𝑇 averaged over all 𝜑𝜑

• d 𝑒𝑒𝜎𝜎𝑡𝑡𝑡𝑡dΩ𝜑𝜑

= d 𝑒𝑒𝜎𝜎dΩ𝜑𝜑

𝑇𝑇ℎ𝜈𝜈

= d 𝑒𝑒𝜎𝜎dΩ𝜑𝜑

ℎ𝜈𝜈−ℎ𝜈𝜈′

ℎ𝜈𝜈= 𝑟𝑟02

2ℎ𝜈𝜈′

ℎ𝜈𝜈

2 ℎ𝜈𝜈ℎ𝜈𝜈′

+ ℎ𝜈𝜈′

ℎ𝜈𝜈− sin2 𝜑𝜑 ℎ𝜈𝜈−ℎ𝜈𝜈′

ℎ𝜈𝜈

• 𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 = 2𝜋𝜋 ∫𝜑𝜑=0𝜋𝜋 𝑒𝑒𝜎𝜎𝑡𝑡𝑡𝑡

dΩ𝜑𝜑sin𝜑𝜑 d𝜑𝜑

𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 = 2𝜋𝜋𝑟𝑟022(1+𝛼𝛼)2

𝛼𝛼2(1+2𝛼𝛼)− 1+3𝛼𝛼

1+2𝛼𝛼 2 −1+𝛼𝛼 2𝛼𝛼2−2𝛼𝛼−1

𝛼𝛼2 1+2𝛼𝛼 2 − 4𝛼𝛼2

3(1+2𝛼𝛼)3− 1+𝛼𝛼

𝛼𝛼3− 1

2𝛼𝛼+ 1

2𝛼𝛼3ln(1 + 2𝛼𝛼)

• See Attix Fig. 7.6

– 𝑒𝑒𝜎𝜎 = 𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 + 𝑒𝑒𝜎𝜎𝑠𝑠

•�𝑇𝑇ℎ𝜈𝜈

= 𝑒𝑒𝜎𝜎𝑡𝑡𝑡𝑡𝑒𝑒𝜎𝜎

or �𝑇𝑇 = ℎ𝜈𝜈 𝑒𝑒𝜎𝜎𝑡𝑡𝑡𝑡𝑒𝑒𝜎𝜎

– See Attix Fig. 7.7

15

• Compton mass energy-transfer coefficient𝜎𝜎𝑡𝑡𝑟𝑟𝜌𝜌

=𝑁𝑁𝐴𝐴𝑍𝑍𝐴𝐴 𝑒𝑒𝜎𝜎𝑡𝑡𝑟𝑟 =

𝜎𝜎𝜌𝜌�𝑇𝑇ℎ𝜈𝜈

16Attix Fig. 7.7

Other differential KN x-sec's

Differential KN x-sec. for electron scattering at angle 𝜃𝜃 per unit solid angle & per electron:d 𝑒𝑒𝜎𝜎dΩ𝜃𝜃

=d 𝑒𝑒𝜎𝜎dΩ𝜑𝜑

(1 + 𝛼𝛼)2(1 − cos𝜑𝜑)2

cos3 𝜃𝜃

• 𝑒𝑒𝜎𝜎 = ∫𝜃𝜃=0𝜋𝜋/2 d 𝑒𝑒𝜎𝜎

dΩ𝜃𝜃dΩ𝜃𝜃

• High forward momentum in the collision causes most of the electrons and most of the scattered photons to be strongly forward-directed when ℎ𝜈𝜈 is large

17Attix Fig. 7.8

Strong forward scattering at high ℎ𝜈𝜈(Refer to Attix Fig. 7.5)

d 𝑒𝑒𝜎𝜎d𝑇𝑇

=𝜋𝜋𝑟𝑟02𝑚𝑚0𝑝𝑝2

ℎ𝜈𝜈′ 2𝑚𝑚0𝑝𝑝2𝑇𝑇ℎ𝜈𝜈 2

2

+ 2ℎ𝜈𝜈′

ℎ𝜈𝜈

2

+ℎ𝜈𝜈′

ℎ𝜈𝜈 3 (𝑇𝑇 − 𝑚𝑚0𝑝𝑝2)2−(𝑚𝑚0𝑝𝑝2)2

• In units of (cm2 MeV-1 e-1)• Probability that a single photon will have a Compton interaction in traversing a layer containing 1

e/cm2, transferring to that electron a KE between 𝑇𝑇 & 𝑇𝑇 + d𝑇𝑇• Energy distribution of the electrons averaged over all scattering angles 𝜃𝜃

