T04- The Exposure PDF

Summary

This document covers various theoretical and practical aspects of radiation exposure, including the relationship between factors like mAs, kVp, and SID. It also touches upon the inverse square law, technical parameters, and their effect on exposure. Additionally, it examines how different situations and anatomical considerations (including patient positioning and body composition) influence the exposure.

Full Transcript

Topic 4 THE EXPOSURE Topic 4 Objectives 1. state the factors that control or influence the radiation exposure 2. Explain the EIV and calculate the new mAs using the EIV 3. explain how and why various technical parameters affect exposure, attenuation, transm...

Topic 4 THE EXPOSURE Topic 4 Objectives 1. state the factors that control or influence the radiation exposure 2. Explain the EIV and calculate the new mAs using the EIV 3. explain how and why various technical parameters affect exposure, attenuation, transmission 4. Explain the mA-time relation, ISL, mAs-kVp relation, 15% rule and mAs distance relation 5. Explain the effect grids, collimation, casts compensating filters and patient body types affect the radiation exposure 6. Understand and state the effect of kVp and SID on exposure 7. Explain the effect that different pathologies have on Exposure and how to adjust the technical factors 8. state the purpose, composition and types of filtration used for X-ray tubes 9. define "Anode Heel Effect” and describe its influencing factors and applications 10. state the effect of processing and viewing conditions on exposure 11. how to adjust the mAs and/or kVp according to grid used Informative Videos Subject contrast https://www.youtube.com/watch?v=_EcKx0SY3cg Exposure & object & IR https://www.youtube.com/watch?v=zJ-zs5lpJY8&t=158s Exposure Indicator – Referred to as Exposure Deviation in GE x-ray machines https://www.youtube.com/watch?v=zY58Ls3iWqo Anode Heel Effect: https://www.youtube.com/watch?v=fREyzdwxCjs The Exposure In radiology, the exposure is defined as “the number of individual X-ray units (quantity or intensity of photons) that emerge from the X-ray tube and X-rays were invented on November 8, 1895, by the German professor subsequently reach the IR” Wilhelm Roentgen marking the beginning of medical imaging The exposure The exposure is directly proportional to mA, time, and kVp2, and inversely proportional to SID2 Factor used to control Exposure is the mAs mAs α Intensity of radiation (mR) α Exposure o mA α To filament α # of é toward the target o time α total # é toward target o mA x Time = mAs mAs doubles  intensity doubles  Exposure doubles Comparable Exposures mAs relation: mAsN = mAsO 100 mAsn = 100 mAso mA x time = mAN X secN = mAO X secO 100 mAn X 1 secn = 50 mAo X 2 seco 8 What affects the choice of mA and time? The need to ↓ exposure time whenever patient motion is a problem The kVp needed, which depends on: o penetrating power o image contrast o ALADA - high kVp technique SID & SOD Equipment condition 9 SID & kVp have a direct effect on exposure, but they are not used to control it SOD SID Source to Image Distance (SID) Source to Object Distance (SOD) OID Object to Image Distance (OID) 11 Source to Image Distance (SID) SID has a direct effect on exposure, but it is not a controlling factor, because it also affects magnification, & skin dose 12 The inverse Square Law The intensity of the beam (exposure) is inversely proportional to the square of the Distance (SID) 13 The Inverse Square Law Dispersion or Divergence of the x-ray beam X-rays disperse or diverge from the target Total # of photons in the beam is constant at all distances Total # of photons/mm2 α 1/distance 14 Clinical Application of the ISL Radiation Protection in mobile radiography