A LEVEL PHYSICS MEDICAL IMAGING SLIDES.pptx

A LEVEL PHYSICS OCR SPECIFICATION

1. A2 Unit G485: Fields, Particles and Frontiers of Physics Module 4: Medical Imaging

2. Module Content X-rays The interaction of X-rays with matter X-ray Intensity 3D X-rays Radioactive Tracers & the Gamma Camera Magnetic Resonance MRI Non-Invasive Techniques in diagnosis Ultrasound Acoustic Impedance Doppler Scans

3. Unit 5.4.1 X-Rays Assessable Learning Outcomes Describe the nature of X-rays Describe in simple terms how X-rays are produced. Describe how X-rays interact with matter Photoelectric Compton Effect Pair Production Define intensity as the power per unit cross-sectional area. Select and use the equation I=I e - μ x Describe the use of X-rays in imagine internal body structures including the use of image intensifiers and of contrast media. Explain how soft tissues like the intestines can be imaged using barium meal. Describe the operation of a Computerised Axial Tomography (CAT) scanner Describe the advantages of a CAT scan compared with an X-ray image

4. a) Describe the nature of X-Rays What does this mean? It is asking if X-rays are waves or particles. X-rays are electromagnetic waves (c) whose spectrum covers a range from about a hundredth to a hundred thousandth of the wavelength of light (a)

5. b) Describe in simple terms how x-rays are produced What does the diagram show? X-rays are produced by bombarding tungsten with high energy electrons. (general statement)

6. b) Describe in simple terms how x-rays are produced In an X-ray tube, electrons are emitted from a from a heated filament and accelerated through a high potential difference towards a tungsten anode. When the electrons smash into the anode, the decelerate and some of their E K will be converted to EM Energy as X-ray photons. This emission is continuous.

7. b) Describe in simple terms how x-rays are produced. X-rays are also produced when beam electrons knock out other electrons from the inner shells of the tungsten atoms. Electrons in the atoms’ outer shells move into the vacancies in the lower energy levels, and release energy in the form of X-rays.

8. b) Describe in simple terms how x-rays are produced. Only about 1% of the electrons’ E K is converted into X-rays while the rest is converted into heat. To avoid overheating the anode is set to rotate at about 3000 rpm. If the anode is stationary, a coolant can be used to ensure overheating does not take place.

9. b) Extra: Energy of an X-ray Photon An electron, charge e, accelerated through a potential difference V gains energy eV. The energy of an X-ray photon can not be more than this, so the E max of an X-ray will also be eV. Since photon energy E = hf We get E max = hf max = eV f max =   In terms of wavelength λ , λ min = Since minimum frequency occurs for minimum wavelength  

10. X-ray production Emission of electrons from the cathode is caused by thermionic emission. Only about 1% of energy is transferred to X-rays. The intensity of the beam is determined by the filament current. The energy of the X-ray photons is determined by the anode voltage. A higher voltage gives higher-energy, shorter wavelength X-rays which are more penetrative.

11. Question 1 Find the f max and λ min for X-rays produced by a tube across which the p.d . is 90,000 V.

12. Question 2 An X-ray tube has an efficiency of 60%. It uses a voltage of 90,000 V and the current through the 24 mA. Calculate Electrical power supplied Power of the X-ray beam.

13. Question 3 An X-ray tube at a dentists uses an accelerating voltage of 70,000 V . Calculate the E k of an electron accelerated through this voltage in a vacuum. If all the energy is converted into a photon of X-radiation, what is the frequency. Calculate the wavelength in question. Thus, the kinetic energy of an electron accelerated by a potential difference of 50 kV is 50 keV . One eV is a very small amount of energy, as there are 6.24 × 10 18  eV/J.

