Radiation Quality

Chapter 4

X-ray Intensity

• Intensity: the amount of energy present per unit time per unit area perpendicular to the beam direction at the location of interest.The number of photons reaching the detector per

second is a measure of beam intensity (photons/cm2).Exposure Rate: (mR/hr); dose rate: (cGy/min)

• Intensity of photon beam at the tumorDepends primarily on the original strength of the

beam.Reduced by beam divergence and attenuation.

Beam Divergence

• The area over which radiation spreads, is proportional to the square of the distance from the source.

• Inverse square law: intensity is inversely proportional to the square of the distance from the source.

I1 / I2 = D22 / D1

2

Or

I2 = I1(D1/D2)2

Beam attenuation

• Attenuation: the removal of energy from the beam.• X-rays interact with charged particles through the

electromagnetic fields associated with the electric and magnetic fields of electrons and nuclei. When a beam passes through matter, energy is removed from

the beam.

• Photons will either… Interact with the filter/attenuator Be absorbed by the material

They deposit all their energy in the filter Direction changed or scattered Unaffected by the filter

Transmission

• Transmission: the ratio of beam intensity I to IO.

• As the filter thickness increases, the intensity of the attenuated beam drops.

• Transmission = I/ IO

IO: initial intensity at the detector before filtration

I: the final beam intensity after filtration

Transmission

• Photon source with single energy- attenuation of the beam:I = IOe-μx

• x = thickness of the filter• μ = linear attenuation coefficient, (length-1)• e = base of natural logarithm (2.718)

• Each millimeter of thickness added to the filter reduces the beam by a constant percent

Transmission Example

• Attenuation coefficient (μ) = 0.2 mm-1

• Thickness (x) = 3mm• IO = 2000 photons

I = IOe-μx

I = 2000 * e(-0.2 *3)

I = 2000 * 0.549I = 1098 photons

Linear Attenuation Coefficient

• Linear attenuation coefficient (μ ): a function of the filter material and the energy of the photon beam.Represents a probability per unit thickness

that any one photon will be attenuated.

• Half-Value Layer determined by μ

Monoenergetic / Homogenous

• Monoenergetic/homogenous: all photons in the beam have the same energyμ: remains unchanged for all filter thicknesses or

number of photons removed (number of photons change)

Higher μ higher probability of interaction smaller HVL

• More easily reduced in intensity

• An exponential function produces a straight line on semi-log graph paper.

Polyenergetic / Heterogenous

• Polyenergetic/heterogenous: broad range of photon energies (bremsstrahlung).μ: each energy has a different value.

• On semi-log graph paper, the slope changes as filter is added due to beam hardening.

• Beam Hardening: the effective energy of the beam increases as it passes through the filter.Only occurs in a polyenergetic/heterogeneous beam.

Half-Value Layer

• Half-value layer (HVL): the thickness required for a particular material to cut the beam’s intensity in half. HVL = 0.693/ μUsed to describe the beam’s penetrabilityConvenient to characterize different bremsstrahlung

beams using their attenuation characteristics.• Materials used to specify beams HVL changes

with energy rangeDiagnostic/superficial: mm AlOrthovoltage: mm CuMeV: mmPb

Homogeneity Coefficient

• Homogeneity Coefficient: HC = 1st HVL/2nd HVL

• As HC 1: the more alike individual beam energy are.

• 1st HVL = 2nd HVL = 3rd HVL… monoenergetic beam

Equivalent Energy

• Equivalent Energy: represents the average energy of the beamThe energy of the monoenergetic beam that

would have an HVL equal to the first HVL of the bremmsstrahlung beam in question.

Attenuation Coefficients

• Linear attenuation coefficient (μ): gives the probability that a given photon will be attenuated in a unit thickness of a particular attenuator.– Photon energy & material dependant

• Mass attenuation coefficient (μ/ρ): probability of interaction per unit mass length when μ/ρ << 1, (cm2/g)– Decreasing the density of the material will cause

much less attenuation.