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Physics A Level CIE > Chapter 13 Medical Physics > Flashcards

Chapter 13 Medical Physics Flashcards

1
Q

The piezoelectric effect is defined as:

A

The ability of particular materials to generate a potential difference (p.d.) by transferring mechanical energy to electrical energy

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2
Q

A transducer is any device that

A
  • converts energy from one form to another
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3
Q

Piezoelectric Crystals

A
  • At the heart of a piezoelectric transducer is a piezoelectric crystal
  • Piezoelectric crystals are materials which produce a p.d. when they are deformed
    • This deformation can be by compression or stretching
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4
Q

If a p.d. is applied to a piezoelectric crystal, then it

A

deforms, and if the p.d. is reversed, then it expands

  • If this is an alternating p.d. then the crystal will vibrate at the same frequency as the alternating voltage
  • Crystals must be cut to a certain size in order to induce resonance
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5
Q

One of the most common piezoelectric crystals is

A

quartz, which is made from a lattice of silicon dioxide atoms

  • When the lattice is distorted, the structure becomes charged creating an electric field and, as a result, an electric current
  • If an electric current is applied to the crystal, then this causes the shape of the lattice to alternate which produces a sound wave
  • Due to the conventional direction of electric current, it will flow from the positive to the negative region of the crystal
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6
Q

A molecule in a quartz crystal. When the compression and stretching alternates, an alternating e.m.f. is induced

A
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7
Q

Applications of the Piezoelectric Transducer

  • Microphone
A
  • A piezoelectric microphone detects pressure variations in sound waves
  • These can then be converted to an electrical signal for processing
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8
Q

Applications of the Piezoelectric Transducer

  • Ultrasound
A
  • In a piezoelectric transducer, an alternating p.d. is applied to produce ultrasound waves and sent into the patient’s body
  • The returning ultrasound waves induce a p.d. in the transducer for analysis by a healthcare professional
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9
Q
  • An ultrasound is defined as:
A

A high frequency sound above the range of human hearing

  • This is above 20 kHz, although in medical applications the frequencies can be up to the MHz range
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10
Q

An ultrasound transducer is made up of a

A
  • piezoelectric crystal and electrodes which produce an alternating p.d.
  • The crystal is heavily damped, usually with epoxy resin, to stop the crystal from vibrating too much
    • This produces short pulses and increases the resolution of the ultrasound device
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11
Q

The structure of an ultrasound transducer

A
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12
Q

A piezoelectric crystal can act as both a receiver or transmitter of

A

ultrasound

  • When it is receiving ultrasound, it converts the sound waves into an alternating p.d.
  • When it is transmitting ultrasound, it converts an alternating p.d. into sound waves
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13
Q

Generation:

An alternating p.d. is applied across a piezo-electric crystal, causing it to

A
  • change shape
  • The alternating p.d. causes the crystal to vibrate and produce ultrasound waves
  • The crystal vibrates at the frequency of the alternating p.d., so, the crystal must be cut to a specific size in order to produce resonance
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14
Q

Detection:

When the ultrasound wave returns

A
  • the crystal vibrates which produces an alternating p.d. across the crystal
  • This received signal can then be processed and used for medical diagnosis
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15
Q

The frequency of the ultrasound is important because

A
  • The higher the frequency of the ultrasound, the higher the resolution and the smaller structures that can be distinguished
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16
Q

The ultrasound gives two main pieces of information about the boundary:

A
  • Depth: the time between transmission and receipt of the pulse (the time delay)
  • Nature: amount of transmitted intensity received (will vary depending on the type of tissue)
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17
Q
  • In an ultrasound scanner, the transducer sends out a beam of sound waves into the body
  • The sound waves are
A
  • reflected back to the transducer by boundaries between tissues in the path of the beam
    • For example, the boundary between fluid and soft tissue or tissue and bone
  • When these echoes hit the transducer, they generate electrical signals that are sent to the ultrasound scanner
  • Using the speed of sound and the time of each echo’s return, the scanner calculates the distance from the transducer to the tissue boundary
  • These distances can be used to generate two-dimensional images of tissues and organs
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18
Q
  • The acoustic impedance, Z, of a medium is defined as:
A

The product of the speed of the ultrasound in the medium and the density of the medium