14. Question 4: Draw a diagram of an X-ray tube.

15. Question 5: A crystal is irradiated by X-rays of wavelength 2.7 x 10 -11 m, and the first order image is formed at an angle of 15°. Using the diffraction formula n λ = d sin θ . Calculate the spacing between the atoms in the crystal. d = wavelength/sin 15

16. Extra: Variation of Intensity Not all X-ray wavelengths have the same wavelength, which results in a variety of intensities. (will look into intensity in more detail later) A graph of variation is as follows:

17. The graph shows the intensity beginning at the minimum wavelength. Intensity then falls away at greater wavelengths, with the exception of certain peaks. These are characteristic of the elements in the anode. This occurs due to the wave-particle duality of X-rays.

18. c) Describe how X-rays interact with matter. (Photoelectric effect) In the same way as UV, X-rays are capable of causing the emission of photoelectrons. E UV photon = Work Function + E k photoelectron Where the work function is the energy required to liberate an electron from the photocathode. Typical values = 4.6 eV = 2.9 eV + 1.7 eV

19. c) Describe how X-rays interact with matter. (Photoelectric effect) These values are very different when looking at X-rays. Using the worked example of 90,000 V, we would have 90000 eV = 2.9 eV + 89997 eV The work function is so small it can be negated. The emitted photoelectrons when produced using X-rays have a E k maximum = the photon energy of the X-rays.

20. c) Describe how X-rays interact with matter. (Photoelectric effect) These emissions cause ionisation in the same way as beta particles. Measuring the energy of the electrons is a method of detecting and measuring the energy of X-rays.

21. c) Describe how X-rays interact with matter. (Pair Production) If the X-ray emissions are of high frequency, it is possible for a photon of X-rays to collide with a particle. This spontaneously produces a positron and an electron. Energy of X-ray photon = + Mass of both positron and electron is 0.00555 u. Where 1 u = 931 MeV. Then mass of photon can be given by 2 x 0.00055 u x 931 MeV u -1 = 1.02 MeV  

22. Because of having such a high potential difference, the effect will only be seen in high voltage systems or in those with high energy gamma rays.

23. c) Describe how X-rays interact with matter. (Compton Effect) This states that deflected X-ray photons have a longer wavelength than the initial wavelength. The degree of change in wavelength varies with the angle of deflection.

24. Question 1: When a photon of wavelength λ is deflected through an angle θ by a stationary electron, the change in the wavelength of the photon is ) Calculate the wavelength of the emerging photon at an angle of 30° when the incident photon is gamma radiation from cobalt-60 with an energy of 1.3 MeV.  

25. Question 2 a: An X-ray photon of energy 1.22 MeV generates an electron positron pair. The energy required for this is 1.02 MeV. How will the law of conservation of energy apply? What becomes of the 0.02 MeV difference between the two values? It becomes kinetic energy of the positron and electron.

26. Question 2 b: If the X-ray photon is of even higher energy, say 1.62 MeV, how does it require modification to your answer in 2 a) ? What would happen to the speed? The electron and positron would gain speeds greater than c.

27. Question 3 Calculate the loss of energy of the photon from question 1. Put this energy into electronvolts . What happens to this energy?? 330 keV . It becomes extra kinetic energy of the scattered electron.

28. d) Define intensity as the power per unit cross-sectional area. You have to be careful in defining X-ray intensity, as it can vary.

29. High-Energy EM Radiation, UV, X-ray and Gamma rays are dangerous. For this reason it is important to limit the exposure to patients and radiographers. To do so, any dose of X-rays needs to be assessed. To do this, you need to know: - Intensity - Amount of absorption - Exposure time

30. d) cntd : X-ray absorption Collimated: When these X-ray beams pass through any substance, the intensity decreases with distance. Amount of absorption varies with frequency. Low frequency (low energy) X-rays are mostly absorbed by causing photoelectrons. Slightly higher frequencies = compton effect Even higher = pair-production.

31. e) Select and use the equation I=I e - μ x Above is the equation for Intensity I is the Initial Intensity μ is the constant called the attenuation coefficient X is the distance through the medium.