• |𝑇𝑇𝑚𝑚𝑎𝑎𝑚𝑚 𝜃𝜃=0° (head−on collision) = 2 ℎ𝜈𝜈 2

2ℎ𝜈𝜈+𝑚𝑚0𝑐𝑐2= ℎ𝜈𝜈

1+𝑚𝑚0𝑐𝑐22ℎ𝜈𝜈

= ℎ𝜈𝜈 1 + 𝑚𝑚0𝑐𝑐2

2ℎ𝜈𝜈

−1

ℎ𝜈𝜈≫1ℎ𝜈𝜈 − 0.256 MeV

18Attix Fig. 7.9

PHOTOELECTRIC EFFECT

Most important interaction of low-E photons with high-Z matter Incident photon can give up all of its ℎ𝜈𝜈 in colliding with a tightly bound electron (e.g.,

inner shell electrons bounded by potential energy 𝐸𝐸𝑏𝑏) especially of high-Z atom

19Attix Fig. 7.10

Kinematics

𝑇𝑇 = ℎ𝜈𝜈 − 𝐸𝐸𝑏𝑏 − 𝑇𝑇𝑎𝑎 = ℎ𝜈𝜈 − 𝐸𝐸𝑏𝑏

– KE given to the recoiling atom 𝑇𝑇𝑎𝑎 ≅ 0

• Can occur only when ℎ𝜈𝜈 > 𝐸𝐸𝑏𝑏• More likely occur as ℎ𝜈𝜈 is smaller• 𝑇𝑇 is independent of scattering angle 𝜃𝜃

20

Photoelectric interaction x-sec.

No simple equation for the differential photoelectric x-sec. X-sec. is available based on experimental results supplemented by theoretically assisted

interpolations

Directional distribution of photoelectrons per unit solid angle (See Attix Fig. 7.11)• Emitted to sideways along the direction of the photon's electric field for low photon energies• Emitted toward smaller angles with increasing photon energy

21Attix Fig. 7.11

Theoretical calculation results

Photoelectric x-sec. per atom (integrated over all angles of photoelectron emission)

• 𝑎𝑎𝜏𝜏 ≅ 𝑘𝑘 𝑍𝑍𝑛𝑛

(ℎ𝜈𝜈)𝑚𝑚

– In units of (cm2/atom)– 𝑘𝑘 = constant– 𝑛𝑛 ≅ 4 at ℎ𝜈𝜈 = 0.1 MeV, gradually rising to about 4.6 at 3 MeV– 𝑚𝑚 ≅ 3 at ℎ𝜈𝜈 = 0.1 MeV, gradually decreasing to about 1 at 3 MeV

For ℎ𝜈𝜈 ≤ 0.1 MeV

• 𝑎𝑎𝜏𝜏 ∝𝑍𝑍4

(ℎ𝜈𝜈)3

Mass attenuation coefficient

• 𝜏𝜏𝜌𝜌

= 𝑁𝑁𝐴𝐴𝑍𝑍𝐴𝐴 𝑒𝑒𝜏𝜏 = 𝑁𝑁𝐴𝐴𝑍𝑍

𝐴𝐴𝑎𝑎𝜏𝜏𝑍𝑍⇒ 𝜏𝜏

𝜌𝜌∝ 𝑍𝑍3

(ℎ𝜈𝜈)3

• See Attix Fig. 7.13

22

23Attix Fig. 7.13

∝ (ℎ𝜈𝜈)−3

Two K-shell electrons cannot participate in the PE effect when ℎ𝜈𝜈 < (𝐸𝐸𝑏𝑏)𝐾𝐾= 88keV

∝ (ℎ𝜈𝜈)−3

(𝐸𝐸𝑏𝑏)𝐿𝐿1= 15.9, (𝐸𝐸𝑏𝑏)𝐿𝐿2= 15.2, (𝐸𝐸𝑏𝑏)𝐿𝐿3= 13.0 keV

𝑍𝑍𝑃𝑃𝑏𝑏𝑍𝑍𝐶𝐶

=826≈ 10 ⇔

𝜏𝜏/𝜌𝜌 𝑃𝑃𝑏𝑏

𝜏𝜏/𝜌𝜌 𝑃𝑃𝑏𝑏

Energy-transfer x-sec.

