10Attenuation Reduction in X-ray intensity with thickness 100 # interaction is proportional to the # photons, so the curve is exponential decayNumber of photons x 1015503Thickness/cm63
11Attenuation coefficient I = I0e-μxwhere I0 = the original intensityx = distance through absorberμ = attenuation coefficient
12Half-value thickness x½ Thickness required to reduce the intensity to half its original intensityI = I0/2 = I0e-μx½Taking logsx½ = ln2/μ = 0.693/μHalf value thickness depends on the energy of the X-rays and the substance
13ExampleA parallel beam of X-rays of intensity 0.2 kW.m-2 is passed through 5 mm of a material of half-value thickness 2 mm. Calculate the intensity of the beam.HL Physics, C. Hamper, Pearson 2009
14Example First calculate the attenuation coefficient μ A parallel beam of X-rays of intensity 0.2 kW.m-2 is passed through 5 mm of a material of half-value thickness 2 mm. Calculate the intensity of the beam.First calculate the attenuation coefficient μμ = 0.693/x½ = 0.693/2 = 0.35 mm-1I = I0e-μx = 0.2e-0.35x5 = kW.m-2HL Physics, C. Hamper, Pearson 2009
16Taking an X-ray picture Have to choose X-rays of the right intensity/energyX-rays are dangerous (ionising) – need to keep exposure time and intensity to a minimumSimplest way is to place broken part on photographic film (which is enclosed in a light proof box
17Black areas are where lots of photons hit the photographic film and “exposed” it Few X-ray photons travelled through and hit paper
18Intensifying screenA screen of fluorescent material which gives out visible light when X-rays hit it is put on either side of the photographic film. This means X-rays of lower intensity can be used.
19Barium MealTo help see soft-tissue such as the gut, it can be filled with a material with is opaque to X-rays like barium sulphate. Drinking a “barium meal”
20Digital imagesA array of photosensitive diodes works in a similar way to a CCD. They gain charge when exposed to X-rays. This produces a p.d. which can be converted to a digital signal.
21Digital imagesThe digital image can be enhanced and coloured electronically – as in airport security.
22TomographyFor parts deep in the body a “slicing” technique called tomography can be used
36Ultrasound production An alternating p.d. of frequency >20 kHz is applied to a quartz crystal causing it to vibrate
37Ultrasound detectionA sound wave causes a crystal to vibrate, producing an alternating p.d.
38ReflectionsBecause the reflections are analysed, the transmitter and receiver are in the same place. The same crystal can be used as transmitter and receiver if pulsed signals are used that are short enough so the reflected wave returns before a new pulse is produced.
39ExampleIf the pulse length is 10-6 s and the speed of sound in body tissue is 1500 m.s-1 then the minimum distance the wave can travel before the pulse has finished transmitting is 1500 x 10-6 = 1.5 mm. So it could be reflected off something at a depth of 0.75 mm. This is fine!
40Frequency?Need to use a short wavelength to avoid the wave spreading out from diffraction. Smallest object a doctor might be interested in is around a few mm, so wavelength needs to be a bit less. If λ = 1mm and velocity = 1500 m.s-1 then f = 1500/10-3 = 1.5 MHz.
41FrequencyHigher frequencies would give higher resolution but are absorbed easier (higher attenuation). Operator adjusts frequency and pulse length to get the best image for organs of different depth and size.
42Acoustic impedanceWhen ultrasound is incident on a boundary between two media (materials), part of the wavefront is reflected and part refracted.
43Acoustic impedance - ZThe percentage reflected depends upon the relative acoustic impedance of the two media.Acoustic impedance (Z) = ρcwhere ρ = density of mediumc = the velocity of the ultrasoundUnit of Z = kg.m-2.s-1
44Acoustic impedanceThe greater the difference in acoustic impedance, the greater the % reflection.
45Acoustic impedanceThe greater the difference in acoustic impedance, the greater the % reflection. The difference between air and skin is great. To prevent all of the ultrasound being reflected by the skin, gel is used to fill the gap between the transmitter and the skin.
46A - scans A plot of strength of reflected beam against time Signal strengthTime
47A - scansorganprobeThe depth and thickness of the organ can be deduced from the times of the reflected pulsetissuegel
48A - scans Notice the 2nd pulse is smaller due to attenuation organ probeNotice the 2nd pulse is smaller due to attenuationtissuegel
49A - scansorganprobeThe last pulse is large as most of the ultrasound is reflected by the last boundarytissuegel
50A - scansorganprobeTo gain more information the probe can be moved up and down to reveal the size, shape and changes in thickness of the organ.tissuegel
51A - scansorganprobeTo gain more information the probe can be moved up and down to reveal the size, shape and changes in thickness of the organ.tissuegel
52B - scansAn A scan gives information but not an image. B – scans converts the signal into a dot whose brightness corresponds to the strength of the signal. By sweeping the probe across the organ an image can be produced.
533-D ultrasoundAn array of probes moved around the patient can be used to construct a 3D image using a computer.