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Physics Year 13 > Chapter 27 - Medical Imaging > Flashcards

Chapter 27 - Medical Imaging Flashcards

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1
Q

define X-rays

A
  • EM waves with wavelength 10^-8 —> 10^-13 m
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2
Q

what are some key features of X-rays

A
  • travel at speed of light
  • can be polarised
  • can be diffracted by atoms in crystals
  • have 10-10000x more energy (per photon) than visible light
  • can be harmful to cells
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3
Q

describe the general way that X-rays are produced and what we can say about energy transfer

A
  • X-rays are produced when fast-moving electrons are decelerated by interactions with atoms of a metal such as tungsten

Ek of electrons = Energy of photons (if a photon is released)
electrons and photons interact one-to-one

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4
Q

explain the structure of an X-ray tube

A
  • consists of an evacuated tube (so that electrons don’t interact with gas particle) containing two electrodes
  • external power supply produces large accelerating P.D. between the electrodes
  • cathode (negative) also has a low voltage supply over it, the cathode is a filament from which electrons are produced by thermionic emission as it is heated
  • the electrons accelerate to the positive anode, this is a piece of high bpt metal such as tungsten
  • x-ray photons are produced when the electrons are decelerated when hitting anode
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5
Q

what is the general efficiency of an X-ray tube and what often happens at the anode because of this

A
  • generally only about 1% of the electrons’ energy is converted to X-rays, the rest is dissipated as thermal energy
  • so often the anode needs to be cooled or rotated
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6
Q

how can we calculate the wavelength of the X-rays from the accelerating P.D.

A 
W = VQ
E = eV
hf = eV
hc/lambda = eV
lambda = hc/eV

wavelength = hc/eV

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7
Q

what is the effect of increasing the number of electrons but with a constant energy

A
  • intensity of X-rays increases because more photons are produced but their wavelength remains cosntant
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8
Q

what generally occurs when X-rays interact with matter

A
  • X-ray photons interact with the atoms of the matter they pass through, they can be scattered or absorbed, decreasing intensity
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9
Q

what is attenuation

A

Attenuation is the decrease in intensity of Electromagnetic radiation as it passes through matter

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10
Q

name the 4 attenuation mechanisms, which one does not involve the electron inside the atom

A
  • Simple scatter
  • Photoelectric effect
  • Compton scattering
  • pair production - this one does not involve electrons inside the atom
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11
Q

explain what occurs in simple scattering and for which energies of photons it’s most prominant

A
  • where Ephoton = 1 - 20 keV
  • X-ray photon interacts with electron in the atom but has less energy than is required to remove it (E < work function)
  • the photon ‘bounces off’/is scattered in an elastic collision
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12
Q

explain what occurs in the photoelectric effect (for X-rays and matter) and for which energies of photons it’s most prominant

A
  • where Ephoton < 100 keV
  • X-ray photon is absorbed by an electron, this gives the electron sufficient energy to escape from the atom
  • this is the dominant mechanism in X-rays (medical) because V = 30-100 kV
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13
Q

explain what occurs in Compton scattering and for which energies of photons it’s most prominant

A
  • Where Ephoton = 0.5 - 5 MeV
  • incident X-ray photon interacts with an electron, electron is ejected from the atom, photon is scattered with reduced energy
  • both energy and momentum are conserved
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14
Q

explain what occurs in pair production and for which energies of photons it’s most prominant

A
  • onle occurs where Ephoton > 1.02 MeV
  • X-ray photons interacts with the nucleus of an atom, it disappears and its energy is used to produce an electron and a positron
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15
Q

what is the amount of attenuation dependent on

A
  • for a given energy of photon, attenuation is dependent on thickness of material and type of material
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16
Q

what is the equation for calculating the intensity of X-rays after having passed through a medium

A 
I = Io (e^-(mew)(x))
I = final intensity
Io = initial intensity
Mew = attenuation coefficient
x = thickness of material
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17
Q

what make better contrast media in medical x rays and why

A

photoelectric effect is the dominant attenuation mechanism

  • in the photoelectric effect, mew is direct prop to Z^3
  • therefore higher mass number elements are better for this
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18
Q

why do we sometimes have to use contrast media, what are some common contrast media