𝑇𝑇ℎ𝜈𝜈

=ℎ𝜈𝜈 − 𝐸𝐸𝑏𝑏ℎ𝜈𝜈

• First approximation to the total fraction of ℎ𝜈𝜈 that is transferred to "all" electrons• Part or all of 𝐸𝐸𝑏𝑏 is converted to electron KE through the Auger effect

Disposal mechanisms of 𝐸𝐸𝑏𝑏

• Fluorescence x-ray– Inner shell vacancies due to PE, IC, EC, or charged-particle collision are promptly filled by another electrons

falling from less tightly bound shells– This transition is sometimes accompanied by the emission of a fluorescence x-ray with an energy equal to

the difference in potential energy between the donor and recipient levels (e.g., ℎ𝜈𝜈𝐾𝐾, ℎ𝜈𝜈𝐿𝐿) with a probability of fluorescence yield (See Attix Fig. 7.14)

• 𝑌𝑌𝐾𝐾 > 𝑌𝑌𝐿𝐿• 𝑌𝑌𝑀𝑀 ≈ 0

24

25Attix Fig. 7.14

0.42

Cu

– ℎ�̅�𝜈𝐾𝐾 rather than ℎ𝜈𝜈𝐾𝐾 because of several energy levels in the L or higher shells• ℎ�̅�𝜈𝐾𝐾 < (𝐸𝐸𝑏𝑏)𝐾𝐾; See Attix Fig. 7.15

– 𝑃𝑃𝐾𝐾 = 𝜏𝜏𝐾𝐾𝜏𝜏

= the fraction of all PE interactions that occur in the K-shell for photons of ℎ𝜈𝜈 > (𝐸𝐸𝑏𝑏)𝐾𝐾

– 𝑃𝑃𝐿𝐿 = 𝜏𝜏𝐿𝐿𝜏𝜏

for photons where (𝐸𝐸𝑏𝑏)𝐿𝐿1< ℎ𝜈𝜈 < (𝐸𝐸𝑏𝑏)𝐾𝐾

– 𝑃𝑃𝐾𝐾𝑌𝑌𝐾𝐾 = the fraction of all PE events in which a K-fluorescence x-ray is emitted by the atom– 𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿 = the fraction of all PE events in which an L-fluorescence x-ray is emitted by the atom

– 𝑃𝑃𝐾𝐾𝑌𝑌𝐾𝐾ℎ�̅�𝜈𝐾𝐾 = the mean energy carried away from the atom by K-fluorescence x-rays per PE interaction– 𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿ℎ�̅�𝜈𝐿𝐿 ≤ 𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿(𝐸𝐸𝑏𝑏)𝐿𝐿1

26

27Attix Fig. 7.15

• Auger effect– 𝐸𝐸𝑏𝑏 disposal mechanism alternative to the fluorescence x-ray– An atom ejects one or more Auger electrons with sufficient KE simultaneously in a kind of chain reaction

• Contributing to the kerma• Exchanging one energetically "deep" inner vacancies for a number of relatively shallow outer-shell

vacancies (these vacancies are finally neutralized by conduction-band electrons)

– Let's consider a possible scenario for a K-shell vacancy (Auger chain reaction or shower)• ℎ𝜈𝜈𝐾𝐾 = (𝐸𝐸𝑏𝑏)𝐾𝐾−(𝐸𝐸𝑏𝑏)𝐿𝐿: the remaining (𝐸𝐸𝑏𝑏)𝐾𝐾−ℎ𝜈𝜈𝐾𝐾 will become electron KE• Auger effect ejecting an M-shell electron for example

