# Radiographic Imaging and Instrumentation I Lecture 7 PDF

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Uploaded by EasiestHarmony86

OneClass

2023

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## Summary

This lecture provides an overview of Radiographic Imaging and Instrumentation I, focusing on Module 1: X-Ray Production. It covers key concepts like different emission spectra, mAs, kVp, filtration, target material and inverse and direct square laws. The lecture is well-organized and easy to follow for students of medical imaging programs.

## Full Transcript

Radiographic Imaging and Instrumentation I Module 1: X-Ray Production Lecture 7: Production of X-Ray Photons 1 What are we learning today? • Different Emission spectrums • mAs • kVp • Filtration • Target material • Inverse & Direct Square Laws X-Ray Emission Spectrum • Total no. of x-rays emit...

Radiographic Imaging and Instrumentation I Module 1: X-Ray Production Lecture 7: Production of X-Ray Photons 1 What are we learning today? • Different Emission spectrums • mAs • kVp • Filtration • Target material • Inverse & Direct Square Laws X-Ray Emission Spectrum • Total no. of x-rays emitted from the x-ray tube could be determined by emission spectrum • Graph • Total no. of x-rays emitted = area under the curve of the emission spectrum X-Ray Emission Spectrum X-Ray Emission Spectrum • General shape of emission spectrum. • Relative position on the x-axis can change. • The larger area under the curve → the higher the intensity or quantity. Intensity • # of X-rays in the useful beam • Bushong uses intensity and exposure interchangeably, so we will too • Units for exposure: Roentgen or C/kg Relating Technical Factor Selection to Beam Output • mAs affects radiation quantity ONLY by controlling number of projectile electrons used for x-ray production • kVp affects radiation quantity, average beam energy AND maximum photon energy Relating Technical Factor Selection to Emission Spectra • Spectrum amplitude indicates quantity of photons • Location of peak amplitude (‘hump’) indicates most common photon energies (and therefore approx. average) • Highest energy at the end of continuous portion is determined by kVp setting Effect of mAs 20mAs to 40 mAs while all the other conditions remain the same ……. ≈ Twice as many electrons will flow from cathode to anode ≈ Twice as many x-rays will be produced Relate it to the emission spectrum and area under the curve Change in mAs As kVp increases… – Efficiency of Brems and Characteristic production increases, so more x-ray photons produced for same number of electrons crossing the tube (mAs) – Kinetic energy of electrons increases, so larger photon energies released during Brems production, so average photon energy increases – Kinetic energy of electrons increases so Emax increases Effect of kVp • kVp ↑ → the area under the curve increases • Area increase ≈ square of the factor by which the kVp was increased • X-ray quantity will increase by the square of this factor Example • Suppose an initial exposure was made using 62 kVp and 10mAs and the total area under the curve was 6.3 cm2 which represents an x-ray quantity (Exposure) of 215 mR • What area under the curve and x-ray quantity would be if the tube voltage was raised to 70 kVp? How to solve… • kV1 = 62 kV • kV2 = 70 kV • Area1 = 6.3 cm • Exposure1 = 215 mR Area2 = (kV2/kV1)2 x Area1 = (70kV/62kV)2 x 6.3cm = 8.0cm Could then use ratios to determine exposure… 215mR/6.3cm = Exposure2/8.0cm Exposure2 = 273mR Or we can substitute… • Since Area represents quantity which we express as exposure…. Exposure2 = (kV2/kV1)2 x Exposure1 Effect of kVp • Relative distribution of emitted x-ray energy shifts to the right • Emax shifts to the right = kVp set on the control panel • Peak of curve shifts to the right • Higher average energy Effect of kVp change Effect of Filtration • Added filtration • Absorbs low energy photons • Therefore, Brems is further reduced on the left Effect of Filtration • ↓ beam intensity • ↑ average beam energy • Sometimes called Beam Hardening Added Filtration Bushong p.131 10TH Effect of Target Material • Atomic number • Z ↑ → efficiency of Brems ↑ • Z ↑ → Higher energy x-rays ↑ • What about characteristic x-rays? Effect of Target Material Effect of Generator Type/Voltage Waveform What Changed Here?(light blue to dark blue) Relative number of photons Photon energy in keV What Changed Here? (purple to dark blue) Relative number of photons Photon energy in keV Inverse Square Law • Describes the decrease in intensity as distance from source is increased • Point source • Photons travel in straight lines • Photons diverge as they travel away from the source Inverse Square Law 4 mR 3 ft 27 1 mR 6 ft Elsevier items and derived items © 2009 by Mosby, Inc., an affiliate of Elsevier Inc. Inverse Square Law 2 I1 d 2 = 2 I 2 d1 Elsevier items and derived items © 2009 by Mosby, Inc., an affiliate of Elsevier Inc. or d2 I1 = I2 d1 2 I1 is the intensity at distance d1 from the source and I2 is the intensity at distance d2 from the source. 28 Inverse Square Law • If the exposure measured at 90 cm from the source was 400 mR then calculate the exposure if the object moved to 180 cm away from the source? • Clue: DOG Problem: How do we know how to adjust our technical factors when SID changes? • Inverse square law describes rate of decrease in intensity • Knowing how much the intensity will decrease is good, but what about when we don’t want to change intensity? • Square law is derived from the inverse square law relationship and identifies how to manipulate our factors to keep the intensity of photons reaching our image receptor the same when changing distances Direct Square Law • mAs₁ /mAs₂ = (d₁ /d₂) 2 • Based on: • inverse square describes rate of decrease in intensity • mAs is primary control of intensity • Then rate of mAs increase needed to maintain intensity is inversely proportional to inverse square law • Inverse of the inverse square law = square law From Our Previous Example • Let’s say our initial mAs used was 5mAs. What would need to use when our SID was increased to maintain the X-ray intensity reaching our image receptor? • d₁ = 90 • d₂ = 180 • mAs₁ = 5 • mAs₂ = ? • mAs₁ /mAs₂ = (d₁ /d₂) 2 Summary • Define intensity • What measure do we use • Review of factors affecting beam energy • Change in appearance of the emission spectrum • Inverse & Direct Square Law → Use in lab! Readings • Unit 1.3 in course manual & study guide • Bushberg – Chapters 2 (pages 32-34) & 6 (section 6.1) • Bushong – Chapter 8 • Inverse and direct square laws are discussed in chapter 9, page 147