and radioscopy 15 ISL Problems The radiation exposure-rate measured with a dosimeter at 40’’ from the tube, is 2 R/min. What would be the exposure rate at 50’’ from the source? IO = 2 R/min. IN = (Do2 / Dn2) X Io DO = 40’’ IN = (40”)2 / (50”)2 X 2 R/min. DN = 50’’ IN = (1600” / 2500”) X 2 R/min. IN = ? IN = 1.2 R/min 16 A radioscopic X-ray tube operating at 3 mA and 120 kVp emits a radiation output of 0.36 mR/sec. at 40" from the focal spot. Calculate the radiation output at 20" from the source. IO = 0.36 mR/sec. IN = (DO2 / DN2) X IO DO = 40’’ IN = (40”)2 / (20”)2 X 0.36 mR/sec. DN = 20’’ IN = (1600” / 400”) X 0.36 mR/sec. IN = ? IN = 1.44 mR/sec 17 A patient undergoing radiation therapy receives a skin dose of 5 Rads/minute when the source to skin distance is set at 100 cm. Calculate the skin dose if the radiation source is placed at 150 cm. Solution: Io = 5 Rads/min. In = (Do2 / Dn2) X Io Do = 100 cm In = (100 cm)2 / (150 cm)2 X 5 Rads/min. Dn = 150 cm In = (10,0000 cm2 / 22,500 cm2) X 5 Rads/min. In = ? In = 2.22 Rads/min. mAs-Distance Relation (or Square-Law) The mAs-Distance Relation corrects the Exposure when SID varies A small ↑ is distance results in a very large ↓ in exposure If SID ↑, mAs must be increased to maintain the same exposure to the IR If DN = 2 DO, then mAsN = 4 X mAsO If DN = ½ DO, then mAsN = ¼ X mAsO If DN = 3 DO, then mAsN = 9 X mAsO If DN = ⅓ DO, then mAsN = ⅑ X mAsO 19 mAs- Distance Problems In a given radiographic room, an upright CXR is taken at 10 mAs, 120 kVp at 72" SID on an average size patient. The following day, the same patient comes on a stretcher for another CXR, but this time the patient cannot sit or stand. What mAs should be used to maintain the same exposure as the previous X- ray, if the maximum SID that can be used under the new condition is 40"? mAsO = 10 mAs mAsN = (DN2 / DO2) X mAsO DO = 72’’ mAsN = (40”)2 / (72”)2 X 10 mAs DN = 40’’ mAsN = (1600” / 5184”) X 10 mAs mAsN = ? mAsN = 3.09 mAs If the mAs is 100 at 40" SID, then what is the new mAs at a new SID? mAsN = (DN2 / DO2) X mAsO = (202 / 402) X 100 mAs = 25 mAs mAsN = (DN2 / DO2) X mAsO = (302 / 402) X 100 mAs = 56.25 mAs mAsN = (DN2 / DO2) X mAsO = (502 / 402) X 100 mAs = 156.25 mAs mAsN = (DN2 / DO2) X mAsO = (602 / 402) X 100 mAs = 225 mAs mAsN = (DN2 / DO2) X mAsO = (702 / 402) X 100 mAs = 306.25 mAs mAsN = (DN2 / DO2) X mAsO = (802 / 402) X 100 mAs = 400 mAs 21 A portable X-ray is taken at 50 mAs, 80 kVp at 60" SID. What mAs should be used if the SID increases by a factor of 1.3? Solution: DN = 60” X 1.3 = 78” mAsO = 50 mAs mAsN = (Dn2 / Do2) X mAsO DO = 60’’ mAsN = (78”)2 / (60”)2 X 50 mAs DN = 60” X 1.3 mAsN = (6084/3600) X 50 mAs mAsN = ? mAsN = 84.5 mAs In practice the SID may be changed due to: Need to ↑ sharpness (↓ Ug) Need to ↓ skin dose (↑ SSD) Equipment limitations Room limitation (mobile) ↑ or ↓ Magnification 23 Exposure-kVp Relation: “The change in the intensity of exposure (mR) is directly proportional to the square of the new kVp”. IN (mR) kVpN2 ------------- = --------- IO (mR) kVpO2 24 kVp and Exposure kVp has the greatest effect on Exposure because the energy of the beam increases causing a greater transmission of the primary beam & forward scatter 25 kVp 15% Rule Applicable to mid-range kVp only: 65-90 kVp 15% ↑ in kVp results in 2X exposure (mR) & vice-versa 0.85 X kVp + 2 X mAs = same exposure 1.15 X kVp + 0.5 X mAs = same exposure Quick calculation: 15% = 10% + 5% (half of 10%) 26 15%-Rule Problems An X-ray of the shoulder is taken at 20 mAs, 70 kVp and 48" SID. What kVp would give a comparable exposure if the mAs is doubled? mAsO = 20 mAs mAsN = mAsO X 2 mAsN = 40 mAs mAsN = 20 mAsO X 2 = 40 mAs kVpO = 70 kVp kVpN = 70 kVp X 0.85 kVpN = ? kVpN = 59.5 kVp 27 What kVp would give a comparable