» 𝑇𝑇𝑀𝑀 = (𝐸𝐸𝑏𝑏)𝐾𝐾−(𝐸𝐸𝑏𝑏)𝐿𝐿−(𝐸𝐸𝑏𝑏)𝑀𝑀» Then, two vacancies in the L- & M-shells» Two N-shell electrons fill these vacancies, and the atom emits two more Auger electrons from

the N-shell for example ⇒ Then, the atom has four N-shell vacancies• 𝑇𝑇𝑁𝑁1 = (𝐸𝐸𝑏𝑏)𝐿𝐿−(𝐸𝐸𝑏𝑏)𝑁𝑁 − 𝐸𝐸𝑏𝑏 𝑁𝑁 = (𝐸𝐸𝑏𝑏)𝐿𝐿−2(𝐸𝐸𝑏𝑏)𝑁𝑁• 𝑇𝑇𝑁𝑁2 = (𝐸𝐸𝑏𝑏)𝑀𝑀−2(𝐸𝐸𝑏𝑏)𝑁𝑁

» Thus, the total KE of the 3 Auger electrons: 𝑇𝑇𝐴𝐴 = 𝑇𝑇𝑀𝑀 + 𝑇𝑇𝑁𝑁1 + 𝑇𝑇𝑁𝑁2 = (𝐸𝐸𝑏𝑏)𝐾𝐾−4(𝐸𝐸𝑏𝑏)𝑁𝑁» This process is repeated until all the vacancies are located in the outermost shell» Total KE of all the Auger electrons = (𝐸𝐸𝑏𝑏)𝐾𝐾 – sum of BE of all the final electron vacancies» (𝐸𝐸𝑏𝑏)𝐾𝐾 ends up as electron KE, contributing to the kerma

28

The probability of any other fluorescence x-ray except those from the K-shell being able to carry energy out of an atom is negligible for ℎ𝜈𝜈 > (𝐸𝐸𝑏𝑏)𝐾𝐾

Then, all the rest of the (𝐸𝐸𝑏𝑏)𝐾𝐾 and all of the BE involved in PE interactions in other shells may be assumed to be given to Auger electrons

The mean energy transferred to charged particles per PE event = ℎ𝜈𝜈 − 𝑃𝑃𝐾𝐾𝑌𝑌𝐾𝐾ℎ�̅�𝜈𝐾𝐾

• Photoelectric mass energy-transfer coefficient for ℎ𝜈𝜈 > (𝐸𝐸𝑏𝑏)𝐾𝐾𝜏𝜏𝑡𝑡𝑟𝑟𝜌𝜌

=𝜏𝜏𝜌𝜌ℎ𝜈𝜈 − 𝑃𝑃𝐾𝐾𝑌𝑌𝐾𝐾ℎ�̅�𝜈𝐾𝐾 − (1 − 𝑃𝑃𝐾𝐾)𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿ℎ�̅�𝜈𝐿𝐿

ℎ𝜈𝜈

• For (𝐸𝐸𝑏𝑏)𝐿𝐿1< ℎ𝜈𝜈 < (𝐸𝐸𝑏𝑏)𝐾𝐾𝜏𝜏𝑡𝑡𝑟𝑟𝜌𝜌

=𝜏𝜏𝜌𝜌ℎ𝜈𝜈 − 𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿ℎ�̅�𝜈𝐿𝐿

ℎ𝜈𝜈≈𝜏𝜏𝜌𝜌ℎ𝜈𝜈 − 𝑃𝑃𝐿𝐿𝑌𝑌𝐿𝐿(𝐸𝐸𝑏𝑏)𝐿𝐿1

ℎ𝜈𝜈

29

30Attix Fig. 7.16

The size of K-edge step is less than that in 𝜏𝜏𝜌𝜌

curve

due to the loss of K-fluorescence energy

PAIR PRODUCTION

Absorption process in which a photon disappears and gives rise to an electron & a positron Only occurs in a Coulomb field

• Near an atomic nucleus field– Dominant process– ℎ𝜈𝜈 ≥ 2𝑚𝑚0𝑝𝑝2 = 1.022 MeV

• Near an atomic electron field– Called "triplet production" = 2 electrons + 1 positron = host electron (w/ significant KE) + pair production– ℎ𝜈𝜈 ≥ 4𝑚𝑚0𝑝𝑝2 = 2.044 MeV due to momentum-conservation considerations

31

PP in the nuclear coulomb force field

• Energy conservationℎ𝜈𝜈 = 2𝑚𝑚0𝑝𝑝2 + 𝑇𝑇− + 𝑇𝑇+ = 1.022 MeV + 2�𝑇𝑇

• Average KE of the products

�𝑇𝑇 =ℎ𝜈𝜈 − 1.022 MeV

2• Approximate departure angle of the product relative to the original photon direction