  • This quantity describes how much resistance an ultrasound beam encounters as it passes through a tissue
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19
Q
  • Acoustic impedance can be calculated using the equation:
A

Z = ρc

  • Where:
    • Z = acoustic impedance (kg m-2 s-1)
    • ρ = the density of the material (kg m-3)
    • c = the speed of sound in the material (m s-1)
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20
Q
  • Acoustic impedance can be calculated using the equation:
A

Z = ρc

  • Where:
    • Z = acoustic impedance (kg m-2 s-1)
    • ρ = the density of the material (kg m-3)
    • c = the speed of sound in the material (m s-1)
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21
Q

Z = ρc

This equation tells us

A
  • The higher the density of a tissue, the greater the acoustic impedance
  • The faster the ultrasound travels through the material, the greater the acoustic impedance also
  • This is because sound travels faster in denser materials
    • Sound is fastest in solids and slowest in gases
    • The closer the particles in the material, the faster the vibrations can move through the material
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22
Q

At the boundary between media of different acoustic impedances, some of the wave energy is

A
  • reflected and some is transmitted
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23
Q

The greater the difference in acoustic impedance between the two media, the greater the

A

reflection and the smaller the transmission

  • Two materials with the same acoustic impedance would give no reflection
  • Two materials with a large difference in values would give much larger reflections
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24
Q
  • Air has an acoustic impedance of Zair = 400 kg m-2 s-1
  • Skin has an acoustic impedance of Zskin = 1.7 × 106 kg m-2 s-1
    • The large difference means ultrasound would be significantly
A
  • reflected, hence a coupling gel is necessary
  • The coupling gel used has a similar Z value to skin, meaning that very little ultrasound is reflected
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25
Q
  • The intensity reflection coefficient α is defined as:
A

The ratio of the intensity of the reflected wave relative to the incident (transmitted) wave

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26
Q

intensity reflection coefficient α equation is

A
  • Where:
    • α = intensity reflection coefficient
    • IR = intensity of the reflected wave (W m-2)
    • I0 = intensity of the incident wave (W m-2)
    • Z1 = acoustic impedance of one material (kg m-2 s-1)
    • Z2 = acoustic impedance of a second material (kg m-2 s-1)
  • This equation will be provided on the datasheet for your exam
  • This ratio shows:
    • If there is a large difference between the impedance of the two materials, then most of the energy will be reflected
    • If the impedance is the same, then there will be no reflection
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27
Q

Coupling Medium

A
  • When ultrasound is used in medical imaging, a coupler is needed between the transducer and the body
  • The soft tissues of the body are much denser than air
  • If air is present between the transducer and the body, then almost all the ultrasound energy will be reflected
  • The coupling gel is placed between the transducer and the body, as skin and the coupling gel have a similar density, so little ultrasound is reflected
    • This is an example of impedance matching
28
Q
  • Attenuation of ultrasound is defined as:
A

The reduction of energy due to the absorption of ultrasound as it travels through a material

29
Q

The attenuation coefficient of the ultrasound is expressed in

A

decibels per centimetre lost for every incremental increase in megahertz frequency

  • Generally, 0.5 dB/cm is lost for every 1MHz
30
Q
  • The intensity I of the ultrasound decreases with distance x, according to the equation:
A

I = I0 e−μx

  • Where:
    • I0 = the intensity of the incident beam (W m-2)
    • I = the intensity of the reflected beam (W m-2)
    • μ = the absorption coefficient (m-1)
    • x = distance travelled through the material (m)
31
Q

The absorption coefficient μ, will vary from

A
  • material to material
  • Attenuation is not a major problem in ultrasound scanning as the scan relies on the reflection of the ultrasounds at boundaries of materials
32
Q

X-rays are

A
  • short wavelength, high-frequency part of the electromagnetic spectrum
    • They have wavelengths in the range 10−8 to 10−13 m
  • X-rays are produced when fast-moving electrons rapidly decelerate and transfer their kinetic energy into photons of EM radiation
33
Q

