Name………….
Class………………
Plymstock School Physics
Department
Module G485.4
Medical Imaging
student booklet
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Lesson 44 notes – X-Rays
Objectives
(a) describe the nature of X-rays;
(b) describe in simple terms how X-rays are produced;
Outcomes
Be able to describe the what X-Rays are and their typical wavelengths/frequencies/energy
Be able to describe how X-Rays were discovered
Be able to describe how X-Rays are produced
Be able to use equations to describe the energy needed in production of X-Rays.
X-Ray Production
X-Ray Discovery
•1895
•Wilhelm Roentgen, Professor of Physics in Worzburg, Bavaria
•Accident – he was investigating cathode rays through partially evacuated tubes
•Black paper cover
•But fluorescent screen showed images
Hand mit Ringen (Hand with Rings): print of Wilhelm Röntgen's
first "medical" X-ray, of his wife's hand, taken on December 22,
1895
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Lesson 44 questions – X-Rays
Name……………………………….
Class……………
(
/14)…..%………
ALL
1.
In an X-ray tube, the efficiency of conversion of the kinetic energy of the electrons
into X-rays is 0.15%.
(i)
Calculate the power required in the electron beam in order to produce X-rays
of power 18 W.
power =..................................................... W
[2]
(ii)
Calculate the velocity of the electrons if the rate of arrival of electrons is
7.5 × 1017 s–1.
Relativistic effects may be ignored.
velocity =............................................. . m s–1
[2]
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(iii)
Calculate the p.d. across the X-ray tube required to give the electrons the
velocity calculated in (ii).
p.d. = ........................................................ V
[3]
[Total 7 marks]
2.
The figure below shows a simplified X-ray tube.
Explain briefly, with reference to the parts labelled C and A,
•
how X-rays are generated
•
the energy conversions that occur.
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[Total 7 marks]
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Lesson 45 notes – X-Ray interaction
Objectives
(c) describe how X-rays interact with matter (limited to photoelectric effect, Compton
Effect and pair production);
(d) define intensity as the power per unit cross-sectional area;
(e) select and use the equation I = I0 e−μx to show how the intensity I of a collimated Xray beam varies with thickness x of medium;
Outcomes
Be able to describe the wave nature of X-Rays
Be able to describe the particle nature of X-Rays
Be able to describe 3 predominant reasons for attenuation
Be able to carry out an experiment that models how the intensity of X-Rays attenuates
through materials.
Be able to select and use the equation I = I0 e−μx to show how the intensity I of a
collimated X-ray beam varies with thickness x of medium
Wave Properties
X-Rays are part of the EM Spectrum and so can be reflected, refracted and diffracted.
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Particle nature of X-Rays
The reason that intensity goes down when X-Rays pass through materials (Attenuation)
is because of the following:
Photoelectric effect
Pair production
Compton Effect
Photoelectric effect
Photon energy is absorbed by
electrons.
The electrons move into higher
energy levels.
As they move back down they emit
photons of light depending on how
far they have moved.
If they absorb enough energy
(depending on the work function Φ
of the element) they are emitted
from the atom as a photo-electron
E=Φ+KEmax
hf= Φ+½mv2
Pair Production
X-Ray photon absorbed by a nucleus.
Producing a positron and an electron.
Positron unstable and breaks down into two photons.
Can occur when the X-Ray photon energy is greater than 1.02 MeV, but really only
becomes significant at energies around 10 MeV.
Most X-Ray tubes work at P.Ds of less than 100000V (0.625eV).
Explanation
An electron or positron have a mass of 0.00055u.
1u is equivalent to 931MeV (since 1 u = 1.661 x 10-27 kg E calculated using Einstein’s
equation E=mc2) (check you can calculate this).
So to produce a positron and an electron we must have energy of
2x0.00055x931MeV=1.02meV.
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Compton Effect
Also called Compton Scattering.
It happens when the X-Ray photon is not
completely absorbed by the electron like
in the photoelectric effect.
It’s just deflected off course.
This changes energy of photon and
therefore wavelength.
Explanation of equation
E = hf = hc/λ
E = mec2 = hc/λ
So λ = h/mec
At θ=0 all X-Ray photon energy absorbed (photoelectric effect occurs)
θ=90 impossible (no energy absorbed)
Intensity
Intensity = Power / Cross sectional Area
Attenuation
We have discussed the causes of attenuation.
Now we look at it mathematically:
I = I0 e−μx
μ is the attenuation coefficient for a particular material, I the intensity and x is the
thickness of the material.
e.g: for medical X-Rays, μ is:
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vacuum, 0
Flesh, 100m-1
Bone, 300m-1
The distance for halving the intensity is called the half-value thickness.
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Lesson 45 questions – X-Ray Interaction
Name……………………………….
Class……………
(
/27)…..%………
ALL
1.
An average person in the UK will have at least 30 X-ray photographs taken in their
lifetime.
In order to take an X-ray photograph, the X-ray beam is passed through an
aluminium filter to safely remove low energy X-ray photons before reaching the
patient.
(a)
Suggest why it is necessary to remove these low energy X-rays.
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[1]
SOME
(b)
The average linear attenuation coefficient for X-rays that penetrate the
aluminium is 250 m–1.
The intensity of an X-ray beam after travelling through 2.5 cm of aluminium is
347 W m–2.
Show that the intensity incident on the aluminium is about 2 × 105 W m–2.