�̅�𝜃 ≅𝑚𝑚0𝑝𝑝2�𝑇𝑇

– In units of (radians)– e.g., For ℎ𝜈𝜈 = 5 MeV, �𝑇𝑇 = 1.989 MeV and �̅�𝜃 ≅ 0.26 radians = 15°

32Attix Fig. 7.17

Atomic differential x-sec. (Bethe & Heitler) for the creation of 𝑇𝑇+

d 𝑎𝑎𝜅𝜅 =𝜎𝜎0𝑍𝑍2𝑃𝑃

ℎ𝜈𝜈 − 2𝑚𝑚0𝑝𝑝2d𝑇𝑇+

• 𝜎𝜎0 = 𝑟𝑟02

137= 1

137𝑒𝑒2

𝑚𝑚0𝑐𝑐2

2= 5.80 × 10−28 cm2/electron

• 𝑃𝑃 is dependent upon ℎ𝜈𝜈 & 𝑍𝑍 (See Attix Fig. 7.18)

33Attix Fig. 7.18

Total nuclear PP x-sec. per atom

𝑎𝑎𝜅𝜅 = �𝑇𝑇+

d 𝑎𝑎𝜅𝜅 = 𝜎𝜎0𝑍𝑍2 �0

ℎ𝜈𝜈−2𝑚𝑚0𝑐𝑐2 𝑃𝑃d𝑇𝑇+

ℎ𝜈𝜈 − 2𝑚𝑚0𝑝𝑝2= 𝜎𝜎0𝑍𝑍2 �

0

1𝑃𝑃d

𝑇𝑇+

ℎ𝜈𝜈 − 2𝑚𝑚0𝑝𝑝2= 𝜎𝜎0𝑍𝑍2 �𝑃𝑃

• 𝑎𝑎𝜅𝜅 ∝ 𝑍𝑍2

• 𝑎𝑎𝜅𝜅~ log(ℎ𝜈𝜈)– See Attix Fig. 7.13

– Becomes a constant independent of ℎ𝜈𝜈 for very large ℎ𝜈𝜈 because of electron screening of the nuclear field

• Nuclear PP mass attenuation coefficient𝜅𝜅𝜌𝜌

=𝑁𝑁𝐴𝐴𝜌𝜌 𝑎𝑎𝜅𝜅 =

𝑁𝑁𝐴𝐴𝑍𝑍𝜌𝜌

𝑎𝑎𝜅𝜅𝑍𝑍⇒𝜅𝜅𝜌𝜌∝ 𝑍𝑍

34

PP in the electron field

• Energy conservationℎ𝜈𝜈 = 1.022 MeV + 𝑇𝑇+ + 𝑇𝑇1− + 𝑇𝑇2−

• Average KE of the products

�𝑇𝑇 =ℎ𝜈𝜈 − 1.022 MeV

3• Threshold (See Attix Fig. 7.19)

ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚 = 4𝑚𝑚0𝑝𝑝2

35Attix Fig. 7.19

• Momentum conservation in a moving frame 𝑅𝑅′ with a velocity +𝛽𝛽𝑝𝑝 relative to the laboratory frame 𝑅𝑅

ℎ𝜈𝜈′

𝑝𝑝− 𝑚𝑚𝛽𝛽𝑝𝑝 =

ℎ𝜈𝜈′

𝑝𝑝−

𝑚𝑚0𝛽𝛽𝑝𝑝

1 − 𝛽𝛽2= 0

– Due to the Doppler effect: 𝜈𝜈′ = 𝜈𝜈 1−𝛽𝛽1+𝛽𝛽

• Then we have

𝛽𝛽 =𝛼𝛼

1 + 𝛼𝛼• Energy conservation

ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚′ + 𝑇𝑇1− = ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚

′ + 𝑝𝑝2 𝑚𝑚 − 𝑚𝑚0 = ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚′ + 𝑚𝑚0𝑝𝑝2