Producing X-rays

A
  • At the cathode (negative terminal), the electrons are released by thermionic emission
  • The electrons are accelerated towards the anode (positive terminal) at high speed
  • When the electrons bombard the metal target, they lose some of their kinetic energy by transferring it to photons
  • The electrons in the outer shells of the atoms (in the metal target) move into the spaces in the lower energy levels
  • As they move to lower energy levels, the electrons release energy in the form of X-ray photons
34
Q

When an electron is accelerated, it gains energy equal to the

A
  • electronvolt; this energy can be calculated using:

Emax = eV

  • This is the maximum energy that an X-ray photon can have
35
Q

the maximum X-ray frequency fmax, or the minimum wavelength λmin, that can be produced is calculated using the equation:

A
  • Where:
    • e = charge of an electron (C)
    • V = voltage across the anode (V)
    • h = Planck’s constant (J s)
    • c = speed of light (m s-1)
36
Q

Using X-rays in Medical Imaging

A
  • X-rays have been highly developed to provide detailed images of soft tissue and even blood vessels
  • When treating patients, the aims are to:
    • Reduce the exposure to radiation as much as possible
    • Improve the contrast of the image
37
Q

Reducing Exposure

A
  • X-rays are ionising, meaning they can cause damage to living tissue and can potentially lead to cancerous mutations
  • Therefore, healthcare professionals must ensure patients receive the minimum dosage possible
38
Q

In order to reduce exposure, what are used

A

aluminium filters are used

  • This is because many wavelengths of X-ray are emitted
  • Longer wavelengths of X-ray are more penetrating, therefore, they are more likely to be absorbed by the body
  • This means they do not contribute to the image and pose more of a health hazard
  • The aluminium sheet absorbs these long wavelength X-rays making them safer
39
Q
  • Contrast is defined as:
A

The difference in degree of blackening between structures

  • Contrast allows a clear difference between tissues to be seen
40
Q

Image contrast can be improved by:

A
  • Using the correct level of X-ray hardness: hard X-rays for bones, soft X-rays for tissue
  • Using a contrast media
41
Q
  • Sharpness is defined as:
A

How well defined the edges of structures are

42
Q

Image sharpness can be improved by:

A
  • Using a narrower X-ray beam
  • Reducing X-ray scattering by using a collimator or lead grid
  • Smaller pixel size
43
Q

Bones absorb

A
  • X-ray radiation
    • This is why they appear white on the X-ray photograph
  • When the collimated beam of X-rays passes through the patient’s body, they are absorbed and scattered
44
Q
  • The attenuation of X-rays can be calculated using the equation:
A

I = I0 e−μx

  • Where:
    • I0 = the intensity of the incident beam (W m-2)
    • I = the intensity of the reflected beam (W m-2)
    • μ = the linear absorption coefficient (m-1)
    • x = distance travelled through the material (m)
  • The attenuation coefficient also depends on the energy of the X-ray photons
45
Q

The intensity of the X-ray decays

A
  • exponentially
  • The thickness of the material that will reduce the X-ray beam or a particular frequency to half its original value is known as the half thickness
46
Q

A good contrast is when:

A
  • There is a large difference between the intensities
  • The ratio is much less than 1.0
47
Q

Computed Tomography Scanning

A
  • A simple X-ray image can provide useful, but limited, information about internal structures in a 2D image
  • When a more comprehensive image is needed, a computerised axial tomography (CAT or CT) scan is used
48
Q

The main features of the operation of a CT scan are as follows:

A
  • An X-ray tube rotates around the stationary patient
  • A CT scanner takes X-ray images of the same slice, at many different angles
  • This process is repeated, then images of successive slices are combined together
  • A computer pieces the images together to build a 3D image
  • This 3D image can be rotated and viewed from different angles
49
Q

Advantages of CAT Scans

A
  • Produces much more detailed images
  • Can distinguish between tissues with similar attenuation coefficients
  • Produces a 3D image of the body by combining the images at each direction
50
Q

Disadvantages of CAT Scans

A
  • The patient receives a much higher dose than a normal X-ray
  • Possible side effects from the contrast media
51
Q
  • A radioactive tracer is defined as:
A

A radioactive substance that can be absorbed by tissue in order to study the structure and function of organs in the body