32. e) cntd : Attenuation Constant The attenuation co-efficient for various materials vary with the wavelength of the X-rays being used. Typical values for μ are: For a vacuum: 0 Flesh: 100 m -1 Bone: 300 m -1 Lead: 600 m -1

33. e) cntd : Half-Value Thickness This is the distance through a material through which the X-rays must pass that halves the intensity.

34. e) cntd : Using the Equation Calculate the percentage of the intensity of X-rays not absorbed after passing through 1 cm of flesh, bone and lead. Calculate the half value thickness of bone.

35. (f) describe the use of X-rays in imaging internal body structures including the use of image intensifiers and of contrast media Photographic Film: Requires considerable exposure, produces only a still image. Quality can be improved using a film that is more sensitive to X-rays Put a fluorescent plate behind the film. Use an X-ray absorbing substance as a contrast medium. Giving patients barium meal (barium sulphate) improves contrast. Use an image intensifier Using digital methods instead of film. This includes dots that respond to X-rays. Can be recorded to give a moving image, or printed.

36. X-ray detectors Black and white photographic film; requires a beam of high-intensity. Intensifying screen, which contains a material that absorbs energy from the X-rays and re-emits it as light by fluorescence . This light then produces an image. Fluoroscopic image intensifier . X-rays cause electrons to be emitted from a photocathode. They produce a bright image on a fluorescent screen. This is used when a moving image is needed.

37. e) and f) Questions Suggest a typical value for the intensity of an X-ray beam. Use values from previous examples. 90,000 V 24 mA Efficiency of X-ray tube 60% X-ray tube output efficiency: 0.5% General area of cover: 20 cm 2 – 200 cm 2 .

38. Calculate the intensity on an X-ray plate beneath 3 cm of bone, 15 cm of muscle and 2 cm of lead. Initial intensity: 4.8 x 10 3 W m -2 Attenuation constants: Lead 90 Bone 53 Muscle 6.9 e) and f) Questions

39. January 2011 8a) Describe the use of image intensifiers and contrast media when X-rays are used to produce the images of internal body structures. b) A student suggests an image intensifier uses the photoelectric effect. Explain why this is incorrect . c) i ) Explains how the production of a CAT scan image differs from that of a simple X-ray image. c)ii) Describe the advantages of a CAT scan compared to an X-ray image.

40. Jan 2012 7a) Describe, in simple terms, how X-ray photons are produced in a hospital X-ray machine. b) i ) Explain what is meant by a photon. ii) Explain why an X-ray photon has a greater energy than a photon of visible light.

41. Jan 2012 7)c) Electrons having maximum kinetic energy create the shortest wavelength X-ray photons. Calculate the shortest wavelength of X-ray photons emitted from an X-ray machine operating at 120 kV. 7)d) X-ray photons interact with matter such as in the photoelectric effect. State another interaction mechanism, describing what happens to the X-ray photon interacting with a single atom using the mechanism you have stated.

42. Jan 2013 6)c) A beam of X-rays of intensity 3x10 9 W m -2 is used to target a tumour in a patient. The tumour is situated at a depth of 1.7 cm in soft tissue. The attenuation coefficient of soft tissues is 6.5 cm -1 . Show that the intensity is roughly 5x10 4 W m -2

43. Jun 2010 10)a) State and describe one way in which X-ray photons interact with matter. B) Intensity of an initial beam of X-rays is reduced to 10% of its initial value after passing through 3.00 mm of soft tissue. Calculate the thickness of the soft tissue that reduces the intensity to 50%. C) i ) Explain how imaging intensifiers are used to improve the quality of the X-ray image. (explain how the image is made brighter) C)ii) Explain how contrast media are used to improve the quality of the X-ray image

44. Jun 2011 No exam questions on material to date.

45. Jun 2012 7)c) What does I o represent? c)ii) Bone attentuation : 3.3 cm -1 . Calculate half value thickness.

46. Describe the use of X-rays in imaging internal body structures. 2D X-rays Traditional X-rays show a shadow of the part of the body being imaged. This can be used to show bone breakages. The shadow will be reasonably sharp provided the X-rays come from a point. An extended light source will give a fuzzy image. Draw a diagram to show the difference!