1

1 − 𝛽𝛽𝑚𝑚𝑚𝑚𝑚𝑚2

− 1 = 2𝑚𝑚0𝑝𝑝2

– ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚′ = ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚1−𝛽𝛽𝑚𝑚𝑚𝑚𝑛𝑛1+𝛽𝛽𝑚𝑚𝑚𝑚𝑛𝑛

, 𝛽𝛽𝑚𝑚𝑚𝑚𝑚𝑚 = 𝛼𝛼𝑚𝑚𝑚𝑚𝑛𝑛1+𝛼𝛼𝑚𝑚𝑚𝑚𝑛𝑛

, and 𝛼𝛼𝑚𝑚𝑚𝑚𝑚𝑚 = ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑛𝑛𝑚𝑚0𝑐𝑐2

• Then we have

𝛽𝛽𝑚𝑚𝑚𝑚𝑚𝑚 =45

ℎ𝜈𝜈𝑚𝑚𝑚𝑚𝑚𝑚 = 4𝑚𝑚0𝑝𝑝2

36

• KE of each of the products

𝑇𝑇 =𝛼𝛼2 − 2𝛼𝛼 − 2 ± 𝛼𝛼 𝛼𝛼(𝛼𝛼 − 4)

2𝛼𝛼 + 1– e.g., For ℎ𝜈𝜈 = 10 MeV, 3 keV ≤ 𝑇𝑇 ≤ 8.7 MeV

– 𝑇𝑇 = 2𝑚𝑚0𝑐𝑐2

3for ℎ𝜈𝜈 = 4𝑚𝑚0𝑝𝑝2

X-sec.𝑎𝑎𝜅𝜅 (electrons)

𝑎𝑎𝜅𝜅 (nucleus)≅

1𝐶𝐶𝑍𝑍

• e.g., 𝑎𝑎𝜅𝜅 (electrons)

𝑎𝑎𝜅𝜅 (nucleus)≈ 1% in Pb

• 𝐶𝐶 = 1 for ℎ𝜈𝜈 → ∞• 𝐶𝐶 increases slowly to ~2 with decreasing ℎ𝜈𝜈 to 5 MeV

37

Mass attenuation coefficient𝜅𝜅𝜌𝜌 𝑝𝑝𝑎𝑎𝑚𝑚𝑟𝑟

=𝜅𝜅𝜌𝜌 𝑚𝑚𝑛𝑛𝑐𝑐𝑛𝑛𝑒𝑒𝑎𝑎𝑟𝑟

+𝜅𝜅𝜌𝜌 𝑒𝑒𝑛𝑛𝑒𝑒𝑐𝑐𝑡𝑡𝑟𝑟𝑒𝑒𝑚𝑚

PP energy-transfer coefficient𝜅𝜅𝑡𝑡𝑟𝑟𝜌𝜌

=𝜅𝜅𝜌𝜌

ℎ𝜈𝜈 − 2𝑚𝑚0𝑝𝑝2

ℎ𝜈𝜈

38

RAYLEIGH SCATTERING

Also called "coherent" scattering because the photon is scattered by the combined action of the whole atom

Elastic scattering Contributes to nothing to kerma or dose Small-angle scattering

• Can only be detected in narrow-beam geometry• Dependent upon both 𝑍𝑍 & ℎ𝜈𝜈

X-sec.

• 𝑎𝑎𝜎𝜎𝑅𝑅 ∝𝑍𝑍2

(ℎ𝜈𝜈)2

• 𝜎𝜎𝑅𝑅𝜌𝜌∝ 𝑍𝑍

(ℎ𝜈𝜈)2

– See Attix Fig. 7.13

39

ℎ𝜈𝜈 (MeV) 0.1 1 10

Al 15° 2° 0.5°

Pb 30° 4° 1.0°

ℎ𝜈𝜈 (MeV) 0.01 0.1 1

C 0.07 0.02 0

Cu 0.006 0.08 0.007

Pb 0.03 0.03 0.03

𝜎𝜎𝑅𝑅𝜇𝜇 ⇒

PHOTONUCLEAR INTERACTIONS

An energetic ℎ𝜈𝜈 > a few MeV enters and excites a nucleus, which then emits a proton or neutron

(𝛾𝛾, 𝑝𝑝) interaction• Contributes directly to the kerma• < 5% of the kerma due to pair production• Negligible in dosimetry considerations

