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Physics Applied to Radiology RADI R250 -- Fall 2003

Physics Applied to Radiology RADI R250 -- Fall 2003. Chapter 12. 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

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Physics Applied to Radiology RADI R250 -- Fall 2003

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  1. Physics Applied to RadiologyRADI R250 -- Fall 2003 Chapter 12

  2. Quality vs Quantity (mAs) • mAs controls QUANTITY D mAs = direct µD photon # ½mAs = ½photon # 2x mAs = 2x photon # D mAs = direct µD 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

  3. intensity (or dose) kVp Quality vs Quantity (kVp) • kVp controls QUALITY • penetration & transmission • effective energy in beam (keV) increases with increasing kVp • kVpalso effects quantity • photon # in beam increases with increasing kVp • kVp has a direct exponential relationship with intensity, ½kVp  ¼ I 2x kVp  4x I

  4. 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?

  5. Quantity Filtration (inverse) Target (direct to Z) Waveform (direct) SID (inverse square) Quality Filtration (direct) Target (direct to Z) Waveform (direct) Quantity vs. Quality (other factors that affect magnitude of)

  6. 1st HVL 2nd HVL 3rd HVL 4th HVL 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

  7. 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

  8. Determining HVL • read exp. with no added AL • IO = 310 mR • determine ½ of exp. • ½ IO = 310 ÷ 2 = 155 mR • Find ½ IOon graph • Determine AL thickness • HVL = ~ 3.2 mm Al

  9. Other layer values • quarter valuelayer • thickness of Al that results in reducing the original intensity of the beam to1/4 IO(310 to 77.5) QVL= 7.9 mm Al TVL=13.5 mm Al • tenth value layer • thickness of Al that results in reducing the original intensity of the beam to .1 IO (310 to 31)

  10. Quarter Value Layer vs. 2nd HVL • HVL • mm Althat reducesIO to1/2 • quarter value layer • mm Al that reduces IO to1/4 HVL = 3.2 mm Al QVL=7.8 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

  11. Additional HVL • 3rd HVL • added thickness of Al that results in the intensity of the beam going from 1/4to1/8 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

  12. Modifiers of HVL 1. filtration: ­ filtration ­ HVL 2. kVp: ­ beam energy ­ HVL

  13. 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

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

  15. 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

  16. 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

  17. tube thick thin body part Filtration (cont.) wedge • added filtration types 3. compensating filters • matched to body part to improve image • wedge; trough 4. compound filters (aluminum + copper) Al (Z=13) good for low E removal Cu (Z=29) better for high E • reduces filter thickness • Cu faces tube

  18. low filtration higher filtration Din Filtration # of photons 25 50 75 100 photon energy (keV) • éfilt. = é qualityBrems peak moves to R highest Brems at = E • éfilt. = êquantityê amplitude • characteristic pos. no effect, if produced

  19. 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.

  20. DDistance (SID) (no actual effect on spectrum produced) • Inverse Square Law • assumptions: • point source (d = >7x source size) • isotropic emission from source • reduction not due to attenuation • formulae

  21. 100 R d1 100 R d2 100 R 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 ­

  22. 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

  23. 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

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

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