47. 2D imaging can be difficult in certain situations! Tibia blocking the Fibula Ulna blocking the radius X-ray of the chest cavity Radiographer not positioning person correctly These can cause overlap and block out the desired area.

48. Angiograms These are obtained by a method called subtraction technique. X-ray is taken and then digitised. A contrast medium is then injected, and another X-ray is taken and digitised. The first image is “taken away” from the first, so that only differences are shown. Eliminates all the detail that is not required. Computer can be programmed to get rid of patient movement.

49. Computerised Axial Tomography Computers along with X-rays can be used to give 3D images of the body. Images must be taken from different viewpoints. Final image is generally a slice of the body taken horizontally.

50. CAT Scan The X-ray source is shielded. X-ray beam emerges from a point and spreads out through the patient. Fan shaped, very thin beam. Irradiates only very small sections at a time. X-rays are detected by a ring of up to 1000 detectors. Once it completes one revloution , it has moved up 1 cm. This can take slices, or compose a 3D image of an organ.

51. CAT Scans There is a dose of radiation, but this is less than it used to be because of increased sensitivity of the sensors. Advantage: Can be taken quickly, so can have many in one day.

52. Jan 2011 8)c) i ) Explain how the production of a CAT scan image differs from that of a simple X-ray image. ii) Describe the advantages of a CAT scan compared to an X-ray image.

53. Jan 2013 6)d) Describe the operation of a computerised axial tomography scanner. State one of the advantages of a CAT scan image over a conventional X-ray image.

54. Unit 5.4.2 Diagnostic Methods in Medicine Candidates should be able to: (a) describe the use of medical tracers like technetium-99m to diagnose the function of organs; (b) describe the main components of a gamma camera; (c) describe the principles of positron emission tomography (PET); (d) outline the principles of magnetic resonance, with reference to precession of nuclei, Larmor frequency, resonance and relaxation times; (e) describe the main components of an MRI scanner; (f) outline the use of MRI (magnetic resonance imaging) to obtain diagnostic information about internal organs (HSW 3, 4c and 6a); (g) describe the advantages and disadvantages of MRI (HSW 4c & 6a); (h) describe the need for non-invasive techniques in diagnosis (HSW 6a); ( i ) explain what is meant by the Doppler effect; (j) explain qualitatively how the Doppler effect can be used to determine the speed of blood.

55. Radioactive Tracers and the Gamma Camera Two reasons medical tracers can be placed in a body: Diagnose disease or Treat Disease In both cases, several factors must be accounted for: Gamma ray sources must be used! Alpha and Beta would be absorbed and damage the body. Gamma-ray sources will expose patients to some radiation. Half-Life of the source must be long enough to carry out the investigation, but no longer. Source must not be chemically poisonous! Tracer must be possible to monitor. Must be possible to get the material to the part of the body where it is needed.

56. Medical Tracers These are radioactive substances that are used to show tissue/organ function. Benefit over X-rays: Shows both structure and function. Tracers (Technetium-99) are bound to a substance that is used by the body. Tracer is injected/swallowed and moves through the body to the region of interest. Once the substance is used up, the radiation emitted is recorded (Gamma Camera/PET scanner)

57. Technetium-99 Emits gamma radiation Has a half life of 6 hours Long enough to be recorded, short enough as not to cause any damage to the body. Decays to a stable isotope.

58. What can tracers show? Areas of damaged tissue in the heart by detecting areas of decreased blood flow. This can reveal coronary artery disease and damaged or dead heart muscle caused by heart attacks. They can identify active cancer tumours by showing metabolic activity; cancers will take up more tracer. Can show blood flow in the brain. Helps research and treat neurological conditions such as Parkinson’s, Alzheimer’s, epilepsy, depression etc.