(𝛾𝛾,𝑛𝑛) interaction• Neutrons can cause biological consequences to patients

– Allowable neutron levels should be regulated in radiotherapy x-ray beams (e.g., Linacs)

• Neutrons can activate accelerator hardware

40

TOTAL COEFFICIENTS

Mass attenuation coefficient (neglecting photonuclear interactions)𝜇𝜇𝜌𝜌

=𝜏𝜏𝜌𝜌

+𝜎𝜎𝜌𝜌

+𝜅𝜅𝜌𝜌

+𝜎𝜎𝑅𝑅𝜌𝜌

Mass energy-transfer coefficient (for ℎ𝜈𝜈 > (𝐸𝐸𝑏𝑏)𝐾𝐾)𝜇𝜇𝑡𝑡𝑟𝑟𝜌𝜌

=𝜏𝜏𝑡𝑡𝑟𝑟𝜌𝜌

+𝜎𝜎𝑡𝑡𝑟𝑟𝜌𝜌

+𝜅𝜅𝑡𝑡𝑟𝑟𝜌𝜌

=𝜏𝜏𝜌𝜌ℎ𝜈𝜈 − 𝑃𝑃𝐾𝐾𝑌𝑌𝐾𝐾ℎ�̅�𝜈𝐾𝐾

ℎ𝜈𝜈+𝜎𝜎𝜌𝜌

�𝑇𝑇ℎ𝜈𝜈

+𝜅𝜅𝜌𝜌ℎ𝜈𝜈 − 2𝑚𝑚0𝑝𝑝2

ℎ𝜈𝜈

Mass energy-absorption coefficient𝜇𝜇𝑒𝑒𝑚𝑚𝜌𝜌

=𝜇𝜇𝑡𝑡𝑟𝑟𝜌𝜌

(1 − 𝑔𝑔)

• 𝑔𝑔 = the average fraction of secondary-electron energy that is lost in radiative interactions (bremsstrahlung & in-flight annihilation)

• e.g., In Pb with ℎ𝜈𝜈 = 10 MeV, 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌

= 0.74 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌

• For low 𝑍𝑍 & ℎ𝜈𝜈, 𝑔𝑔 → 0 or 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌

= 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌

41

Coefficients for compounds and mixtures (using the Bragg rule)

• 𝜇𝜇𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

= 𝜇𝜇𝜌𝜌 𝐴𝐴

𝑓𝑓𝐴𝐴 + 𝜇𝜇𝜌𝜌 𝐵𝐵

𝑓𝑓𝐵𝐵 + ⋯

• 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

= 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐴𝐴

𝑓𝑓𝐴𝐴 + 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐵𝐵

𝑓𝑓𝐵𝐵 + ⋯

– 𝑓𝑓𝑗𝑗 = the weight fraction of the separate element 𝑗𝑗

If radiative losses are small:

• 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

≅ 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌 𝐴𝐴

𝑓𝑓𝐴𝐴 + 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌 𝐵𝐵

𝑓𝑓𝐵𝐵 + ⋯ ≅ 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐴𝐴

(1 − 𝑔𝑔𝐴𝐴)𝑓𝑓𝐴𝐴 + 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐵𝐵

(1 − 𝑔𝑔𝐵𝐵)𝑓𝑓𝐵𝐵 + ⋯

• 𝜇𝜇𝑒𝑒𝑛𝑛𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

= 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐴𝐴

1 − 𝑓𝑓𝐴𝐴𝑔𝑔𝐴𝐴 − 𝑓𝑓𝐵𝐵𝑔𝑔𝐵𝐵 − ⋯ 𝑓𝑓𝐴𝐴 + 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝐵𝐵

1 − 𝑓𝑓𝐴𝐴𝑔𝑔𝐴𝐴 − 𝑓𝑓𝐵𝐵𝑔𝑔𝐵𝐵 − ⋯ 𝑓𝑓𝐵𝐵 + ⋯ =𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

1 − 𝑓𝑓𝐴𝐴𝑔𝑔𝐴𝐴 − 𝑓𝑓𝐵𝐵𝑔𝑔𝐵𝐵 − ⋯ ≡ 𝜇𝜇𝑡𝑡𝑡𝑡𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚

(1 − 𝑔𝑔𝑚𝑚𝑚𝑚𝑚𝑚)

42