A LEVEL PHYSICS MEDICAL IMAGING SLIDES.pptx

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

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

Unit 5.4.1 X-Rays
Assessable Learning Outcomes
a) Describe the nature of X-rays
b) Describe in simple terms how X-
rays are produced.
c) Describe how X-rays interact with
matter
1. Photoelectric
2. Compton Effect
3. Pair Production
d) Define intensity as the power per
unit cross-sectional area.
e) Select and use the equation I=I0e-μx
f) Describe the use of X-rays in
imagine internal body
structures including the use of
image intensifiers and of
contrast media.
g) Explain how soft tissues like
the intestines can be imaged
using barium meal.
h) Describe the operation of a
Computerised Axial
Tomography (CAT) scanner
i) Describe the advantages
of a CAT scan compared
with an X-ray image

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)

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)

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 EK will be converted to
EM Energy as X-ray photons. This emission is
continuous.

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.

are produced.
• Only about 1% of the electrons’ EK 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.

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 Emax of an X-ray will also be eV.
Since photon energy E = hf
We get
Emax = hfmax = eV
fmax =
𝑒𝑉
ℎ
In terms of wavelength λ,
λmin=
ℎ𝑐
𝑒𝑉
Since minimum frequency
occurs for minimum wavelength

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.

Question 1
Find the fmax and λmin for X-rays produced by a tube
across which the p.d. is 90,000 V.

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
a) Electrical power supplied
b) Power of the X-ray beam.

Question 3
• An X-ray tube at a dentists uses an accelerating
voltage of 70,000 V .
• Calculate the Ek 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 × 1018 eV/J.

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

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 nλ=d sin θ .
• Calculate the spacing between the atoms in the
crystal.
• d = wavelength/sin 15

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:

• 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.

matter. (Photoelectric effect)
• In the same way as UV, X-rays are capable of causing
the emission of photoelectrons.
• EUV photon = Work Function + Ek 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

• 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 Ek maximum = the photon
energy of the X-rays.

• 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.

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 = −1
0
𝑒 + +1
0
𝑒
• 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

• 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.

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.

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
• ∆λ=(h/𝑚𝑒𝑐)(1 − cos ∅)
• 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.

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.

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.

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.

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.

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

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.

e) Select and use the equation I=I0e-μx
• Above is the equation for Intensity
• I0 is the Initial Intensity
• μ is the constant called the attenuation
coefficient
• X is the distance through the medium.

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

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

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.

(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.

X-ray detectors
1. Black and white photographic film; requires a beam
of high-intensity.
2. 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.
3. 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.

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 cm2 – 200 cm2.

• 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 103 W m-2
• Attenuation constants:
▫ Lead 90
▫ Bone 53
▫ Muscle 6.9
e) and f) Questions

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.

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.

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.

Jan 2013
• 6)c) A beam of X-rays of intensity 3x109 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 5x104 W m-2

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

Jun 2011
• No exam questions on material to date.

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

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!

• 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.

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.

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.

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.

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.

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.

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.

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.

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.

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)

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.

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.

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.

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!

In Use
• Used to diagnose
▫ Thyroid
▫ Liver
▫ Brain
▫ Kidneys
▫ Lungs
▫ Spleen
▫ Heart
▫ Circulatory System

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.

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.

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.

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.

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

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.

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
▫ 4.25X107 X B
 Where B is the total magnetic flux density.

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, fR , 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.

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.

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)

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

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.

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 fL
emitted by a nucleus at the point.
• The scanning of the next point is carried out at a
slightly different frequecy.

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.

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.

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.

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

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)

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…..

….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.

Larmor
• All the protons precess at the same frequency.
• The value of f depends on the strength of the
magnetic field, B0.

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.

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.

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.

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

(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

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.

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

(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
▫ 𝑐 = 𝑓𝜆

(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.

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.

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.

• 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
𝜆′ =
𝑐
𝑓
−
𝑣
𝑓
=
𝑐 − 𝑣
𝑓
• The new frequency, f’, will therefore become
𝑓′ =
𝑐
𝜆′
=
𝑐
𝑐 − 𝑣
× 𝑓

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

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.

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

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 = ρc;
(g) calculate the fraction of reflected intensity using the
equation
𝐼𝑟
𝐼0
=
𝑍2−𝑍1
2
𝑍2+𝑍1
2
(h) describe the importance of impedance matching;
(i) explain why a gel is required for effective ultrasound
imaging techniques.

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.

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.

• 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.

PZT Crystal
Unstressed
Stretched
Compressed

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.

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.

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.

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.

e) Describe the difference between A
scan and B scan.
A Scan – Range Measurement
The Amplitude scan sends a short pulse of ultrasound into
the body simultaneously with an electron beam sweeping
across the Cathode Ray Oscilloscope (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.

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.

(f) calculate the acoustic impedance
using the equation Z = ρc
• 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 = ρc

(f) calculate the acoustic impedance
using the equation Z = ρc
• 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
𝐼𝑟
𝐼0
=
𝑍2 − 𝑍1
2
𝑍2 + 𝑍1
2

Material c (m/s) ρ (kg m-3) Z (kg m-2 s-1)
Air 340 1.3 440
Bone 3500 1600 5.6 x106
Muscle 1600 1000 1.6x106
Soft Tissue 1500 1000 1.5x106
Fat 1400 1000 1.4x106
Blood 1600 1000 1.6x106
Calculate
𝐼𝑟
𝐼0
for US entering
a) Fat from blood
b) Bone from Soft Tissue
c) Air to Bone

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.

A LEVEL PHYSICS MEDICAL IMAGING SLIDES.pptx

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