A
  • soft tissues have low absorption/attenuation coefficients, so contrast mediums with high attenuation coefficients are used to make internal structures more clear
  • they allow you to view the outline of the internal structures/soft tissues
  • Barium and Iodine are common elements that are used because in X-ray imaging
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19
Q

where are barium and iodine generally used as a contrast media

A
  • iodine in liquids and the blood

- barium in digestive systems

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20
Q

how can x-rays be used for therapeutic purposes

A
  • high energy X-ray photons can be used in cancer therapy

- they can kill cancer cells through Compton scattering and pair production

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21
Q

what are the advantages and disadvantages of X-ray scans

A

Adv:

  • cheap, easy
  • shows internal structures

Disadv:

  • the image produced is only 2D so overlapping structures can’t be differentiated
  • not good for imaging soft tissue
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22
Q

briefly explain what a CAT scan is

A

CAT scanners take a large number of X-rays from different angles and compile them into one large 3D image using software

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23
Q

what is the structure of CAT scanner

A
  • patient lies on an examination table that is able to move/slide back and forth into the scanner
  • the scanner is a tube/gantry that contains an X-ray tube on one side and some X-ray detectors opposite
  • these scanners/detectors can rotate within the gantry/tube
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24
Q

explain the process of taking a CAT scan, what moves etc.

A
  • X-ray tube produces a fan-shaped beam of X-rays about 1-10mm thick
  • this irradiates the patient as a normal X-ray would and the detectors opposite record the results and send it to a computer
  • as the X-ray tube rotates the table moves, approx. 1cm per rotation
  • then the process repeats giving X-ray images of ‘slices’ of the body
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25
Q

how long does a typical CAT scan last and what path does the X-ray tube follow

A
  • approx. 10-30 mins

- a spiral path

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26
Q

how are the images from a CAT scan analysed

A
  • radiographers can view each 2D ‘slice’ individually

- or they can be compiled by software into one big 3D image

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27
Q

what are the advantages and disadvantages to using CAT scans

A

Adv:

  • can create a 3D image
  • better for distinguishing between soft tissues with similar attenuation coefficients

Disadv:
- gives a higher radiation dose than normal X-rays

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28
Q

Briefly explain what a gamma camera is

A

The gamma camera is a detector of gamma photons emitted from radioactive nuclei injected into the patient

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29
Q

why is gamma radiation used in PET scanning

A

Gamma radiation is used because:

  • it’s not very ionising so will not cause much harm to the patient’s cells
  • it’s the most penetrative form of radiation so it can easily pass through the patient’s skin/flesh to the ‘camera’
30
Q

what feature must the Gamma radioisotopes have that are used in a gamma camera

A

They must have a short half-life:

  • this gives them a high initial activity so a ‘picture’ can be taken and only small amounts of the isotope are required to be ingested
  • they also decay quickly, minimising radiation exposure time
31
Q

where are gamma radioisotopes for gamma cameras generally produced and why

A

they tend to be produced on-site as they have short half lives so cannot be stored or transported for long periods of time

32
Q

what are two examples of gamma radioisotopes used in the gamma camera

A

Tc-99m = half life of 6 hours

F-18

33
Q

what are medical tracers why are they required

A
  • medical tracers are chemicals that are bonded to the radioisotope that can target a particular area of the body or organ in order to make sure the isotope reaches the correct place
  • the concentrations of this tracer can be detected to look for irragularities
34
Q

what is the first step of detecting gamma photons, what is the collimator

A
  • the collimator is a honeycomb structure of long thin tubes made of lead
  • emitted gamma photons move towards the collimator
  • only those that are perfectly parallel with it can pass through
  • the rest are absorbed
35
Q

what is the second step of detecting gamma photons, what is the scintillator

A
  • the scintillator is a material such as sodium iodide that absorbs single gamma photons and releases thousands of visible light photons
  • only those gamma photons which make it through the collimator interact with the scintillator, approx 1 in 10
36
Q

what is the third step of detecting gamma photons, what is the photomultiplier tube