59. Detecting Gamma Radiation The gamma camera is basically a detector of gamma photons emitted by a source inside a patient. A block of lead with tens of thousands of vertical holes is located near the body. These collimate the beams so only vertically travelling photons are detected: Ensures accuracy! They then enter a large crystal of Sodium Iodide. This scintillates when it absorbs a gamma photon. This means gives off light! Behind this is a layer of photomultiplier tubes. These multiply the effect of the tiny flash,and gives an electrical pulse output.

60. Advantage of Gamma Camera! No radiation from parts of the body where there are no radioactive material. Doctor can look at a specific part of the body!

61. Draw a gamma camera.

62. In Use Used to diagnose Thyroid Liver Brain Kidneys Lungs Spleen Heart Circulatory System

63. In Use Radioactive nuclide technetium-99 is used. Nucleus decays from its excited state to its ground state with the emission of 140 keV photon, with a half-life of 6 hours. This can be incorporated into many different kinds of molecules. Example: Iodine compound containing this can be administered to test the function of the thyroid gland.

64. 3D Images from Gamma Camera Some cameras have more than one scintillating crystal; two at right angles to each other. These are around the patient, and allows 3D images to be found.

65. Materials with a longer Halflife Methods are being developed for administering radioactive material with a longer half-life to people with certain cancers. These will be attached to cancerous cells, thereby destroying them without giving dangerous doses of radiation to healthy tissue.

66. Positron Emission Tomography Extension of gamma-ray photography. Detects abnormal chemical/metabolic activity Radiolabelled glucose is injected into the patients bloodstream, from which it is absorbed into the tissues, which need glucose for respiration. Positron Emission = annihilation of positron and electron and 2 511 keV gamma photons are released in opposite directions simultaneously. If these two are detected simultaneously, the position can be detected.

67. PET Scanning in more detail! Patient is injected with a substance used by the body. This has a positron-emitting radiotracer with a short half-life. Patient is left for a time so that the radiotracer can move through the body. Positrons ( β + ) emitted from the radioisotope collide with electrons in the organs; results in annihilation which gives off high energy gamma rays. Detectors record these emissions and map a slice. Distribution of radioactivity matches up with metabolic activity because the substance is being used by the body.

68. Magnetic Resonance! This depends on a property of nuclei called spin . Gyroscope: Principle: As long as its disc remains spinning rapidly the direction of the spin axis will stay pointing in the same direction independently of any movement of its support. Direction of a gyroscope can be altered by applying a torque to it. The effect of a torque on a gyroscope is to make the gyroscope precess . This means the top of the axle moves round in a horizontal circle.

69. Nuclear Precession Many biological molecules have hydrogen atoms within them. The hydrogen atom has a single proton, which spins on its axis. This spin gives the proton a very small magnetic property called its magnetic moment . When in a magnetic field, the proton experiences a torque, and so it precesses . The frequency of precession is called the Larmor Frequency. Larmor frequency 4.25X10 7 X B Where B is the total magnetic flux density.

70. Nuclear Resonance When the applied magnetic flux density is altered, resonance occurs. The magnetic field is altered by using an alternating current of radio frequency, f R , in coils placed in the magnetic field. When the radio frequency is equal to the Larmor frequency, a relatively large amount of energy is absorbed by the proton. This causes the proton to flip over to its higher energy state.

71. Relaxation Once a proton has gained this energy, it is in a semi-stable state. Will not remain in this state, but will relax back. This is the key to MRI. Relaxation time depends on the magnetic field at the position of any proton. During relaxation, the energy previously gained is lost in the form of radio-waves, which can be detected, amplified and interpreted. Relaxation times for H nuclei in water are long (2 s). Brain tissues = 200 ms Tumour has a relaxation time between each of these two.

