Physics Applied to Radiology RADI R250 -- Fall 2003

1
Physics Applied to RadiologyRADI R250 -- Fall
2003

• Chapter 12

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Quality vs Quantity (mAs)

• mAs controls QUANTITY
• D mAs direct µD photon
• ½ mAs ½ photon
• 2x mAs 2x photon
• D mAs direct µ D dose
• mAs does not change ability of beam to penetrate
  object so beam quality does not change when mAs
  is changed

photon (or dose)
mAs
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Quality vs Quantity (kVp)

• kVp controls QUALITY
• penetration transmission
• effective energy in beam (keV) increases with
  increasing kVp
• kVp also effects quantity
• photon in beam increases with increasing kVp
• kVp has a direct exponential relationship with
  intensity,
• ½ kVp ? ¼ I
• 2x kVp ? 4x I

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mAs and kVp problem examples

• If a 200mA, 75kVp, 31ms results in a 28 mR
  exposure, determine the new exposure if
• A change to 52ms is made?
• A change to 70kVp is made?
• Both changes are made?

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Quantity vs. Quality (other factors that affect
magnitude of)

• Quantity
• Filtration (inverse)
• Target (direct to Z)
• Waveform (direct)
• SID (inverse square)

• Quality
• Filtration (direct)
• Target (direct to Z)
• Waveform (direct)

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Half Value Layer

• HVL thickness of material (Al) that lowers beam
  intensity to ½ the original value
• measures relative beam E compared to attenuation
  ability quality indicator
• monoenergetic
• HVL always

polyenergetic HVL é

• Dx x-ray beam HVL 4 - 8 cm in tissue
• 3 - 5 mm in Al

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Determining HVL

• Measure radiation without filtration
• Measure radiation passing through equal
  incremental units of Al
• Graph data
• x-axis Al thickness
• y-axis radiation
• Read graph for thickness of AL that I to ½ I0

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Determining HVL

• read exp. with no added AL

IO 310 mR

• determine ½ of exp.
• ½ IO 310 2 155 mR

• Find ½ IO on graph

• Determine AL thickness

• HVL 3.2 mm Al

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Other layer values

• quarter value layer
• thickness of Al that results in reducing the
  original intensity of the beam to 1/4 IO (310 to
  77.5)

QVL 7.9 mm Al
TVL13.5 mm Al

• tenth value layer
• thickness of Al that results in reducing the
  original intensity of the beam to .1 IO (310 to
  31)

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Quarter Value Layer vs. 2nd HVL

• HVL
• mm Al that reduces IO to 1/2
• quarter value layer
• mm Al that reduces IO to 1/4

HVL 3.2 mm Al
QVL7.8 mm Al
HVL2 7.8 3.2 4.6 mm Al

• 2nd HVL
• added thickness of Al to HVL that results in IO
  going from 1/2 IO to 1/4 IO

1st
2nd

• HVL2 QVL - HVL

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Additional HVL

• 3rd HVL
• added thickness of Al that results in the
  intensity of the beam going from 1/4 to 1/8 the
  original intensity
• example
• 77/2 38.5 mR
• 38.5 mR at 12.5 mm Al
• 13.5 - 7.8 5.7 mm Al

1st
2nd
3rd
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Modifiers of HVL

• 1. filtration filtration HVL
• 2. kVp beam energy HVL

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Homogeneity Coefficient (HC)

• numerical value to evaluate range of energies in
  beam
• Homogeneous beam monoenergetic beam
• single keV in beam
• as beam hardens it becomes more homogenous

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HC

• ratio of 1st HVL to 2nd HVL
• monoenergetic
• heterogenic
• error

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Filtration

• attenuation of beam by matter in path of beam
  before it reaches imaged/treated object
• inherent filtration
• tube components in path
• tube housing collimator components, etc.
• 1 mm Al (equivalent)
• D over time due to deposits from anode
• Beryllium tube window (mammography)
• allows low E photons needed for exam

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Filtration (cont.)

• added filtration types
• 1. beam hardening
• inserted Al filter to average E in beam
• removal of more low E than high E
• 2. heavy metal filters
• improves image contrast in contrast exam
• offset PE K-edges of filter contrast media so
  that a contrast media better absorbs photons

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Filtration (cont.)
wedge

• added filtration types
• 3. compensating filters
• matched to body part to improve image
• wedge trough
• 4. compound filters (aluminum copper)
• Al (Z13) good for low E removal
• Cu (Z29) better for high E
• reduces filter thickness
• Cu faces tube

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D in Filtration
of photons
25
50
75
100
photon energy (keV)

• é filt. é quality Brems peak moves to R
• highest Brems at E
• é filt. êquantity ê amplitude
• characteristic pos. no effect, if produced

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Inverse Square Law

• The intensity of a beam of x-rays is inversely
  proportional to the square of its distance from
  the source of the radiation.

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D Distance (SID)

• (no actual effect on spectrum produced)
• Inverse Square Law
• assumptions
• point source (d gt7x source size)
• isotropic emission from source
• reduction not due to attenuation
• formulae

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Reason ISL WorksWhy does intensity as distance
?

• Due to the geometry of beam
• A finite amount of radiation produced at source
  for a given exposure
• Only small area of produced radiation is used
• Isotropic emission causes radiation to spread
  over a larger area as distance
• As distance the area

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ISL Problem

• A technologist receives a dose of 100 mR while
  standing 2 from the x-ray tube. What would the
  dose be if the technologist moved to 3 away?
• I1 d12 I2 d22
• (100 mR)(2')2 I2(3')2
• 400 9 I2
• I2 400/9
• I2 44.44 mR 40 mR

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Factor of Change

• relative measure of how dose will change
• multiplier for I1
• uses only distance values
• D d12/d22
• What is the change in dose that occurs if you
  move from 52" to 24" from the source of exposure?
• D d12/d22
• D 522/242
• D 2704/576 4.4694 4.5 times the dose or
  4.5 I1

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Factor of Change

• Example 2
• What is the factor of change when you move from
  10" to 40" from a source of radiation?