exposure, if the mAs in previous problem is reduced by half? Solution: mAsN = mAso / 2 mAsO = 20 mAs mAsN = 20 mAso / 2 = 10 mAs mAsN = 10 mAs kVpO = 70 kVp kVpN = 70 kVp X 1.15 kVpN = ? kVpN = 80.5 kVp 28 What kVp will produce a comparable exposure to the one producedat 40 mAs, 90 kVp, if the new mAs is: a) reduced to 10 mAs? b) increased to 80 mAs? mAsO = 40 mAs mAsO = 40 mAs mAsN = 10 mAs mAsN = 80 mAs kVpO = 90 kVp kVpO = 90 kVp kVpN = ? kVpN = ? kVpN = 90 kVp X 1.15 X 1.15 kVpN = 90 kVp X 0.85 kVpN = 119 kVp kVpN = 76.5 kVp 29 kVp controls IC, penetrating power & patient dose Choice of optimal kVp is determined by the: o minimum penetrating power required o desired IC, which depends on SC: if SC is high (chest area), an ↑ in kVp will be needed if SC is low (mammo.), a ↓in kVp will be needed o exposure time needed o Need to respect the ALADA principle – high kVp & low mAs technique decreases patient dose o department’s preference 30 A HIGH kVp & LOW mAs TECHNIQUE LOWERS PATIENT DOSE Technical Factors used for exposure A: 20mAs, 70kVp, 100” SID Technical Factors used for exposure B: 10mAs, 80kVp, 100” SID  PD ↓ 31 kVp controls IC, penetrating power & patient dose kVp controls IC If mAs & distance are not modified to obtain the same exposure, then increasing kVp will increase the dose! If mAs &/or distance are adjusted to maintain same exposure, when kVp is modified then kVp is inversely proportional to patient dose 32 kVp controls IC, penetrating power & patient dose Why INCREASING kVp results in a PD reduction only if mAs is modified accordingly to maintain comparable exposures? 1. More x-ray transmission and less absorption. It increases the remnant beam which contributes to form the image and contrast therefore mAs can be decreased, thus reducing PD 2. Higher kVp will increase forward scatter even though there will be less scatter produced o more scatter will be able to exit the body of the patient in the forward direction rather than be absorbed by the surrounding tissue 100 mR = 80 kVp and 50 mAs 100000 photons 100 mR = 80 kVp and 50 mAs 100000 photons 200 mR = 92 kVp and 50 mAs 200000 photons 100 mR = 92 kVp and 25 mAs 50000 photons Patient dose increased Patient dose decreased 33 Internal Factors Affecting Exposure Composition of the Anatomical Part:  Exposure increases if: Atomic number (Z) ↑ Tissue Density (ρ) ↑ : (or mass density) o patient absorption α tissue density o Examples: Chest XR in inspiration: ρ ↓, absorption ↓ & Exp.,↑, should ↓ mAs Chest XR in expiration: ρ ↑, absorption ↑ & Exp. ↓, should ↑ mAs 34 Internal Factors Affecting Exposure Part Thickness & Compression: If thickness ↑, absorption ↑ and Exp. ↓; mAs must be ↑ to maintain Exp. Compression usually ↑ Exp. : ↓ volume by pushing tissues to sides and ↓ thickness; mAs must ↓ to maintain Exp. Compression may sometime ↓ Exp. example: o removal of gas shadow and ↑ ρ will ↓ Exp. Some applications of compression include: GI, IVU, Mammo 35 Internal Factors Affecting Exposure Body Habitus: Skull: o Brachycephalic o Mesocephalic B M D o Dolichocephalic Subject types: o Hypersthenic o Sthenic o Hyposthenic o Asthenic Age / Sex: o Tissue density α 1/age o Tissue density: M > F 36 Internal Factors Affecting Exposure AP Supine AP Upright Radiographic Projections/Positioning: different positions → different superimposition different tissue densities different thicknesses o Examples:  Abdomen supine vs. upright vs. prone vs. decubitus AP RLD  AP ribs Vs. oblique ribs Caliper (thickness) vs. mAs:  Accuracy depends on several other factors such as tissue superimposition, pathology, age, etc. 37 A B C D 38 Lumbo-Sacral Spine Projections AP Lateral Oblique 39 Internal Factors Affecting Exposure 40

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