72. Why use MRI? Quality of info is very high No ionising radiation is required. Radiofrequency is no higher than normal radio, so no concerns about electromagnetic waves. Magnetic field used is 100 time stronger than Earths’. 2 more problems are cost, and that a scan can take 45 mins . (Patient must also be still, not suitable for young children)

73. MRI Scanner The scanner contains multiple components! Main Magnet Additional Magnets Radiofrequency Coil Computer Display We must also look at the advantages and disadvantages!

74. Main Magnet These must produce a magnetic field strength of 1.4 T. Must be over whole patient. Produced by coils carrying huge currents in wires kept at temperatures near absolute zero. Principle of superconductivity: Resistance of a wire is 0 provided temp is low. Liquid He at 4.2 K. Value of magnetic field strength must be constant over 90 cm.

75. Additional Magnets Very accurately calibrated additional magnets are positioned to alter the strength of the magnetic field of the main magnet from place to place. The field needs to be known at all points in a 3D space. A scan is done point to point, and for each point, the field strength and the Larmor frequency are known. Transmitter and Receiver are tuned to the f L emitted by a nucleus at the point. The scanning of the next point is carried out at a slightly different frequecy .

76. Radiofrequency Coil If the RF EM waves were continuous, then we wouldn’t be able to measure the relaxation time. This is why RF waves are emitted in pulses. After each pulse, a coil picks up the emitted RF waves from the patient.

77. Computer Amount of Data is huge. Programming involves Isolating slightly different radiofrequencies Linking them to a point in 3D space. This can only be done if the magnetic field strength is known accurately at all points within the volume being scanned. The relaxation time for that point needs to be measured, relative to the type of material at that point, and the whole used to provide a display.

78. Display Normally on a computer screen, and print-outs can be made. Can give slice, or 3D view. Can be rotated. False colours can be attached.

79. Advantages! No ionising radiation involved High quality image Good distinction between different types of soft tissue. Bone provides no barrier, so all images can be clear. So side effects

80. Disadvantages No metallic objects can be scanned, or they heat up. So those with pacemakers and/or surgical pins can’t receive a scan. Must not have external radio waves. Machines are very expensive. Long time for one scan (3/4 hours)

81. MRI Recap Basics This detects the presence of Hydrogen Nuclei. Since body tissue has a high water content, it can be used to produce an image of the body. Pulses of high intensity & amplitude magnetic fields are applied which results in emission of EM radiation from the nuclei by the…..

82. ….following process Hydrogen nuclei spin , giving them a magnetic field. When an external magnetic field is applied, they rotate around the direction of the field. This rotation is called precession . (They are not all in phase) The frequency of precession is called the Larmor Frequency . Application of a pulse of radio waves at the Larmor frequency causes resonance / flipping – when the nuclei absorb the energy into the precession. When the pulse is removed, the nuclei lose energy, emitting radio waves. This takes place in a short time called the relaxation time .

83. Larmor All the protons precess at the same frequency. The value of f depends on the strength of the magnetic field, B .

84. What next?... The emissions from the hydrogen nuclei are detected by a radio aerial and processed by a computer to give a 3D body image. MRI scanning is non-invasive and does not cause ionisation. Very Expensive in terms of running costs.

85. Controlling Contrast The response of different tissue types can be enhanced by varying the time between pulses. Those with large molecules such as fat are best imaged using rapidly repeated pulses. This technique is used to image the internal structure of the body. Allowing more time between pulses enhances the response of watery substances. This is used for diseased areas.

86. Jan 2013 7)a) An MRI scanner is a valuable item of diagnostic equipment. It is capable of generating 3D images of the patient. Describe the operation of an MRI scanner with particular reference to Larmor Frequency Resonance of the Protons Relaxation Times 7)b) State one disadvantage and one advantage of MRI scan.

