Interaction of X-rays with matter

Basic interactions of X-rays with matter

When electromagnetic radiation (X-rays, gamma rays) passes through matter, the incident photons
interact with the atoms of the material.

1.Coherent Scattering
2.Photo electric interaction
3.Compton Scattering
4.Pair production
5.Photodisintegration
 Coherent Scattering
In coherent scattering (also known as unmodified or elastic scattering) the incident radiation
undergoes a change in direction without a change in wavelength.

▪ This interaction consists of an electromagnetic wave passing

near the electron and setting it into oscillation.
▪ The oscillating electron reradiates the energy at the same
wavelength as the incident electromagnetic wave.
▪ Thus, no energy is changed into electronic motion and no energy
is absorbed in the medium.
▪ The only effect is the scattering of the photon at small angles.
▪ It occurs when the energy of the X-ray is small in relation to the
ionization energy of the atom. It therefore occurs with low
energy radiation (<10keV).
▪ The percentage of radiation that undergoes coherent scattering
is small compared to that of the other basic interactions; in
general, it is less than 5%.
 Photoelectric interaction
➢ An incoming x ray photon knocks out a tightly bound orbital electron (usually K shell). The energy of
the x ray photon must be equal to or greater than the binding energy of the orbital electron.
➢ The photon disappears, giving up all its energy to the electron.
➢ The electron knocked out of it’s orbit is called a photoelectron.
➢ The atom now has an empty orbital electron. The electrons from other orbitals will jump the shells
(i.e. L shell to K shell or M shell to K shell, etc.). This produces characteristic radiation within the
object being imaged. This cascade of electrons continues until the atom has filled all it’s empty
shells.
 Photoelectric interaction

ℎ𝑓 = 𝑊 + 𝐾. 𝐸.
hf = Energy of incident photon
W = electron binding energy,
also called work function= hfo
K.E. = kinetic energy of photoelectron

Ephotoelectron = Eincident photon – Ebinding energy

Example: Eincident photon = 24 keV

- Ebinding energy = 20 keV
Ephotoelectron = 04 keV
 Photoelectric Interaction:
Probability of occurrence

➢ The incident photon must have sufficient energy to overcome the binding energy of the electron.
➢ When the energy of the X-ray is just slightly greater than the binding energy, the probability that the
Photoelectric Effect will occur increases greatly.

➢ The tighter an electron is bound in its orbit, the more likely it is to be involved in a photoelectric reaction. The
Photoelectric Effect is more probable to occur in the intense electric field near the nucleus than in the outer
levels of the atoms, and it is more common in elements with high Z than in those with low Z.

Photoelectric reactions are most likely to occur with low energy photons and elements with high atomic numbers
provided the photons have sufficient energy to overcome the forces binding the electrons in their shells.
 Photoelectric interaction and radiologic contrast
➢ PEA is related to the atomic number of the attenuating medium (Z), the energy
of the incident photon (E) and the physical density of the attenuating medium (p)
by: Z³ ρ / E³.
➢ Photoelectric absorption is about four to six times greater in bone than in an
equal mass of soft tissue.
➢ From the point of view of film quality, the photoelectric effect is desirable. The
photoelectric effect does not produce scatter radiation and it enhances natural
tissue contrast.
➢ From the point of view of patient exposure, though, it is undesirable. Patients
receive more radiation from photoelectric reactions than from any other type of
interaction.

➢ The two radiocontrast agents iodine and barium have ideal K-shell binding
energies for absorption of X-rays: 33.2 keV and 37.4 keV respectively, which is
close to the mean energy of most diagnostic X-ray beams.

➢ Why is Barium used in the X-ray imaging of GIT?

➢ Soft tissue, Z= 7.4; Ba, Z=56

 Compton Scattering
➢ X-ray photon can collide with a loosely bound outer electron much like a billiard ball collides with another
billiard ball. At the collision, the electron receives part of the energy and the remainder is given to a
Compton (scattered) photon, which then travels in a direction different from that of the original X-ray.

➢ The resultant incident photon gets scattered (changes direction) and imparts energy to the electron (recoil
electron).
➢ The scattered photon will have a different wavelength and thus a different energy.

hfi = Er + hfs Er

ℎ
Compton wavelength of electron 2.43×10−12 m
𝑚0 𝑐
The wavelength shift λ′ − λ is at least zero (for θ = 0°) and at most twice the Compton wavelength of the
electron (for θ = 180°).
 Compton Scattering:
Probability of Occurrence
Interaction probability is proportional to
• the the electron density
• physical density of the material
Inversely proportional to
• photon energy
Does not depend on atomic number (unlike photoelectric effect
and pair production)

➢ However, since the Photoelectric Effect is more probable to occur in high Z materials than in low Z
materials, the fraction of X-rays that lose energy by the Compton Effect is greatest in low Z elements.

❖ Scattered radiation degrades the photographic quality.

Radiographic image contrast is less with Compton reactions

than with the photoelectric effect.
 Pair Production

➢ Photon interacts with the strong electric field around the nucleus
➢ it undergoes a change of state and is transformed into two particles (essentially electromagnetic
energy is converted into matter): one electron, one positron

▪ Equivalent energy of an electron or position is

0.51Mev.
▪ Minimum energy of the incident photon must be
1.02MeV for pair production.
▪ Corresponding maximum photon wavelength is
1.2pm. Electromagnetic waves with such wavelength
are gamma rays.
▪ Additional photon energy becomes the kinetic energy
of the electron and positron.