A
  • the visible light photons that are emitted from the scintillator travel through the light guide into a photomultiplier tube
  • the photomultiplier tubes are collections of tubes containing photocathodes and dynodes that convert single visible light photons into electrical voltage
37
Q

how are the detected gamma photons analysed

A
  • outputs of the photomultiplier tube are connected to a computer
  • software calculates where and when the gamma photons hit the scintillator
  • it constructs an image to show the concentrations of the medical tracer at different points in the body
38
Q

how can the quality of image of a gamma camera be improved

A

using narrower/longer/thinner collimators
or
longer scanning time

39
Q

what is a common radioisotope used in PET scans and what is its half life

A

fluorine-18 and 110 mins

40
Q

what is the radiation from fluorine 18, what part do we detect

A

beta +
(18,9)F —> (18,8)O + (0,1)e+ + Ve + gamma
- we do not detect the gamma directly
- we detect the gamma released by the annihilation of the positron with electrons in the body

41
Q

how can Fluorine-18 be made

A
  • in particle accelerators near the site
  • collide a proton with oxygen-18
    (18,8)O + (1,1)p —> (18,9)F + (1,0)n
42
Q

in what ways are CAT scans and PET scans similar and different

A
  • both use EM radiation to produce 2D ‘slices’ through the body that can be formed into a 3D image using software
  • CAT scans use X-rays, PET scans use gamma
43
Q

what are two common medical tracers used in PET scans

A

Fluorodeoxyglucose (FDG)

  • like glucose but an oxygen has been replaced with F-18
  • treated by body like normal glucose
  • shows where areas of high respiration are

Carbon Monoxide:

  • uses C-11 instead of C-12, this is positron emitting
  • used for blood stream
44
Q

what is the structure of a PET scanner

A
  • patient lies on a horizontal examination table
  • they are surrounded by Gamma detectors/cameras, (sodium iodide scintillator and photomultiplier tube)
  • they are injected with one of the medical tracers
45
Q

what occurs in a PET scan and what is detected

A
  • the medical tracer injected is usually positron emitting
  • these positrons annihilate almost immediately with electrons, emitting two gamma photons that travel in opposite directions
  • there are gamma detectors diametrically opposite which are connected to a computer
  • it can calculate the points of annihilation (and therefore where the medical tracer is) by the difference in arrival times by the gamma photons
46
Q

what are some advantages and disadvantages of PET scans

A

Advantages:

  • non-invasive
  • good for detecting cancers, and analysing the heart
  • good for analysing the brain e.g. Alzheimer’s disease

Disadvantages:
- very expensive due to the cost of producing radioisotopes

47
Q

define ultrasound

A

Ultrasound is longitudinal waves with a frequency greater than 20kHz

48
Q

what are the benefits of ultrasound as a medical scanning technique

A
  • non-invasive
  • cheap
  • non-ionising and therefore mostly harmless
49
Q

what are some features of ultrasound as a wave

A
  • it can be reflected, refracted and diffracted in the body just like other waves
50
Q

what frequencies of ultrasound tend to be used in medical imaging

A

1 - 15 MHz

51
Q

what is an ultrasound transducer

A

“an ultrasound transducer is a device to generate and receive ultrasound through the Piezoelectric effect”

52
Q

what is the Piezoelectric effect

A
  • when an emf is induced over the end of some crystals, they can stretch and squash
  • the same occurs in reverse, when they are stretched or squashed they can produce an alternating P.D.
  • strain < 0.1%
53
Q

how is ultrasound in medical scanners produced and why is this frequency chosen

A
  • an alternating P.D. of about 5 MHz is applied across the crystal
  • this makes it rapidly expand/compress, producing ultrasound
  • frequency chosen = natural frequency of crystal
54
Q

in what format is this ultrasound emitted from the ultrasound transducer

A

5000 pulses per second of 5 MHz sound

55
Q

how is the ultrasound detected

A
  • the transducer can also detect ultrasound, the ultrasound can cause the crystal to compress/stretch and this produces an alternating EMF, this can be detected by a circuit
  • most ultrasound scanners use Lead Zirconate Titanate
56
Q

what is an A-scan

A
  • an A-scan is the simplest type of ultrasound scan

- a single transducer is use to record along a straight line through the body

57
Q

how does an A-scan work

A
  • A transducer sends ultrasound pulses into a patient’s body, each pulse is partly reflected and transmitted an any boundary between tissues
  • a reflected ‘echo’ pulse is received at the transmitter
  • this ‘echo’ pulse is at a lower energy than the original pulse, the intensity of this pulse is dependent on the acoustic impedance at tissue boundaries
58
Q