87. Jun 2010 Outline the main principles of the use of MR to obtain diagnostic information about internal organs.

88. (h) describe the need for non-invasive techniques in diagnosis MRI is non-invasive, but it is expensive. We can use different non-invasive techniques instead. Non-invasive techniques do not involve ionising radiation. Methods we will look at: Endoscopy Ultrasound

89. The Endoscope These use optical fibres These can be inserted into body openings. Light is passed down one set of optic fibres. These are arranged so that they have the exact same arrangement in the bundle at the bottom as at the top. This allows the doctor to see what is being examined through an eye piece.

90. Ultrasound These are waves that are above the audible sound frequency range (20 Hz to 20,000 Hz) In medical ultrasounds the frequencies are in the megahertz range. These cause no ionisation. Ultrasound can show both muscle and blood. A low frequency is used as often as possible as high frequency ultrasounds can be destructive to tissues. Will be examined in greater detail later on

91. ( i ) explain what is meant by the Doppler effect The doppler effect is used in Doppler ultrasound scanning. When sound is emitted from a source, the waves spread out in concentric circles, with the distance between the waves being the wavelength. When the frequency of the wave is f , the speed of the wave, c , is given by  

92. ( i ) explain what is meant by the Doppler effect If you move towards a stationary source of sound you will not hear the same frequency as when you were stationary. Extra waves will have passed into your ears. This change in frequency due to the persons movement is called the doppler effect.

93. Numerical Example! Stationary source 200 Hz c = 340 m/s Wavefronts = 1.7 m apart If person is stationary, wavefronts go past at 200/s Imagine you are travelling towards the source at 30 m/s. You will have passed an extra (30/1.7) = (17.6 wavefronts . Altogether, 217.6 wavefronts will have passed. This means the frequency will be 217.6 Hz. You will hear a higher pitch.

94. Source movement When the source of the sound is moving, it creates a different effect. Wavelengths behind the source are larger. Wavelengths in from are shorter. This is the principle of ultrasound.

95. Consider a wave of frequency f with speed c. Time, t, between each wave is 1/f. In this time the wave has moved c/f. If the speed of the source is v, the source has moved v/f. This gives a new wavelength, λ ’, in front of the source as The new frequency, f’, will therefore become  

96. Question Can you think of 2 examples where the wave source is travelling faster than the wave itself?

97. Question Calculate the frequency heard when a train, travelling towards you with a speed of 60 m/s, sounds its whistle of frequency 400 Hz. Velocity of sound is 340 m/s.

98. Jan 2012

99. Jun 2011

100. (j) explain qualitatively how the Doppler effect can be used to determine the speed of blood.

101. Unit 5.4.3 Ultrasound Candidates should be able to: (a) describe the properties of ultrasound; (b) describe the piezoelectric effect; (c) explain how ultrasound transducers emit and receive high-frequency sound; (d) describe the principles of ultrasound scanning; (e) describe the difference between A-scan and B-scan; (f) calculate the acoustic impedance using the equation Z = ρ c ; (g) calculate the fraction of reflected intensity using the equation (h) describe the importance of impedance matching; ( i ) explain why a gel is required for effective ultrasound imaging techniques.  

102. a) Describe the properties of Ultrasound Ultrasound waves are longitudinal with high frequencies ( ≈ > 20,000 Hz, though medical Ultrasound is between 1 to 15 MHz. ) When an ultrasound reaches a boundary, some of it is reflected, and some passes through the material. Those that pass through will undergo refraction if the angle of incidence is not 90°. Reflected waves are detected by an Ultrasound scanner and are used to generate an image.

103. b) Describe the Piezoelectric Effect. This is the effect relied upon to create Ultrasound. This works on the basis of certain crystals contracting upon putting a potential difference across them. An example of this kind of crystal would be Lead Zirconate Titrate. When a high frequency alternating p.d . is applied the crystals deform/oscillate at the frequency of the signal and send out Ultra sound waves.