52
Q

Radioactive isotopes, such as technetium-99m or fluorine-18, are suitable for

A

radioactive tracers

  • They both bind to organic molecules, such as glucose or water, which are readily available in the body
  • They both emit gamma (γ) radiation and decay into stable isotopes
  • Technetium-99m has a short half-life of 6 hours (it is a short-lived form of Technetium-99)
  • Fluorine-18 has an even shorter half-life of 110 minutes, so the patient is exposed to radiation for a shorter time
53
Q
  • Positron Emission Tomography (PET) is:
A

A type of nuclear medical procedure that images tissues and organs by measuring the metabolic activity of the cells of body tissues

54
Q

A common tracer used in PET scanning is a

A

glucose molecule with radioactive fluorine attached called fluorodeoxyglucose

  • The fluorine nuclei undergoes β+ decay – emitting a positron (β+ particle)
55
Q

The radioactive tracer is

A
  • injected or swallowed into the patient and flows around the body
  • Once the tissues and organs have absorbed the tracer, then they appear on the screen as a bright area for a diagnosis
    • This allows doctors to determine the progress of a disease and how effective any treatments have been
  • Tracers are used not only for the diagnosis of cancer but also for the heart and detecting areas of decreased blood flow and brain injuries, including Alzheimer’s and dementia
56
Q

The Process of Annihilation

A
  • When a positron is emitted from a tracer in the body, it travels less than a millimetre before it collides with an electron
  • The positron and the electron will annihilate, and their mass becomes pure energy in the form of two gamma rays which move apart in opposite directions
57
Q
  • Annihilation doesn’t just happen with electrons and positrons, annihilation is defined as:
A

When a particle meets its equivalent antiparticle they are both destroyed and their mass is converted into energy

  • As with all collisions, the mass, energy and momentum are conserved
58
Q

Once the tracer is introduced to the body it has a short

A

half-life, so, it begins emitting positrons (β+) immediately

  • This allows for a short exposure time to the radiation
  • A short half-life does mean the patient needs to be scanned quickly and not all hospitals have access to expensive PET scanners
59
Q

In PET scanning:

A
  • Positrons are emitted by the decay of the tracer
  • They travel a small distance and annihilate when they interact with electrons in the tissue
  • This annihilation produces a pair of gamma-ray photons which travel in opposite directions
60
Q

Detecting Gamma-Rays from PET Scanning

A
  • The patient lays stationary in a tube surrounded by a ring of detectors
  • Images of slices of the body can be taken to show the position of the radioactive tracers
61
Q
  • The detector (for pet scanning) consists of two parts:
A
  • Crystal Scintillator – when the gamma-ray (γ-ray) photon is incident on a crystal, an electron in the crystal is excited to a higher energy state
    • As the excited electron travels through the crystal, it excites more electrons
    • When the excited electrons move back down to their original state, the lost energy is transmitted as visible light photons
  • Photomultiplier -The photons produced by the scintillator are very faint, so they need to be amplified and converted to an electrical signal by a photomultiplier tube
62
Q

Detecting gamma rays with a PET scanner

A
63
Q

Creating an Image from PET Scanning

A
  • The γ rays travel in straight lines in opposite directions when formed from a positron-electron annihilation
    • This happens in order to conserve momentum
  • They hit the detectors in a line – known as the line of response
  • The tracers will emit lots of γ rays simultaneously, and the computers will use this information to create an image
  • The more photons from a particular point, the more tracer that is present in the tissue being studied, and this will appear as a bright point on the image
  • An image of the tracer concentration in the tissue can be created by processing the arrival times of the gamma-ray photons
64
Q

Calculating Energy of Gamma-Ray Photons

A
  • In the annihilation process, both mass-energy and momentum are conserved
  • The gamma-ray photons produced have an energy and frequency that is determined solely by the mass-energy of the positron-electron pair
65
Q
  • The energy E of the photon is given by
A

E = hf = mec2

  • Where:
    • me = mass of the electron or positron (kg)
    • h = Planck’s constant (J s)
    • f = frequency of the photon (Hz)
    • c = the speed of light in a vacuum (m s–1)
66
Q
  • The momentum p of the photon is given by
A
  • Where:
    • me = mass of the electron or positron (kg)
    • h = Planck’s constant (J s)
    • f = frequency of the photon (Hz)
    • c = the speed of light in a vacuum (m s–1)

Physics A Level CIE (14 decks)

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