how are the results from an A-scan analysed

A
  • the pulsed voltage from the ‘echos’ is displayed on a voltage-time plot
  • the average speed of ultrasound in the body and the time between echos can be used to calculate distances between things in the body
59
Q

what is a B-scan and how does it work

A
  • B-scans are ultrasound scans that can produce a 2D image on a screen
  • a transducer is moved over the patient’s skin, it is attached to a high speed computer
  • for each transducer position a computer produces rows of dots on a screen where each dot represents a tissue boundary
  • the brightness of each dot is directly proportional to the intensity of the reflected ultrasound
  • collections of dots combine to produce an image
60
Q

what is a reason for using non-invasive medical diagnosis techniques

A

less chance of infection

61
Q

what occurs to ultrasound when it is incident at a boundary, what is this dependent on

A
  • when a beam of ultrasound is incident at a boundary, some of its intensity is reflected and some of its intensity is refracted
  • the proportion of the intensity that is reflected depends on the two material’s acoustic impedances
62
Q

define acoustic impedance and give the corresponding equation and units

A

“the acoustic impedance, Z, of a substance is defined as the product of the density, p, of the substance and the speed, c, of the ultrasound in the substance”

Z = pc
Z is measured in kgm^-2s^-1

63
Q

give some typical acoustic impedance values

A 
air = 0.0004 (x10^6)
fat = 1.38 (x10^6)
muscle = 1.69 (x10^6)
bone = 7.6 (x10^6)
64
Q

what is the equation for the intensity reflection coefficient, and what does this actually mean

A

Ir/Io = (Z2 - Z1)^2 / (Z2 + Z1)^2

this is the ratio of the reflected intensity of ultrasound to the incident intensity of ultrasound for a collimated beam of ultrasound that is normally incident at a boundary between two surfaces of acoustic impedance Z1 and Z2

65
Q

what is the effect on reflection if the difference in acoustic impedance is greater

A

the proportion of the intensity that is reflected is greater

66
Q

what is coupling gel and why is it needed

A
  • it is a gel that has a similar acoustic impedance to skin
  • if an ultrasound is done where this isn’t used, air bubbles between the transducer and the skin can cause a large amount of the ultrasound to be reflected
  • because air and skin have very different acoustic impedance values
  • the gel both has a similar acoustic impedance to skin and fills in air gaps
67
Q

what is the principle of behind Doppler imaging

A
  • the frequency of ultrasound changes when it is reflected off a moving object
  • Doppler imaging uses the reflection of ultrasound off moving blood cells to assess blood flow
68
Q

how does a colour Doppler scan work

A
  • the transducer is pressed lightly on the skin over the blood vessel
  • transducer sends pulses of ultrasound and receives reflected pulses
  • ultrasound reflected off tissue returns with the same frequency
  • ultrasound reflected off moving objects (blood cells) return with a changed frequency
  • the transducer is connected to a computer which can analyse all of this and produce a colour coded image
69
Q

how does the frequency of the returning ultrasound waves change in relation to the blood cells

A

If:

  • the blood cells are moving towards the transducer, the frequency increases
  • the blood cells are moving away from the transducer, the frequency decreases

change in frequency is directly proportional to the speed of the blood cells

70
Q

what is the equation for the speed of red blood cells in a blood vessel

A

delta(f) = 2fvcos(theta)/c

delta(f) = change in frequency of ultrasound
f = frequency of incident ultrasound
v = speed of blood cells
theta = angle of ultrasound (to the blood flow)
c = speed of ultrasound in blood
71
Q

how would you determine the speed of blood in an artery in the arm

A
  • place ultrasound transducer at an angle to the arm
  • record initial frequency of ultrasound
  • the ultrasound is reflected by moving blood cells and thus its frequency changes
  • record the frequency of the reflected ultrasound and calculate change in frequency
  • use DeltaF = 2fvcos(theta)/c
72
Q

how are X-ray images produced

A
  • X-rays are detected by a film/ scintillation counter
  • shadow forms on the film on exposure to X-rays, giving a negative image of the internal structures from different attenuation coefficients
  • generally attenuation occurs due to the photoelectric effect as medical X-rays are generally E < 100KeV

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