104. Because the process can work in reverse, the same crystal can also act as a receiver of Ultrasound. They will convert sound-waves into alternating p.d’s . Lead Zirconate Titanate The thickness of the crystal is half the wavelength of the ultrasound it produces. Ultrasound of this frequency will make the crystal resonate and produce a large signal. This is heavily damped to produce short pulses and increase the resolution of the device.

105. PZT Crystal Unstressed Stretched Compressed

106. The Ultrasound Transducer This acts as both a transmitter and receiver of ultrasound. It contains Faceplate Piezoelectric Crystal Backing Material Tuning Device Cable The faceplate is curved. This shapes the ultrasound into a narrow beam. The tuning device controls the frequency of the ultrasound waves. To ensure the sound enters the body, a gel is applied between the transducer and the skin.

107. Principles of Ultrasound Scanning. Ultrasound is reflected from surfaces rather than going right through a body. Echoes are used. A boundary between tissue and liquid, or tissue and bone, or air and skin, reflects the waves. Ultrasound sent into the body must be pulsed. One pulse is sent out, and there is a pause until reflected echoes come back to be detected.

108. Numerical Values Speed of Ultrasound waves in muscle : 1600 m/s. Speed of ultrasound in air: 340 m/s. Frequency of Ultrasound: 1 MHz. Calculate the time taken for the ultrasound to travel through 20 cm of muscle.

109. Pulse Repetition Frequency This states that the transmission of pulses cannot be at a frequency that exceeds the maximum time allowance for a reflection. Eg : If a minimum time of 1 ms is allowed for a reflection to be received, frequency must not exceed 1000 Hz.

110. January 2011

111. e) Describe the difference between A scan and B scan. A Scan – Range Measurement The A mplitude scan sends a short pulse of ultrasound into the body simultaneously with an electron beam sweeping across the C athode R ay O scilloscope (CRO) screen. The scanner receives reflected ultrasound pulses that appear as vertical deflections on the CRO screen. Weaker pulses are amplified more to avoid loss of valuable data – Time Gain Compensation Horizontal positions of the reflected pulses indicate the time the echo took to return, and are used to work out distances. A stream of pulses can produce a steady image on the screen due to persistence of vision.

112. e) Describe the difference between A scan and B scan. B Scan – The Brightness value In a brightness scan, the electron beam sweeps down the screen rather than across. The amplitude of the reflected pulses is displayed as the brightness of the spot. You can use a linear array of transducers to produce a 2D image. This array of transducers, as well as a fanning out of US beam across the body, gives the B Scan. Many returning echoes are recorded and sued to build up an image on screen.

113. (f) calculate the acoustic impedance using the equation Z = ρ c If at the first boundary an ultrasound is completely reflected then there will be none left to be reflected at a further boundary. To get multiple reflections from different boundaries depends on the fraction of intensity of the US reflected as transmitted. Acoustic impedance, Z, is used in determining the fraction of the intensity that is refracted at a boundary between two materials of different acoustic impedance. This is defined by the equation Z = ρ c

114. (f) calculate the acoustic impedance using the equation Z = ρ c ρ is the density of the material and c is the speed of sound in the material. The equation for the ratio of the intensity reflected against the incident intensity when US is at a boundary and leaving one material to another with unique Acoustic Impedances is  

115. Material c (m/s) ρ (kg m -3 ) Z (kg m -2 s -1 ) Air 340 1.3 440 Bone 3500 1600 5.6 x10 6 Muscle 1600 1000 1.6x10 6 Soft Tissue 1500 1000 1.5x10 6 Fat 1400 1000 1.4x10 6 Blood 1600 1000 1.6x10 6 Calculate for US entering Fat from blood Bone from Soft Tissue Air to Bone  

116. Impedance Matching You need a coupling medium between the transducer and the body. Soft tissue has a different a.i from air so almost all the US energy is reflected from the body. The coupling medium displaces the air and has an impedance closer to that of soft tissue. This is an example of impedance matching. This coupling medium is usually an oil or a gel.