Pair Annihilation
RELATIVE FREQUENCY OF BASIC INTERACTIONS
How are these interactions related to diagnostic radiology?

➢ Photoelectric Effect is more useful than the Compton Effect

because it permits us to see bones and other tissues in the body.
➢ At 30keV bone absorbs X-rays about 6 times better than soft
tissue due to the Photoelectric Effect.
➢ To make further use of the Photoelectric Effect radiologists often
inject high Z materials, or contrast media, into different parts of
the body.
➢ Pair Production is of no use in diagnostic radiology because of
the high energies needed.
➢ Compounds containing iodine are often injected into the bloodstream to show the
arteries.
➢ Radiologists give barium compounds orally to see parts of the gastrointestinal tract
➢ Since gases are poorer absorbers of X-rays than liquids and solids, it is possible to use
air as a contrast medium.
➢ In a double-contrast study, barium and air are used separately to show the same organ.
➢ If the Photoelectric Effect did not exist and radiologists had to rely on the Compton
Effect, X-rays would be much less useful because the Compton Effect depends only on
the density of the materials.
Photon Beam Attenuation
 Attenuation vs Absorption

When photons interact with matter three things can occur. The photon may be:

➢ Transmitted through the material unaffected

➢ Scattered in a different direction from that traveled by the incident
photon
➢ Absorbed by the material such that no photon emerges
 Attenuation vs Absorption
Radiation a
Source
Detector

d c

➢ Initial photons produced by the source (the “primary”) and the photons created by the initial photon
interaction (the “secondary”).
➢ The “primary” photon will either be absorbed or unaffected by passage through the material as indicated
by “b”.
➢ The “secondary” photons may be absorbed such as “d” in the slide or they may be scattered such as “a”
and “c” which means that their direction has been changed so that they are not detected in the forward
direction. They were not absorbed but they are no longer traveling in the same direction as the incident
photon.
 Attenuation vs Absorption

Attenuation of the photon beam can be considered as a combination of

scattering and absorption.

Attenuation = Scattered + Absorbed

If the photons are scattered or absorbed, they are no longer traveling in

the direction of the intended target.
 Mono-energetic Radiation

• all photons in beam have same energy

• attenuation results in
• Change in beam quantity
• no change in beam quality
• # of photons & total energy of beam changes by same
fraction
 Attenuation Formula
N = No e -mx
where
N = # incident photons OR # photons penetrating absorber of 0 thickness
o
N = # transmitted photons
e = base of natural logarithm (2.718…)
m = linear attenuation coefficient (1/cm); property of
energy
material N N
o
x = absorber thickness (cm)

Mono-energetic radiation beam

 If x=0 (no absorber)
N = No e -mx
where
N = number of incident photons
o
N = number of transmitted photons
e = base of natural logarithm (2.718…)
m = linear attenuation coefficient (1/cm); property of
energy
material N N
o
x = absorber thickness (cm)

X=0
Mono-energetic radiation beam
 Exponential Attenuation

-mx
I = Io e

m is called linear
attenuation coefficient
unit is per cm.
 Linear Attenuation Coefficient

➢ Linear Attenuation Coefficient (LAC) provides a measure of the

fractional attenuation per unit length of material traversed.

µ is the linear attenuation

I = Io e -m x coefficient.
Unit: cm-1
 Attenuation

90% 90% 90% 90%

100 90 81 73 66
 Half Value Layer (HVL)
• Absorber thickness that reduces beam
intensity by exactly 1/2
• Units of thickness
• value of x which makes N equal to No / 2

N = No e -mx X1/2 = HVL = 0.693 / m

N/No= 0.5 = e -mx

Higher HVL means

more penetrating beam
lower attenuation coefficient
 Half Value Layer

A half value layer of any material will permit only 50% or ½ of the incident
radiation to pass.

A second half value layer will permit ½ of the incident radiation (already
reduced by ½) to pass so that only ¼ of the initial radiation (½ x ½) is permitted
to pass.

If “n” half value layers are used, (½)n of the initial radiation is permitted to pass.
“n” may be any number.
 Half Value Layer - Example

The half value layer (HVL) of a material is 2 cm. A

researcher has a piece of the material which is 7 cm
thick. What fraction of the initial radiation will pass
through the piece?
 Mass Attenuation Coefficient
➢ Mass attenuation coefficient
➢ Linear attenuation coefficient divided by density of the material.
𝝁
➢
𝝆

Independent of density of material.

Unit: g-1.cm2

➢ Mass Attenuation Coefficient (MAC) provides a measure of the

fractional attenuation per unit mass of material encountered.
 X-Ray Beam Attenuation

 reduction in beam intensity by Lower Higher

Energy
 absorption (photoelectric) Energy
 deflection (scattering)
 Attenuation alters beam
 quantity
 quality
 higher fraction of low energy photons
removed
 Beam Hardening
Half Value Layer
 Poly-energetic Attenuation

 Curved line on semi-log graph

 line straightens with increasing attenuation
 slope approaches that of monochromatic
beam at peak energy
 Mean energy increases with attenuation
 beam hardening
 Factors Affecting Attenuation

• Energy of radiation / beam quality

• higher energy
• more penetration
• less attenuation
• Matter
• density
• atomic number
• electrons per gram
higher density, atomic number, or electrons per gram
increases attenuation