[3]
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(c)
The X-ray beam at the filter has a circular cross-section of diameter 0.20 cm.
Calculate the power of the X-ray beam from the aluminium filter. Assume that
the beam penetrates the aluminium filter as a parallel beam.
power = .................................................... W
[2]
[Total 6 marks]
MOST
2.
Fig. 1 shows data for the intensity of a parallel beam of X-rays after penetration
through varying thicknesses of a material.
intensity / MW m–2
thickness / mm
0.91
0.40
0.69
0.80
0.52
1.20
0.40
1.60
0.30
2.00
0.23
2.40
0.17
2.80
Fig. 1
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(a)
On Fig. 2 plot a graph of transmitted X-ray intensity against thickness of
absorber.
1.0
0.8
0.6
intensity
/ MW m –2
0.4
0.2
0
0
1.0
2.0
3.0
thickness / mm
Fig. 2
[3]
(b)
(i)
Find the thickness that reduces the intensity of the incident beam by one
half.
thickness = …………….. mm
[1]
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(ii)
Use your answer to (b)(i) to calculate the linear attenuation coefficient μ.
Give the unit for your answer.
 = …………….. unit ………
[4]
[Total 8 marks]
ALL
3.
In order to take an X-ray photograph, the X-ray beam is passed through an
aluminium filter to remove low energy X-ray photons before reaching the patient.
(a)
Suggest why it is necessary to remove these low-energy X-rays.
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[1]
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SOME
(b)
The average linear attenuation coefficient for X-rays that penetrate the
aluminium is 250 m–1. The intensity of an X-ray beam after travelling through
2.5 cm of aluminium is 347 W m–2.
Show that the intensity incident on the aluminium is about 2 × 105 W m–2.
[3]
(c)
The X-ray beam at the filter has a circular cross-section of diameter 0.20 cm.
Calculate the power of the X-ray beam emerging from the aluminium filter.
Assume that the beam penetrates the aluminium filter as a parallel beam.
power = ............................ W
[2]
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ALL
(d)
The total power of X-rays generated by an X-ray tube is 18W.
The efficiency of conversion of kinetic energy of the electrons into X-ray
photon energy is 0.15%.
(i)
Calculate the power of the electron beam.
power = ..................... W
[2]
(ii)
Calculate the velocity of the electrons if the rate of arrival of electrons is
7.5 × 1017 s–1. Relativistic effects may be ignored.
velocity = ..................... m s–1
[2]
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(iii)
Calculate the p.d. across the X-ray tube required to give the electrons
the velocity calculated in (ii).
p.d.= ........................ V
[3]
[Total 13 marks]
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Lesson 46 notes – X-Ray imaging
Objectives
(f) describe the use of X-rays in imaging internal body structures including the use of
image intensifiers and of contrast media (HSW 3, 4c and 6);
(g) explain how soft tissues like the intestines can be imaged using barium meal;
(h) describe the operation of a computerized axial topography (CAT) scanner;
(i) describe the advantages of a CAT scan compared with an X-ray image (HSW 4c, 6).
Outcomes
Be able to describe how X-Rays can be used in the imaging of internal body structures
in 2D and 3D using computerised axial tomography.
Be able to describe the advantages of a CAT scan compared with an X-ray image
(HSW 4c, 6).
Be able to explain how soft tissues like the intestines can be imaged using barium meal.
Use of X-Rays in diagnosis
X-Rays are very penetrating and can pass through the body.
They are absorbed more by dense materials such as bone (so they are useful for
checking the skeleton) and particularly metal.
Because X-Rays can damage cells we need to minimise exposure by:
only using X-Rays when necessary,
shielding other parts of the body,
focussing the X-Rays,
using short exposure times.
CAT scans
Multiple X-Ray images can be combined to give 3D images
This is called a CAT scan. Computerised Axial Tomography (tomos – Greek for slice)
The scans are put together to form the
images using computers.
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The fan shaped X-Rays slice through the parts of the body and are detected using
fluorescent material whose sparks of light are detected by CCDs that you might find in a
digital camera.
Advantages of CAT Scans over normal X Rays
 A major advantage of CT is its ability to image bone, soft tissue and blood
vessels all at the same time.
 Unlike conventional x-rays, CT scanning provides very detailed images of many
types of tissue as well as the lungs, bones, and blood vessels.
 CT examinations are fast and simple; in emergency cases, they can reveal
internal injuries and bleeding quickly enough to help save lives.
 CT imaging provides real-time imaging, making it a good tool for guiding
minimally invasive procedures such as needle biopsies and needle aspirations of
many areas of the body, particularly the lungs, abdomen, pelvis and bones.
 The dose of radiation is low because of the sensitivity of the sensors.
Barium Meal
The gastrointestinal tract, like other soft-tissue structures, does not show up very well
on X-Rays.
Barium salts are radio opaque, (they show clearly on a radiograph). If barium is
swallowed before radiographs are taken, the barium within the oesophagus, stomach or
duodenum shows the shape of the lumina of these organs.
Liquid suspensions of barium sulfate are non-toxic and they usually have a chalky taste
that can be disguised by adding flavours.
A barium meal usually takes less than an hour. The patient ingests gas pellets and citric
acid to expand the stomach. Then about about 709mL of Barium is ingested. The
patient may move or roll over to coat the stomach and oesophagus in barium.
Following these preparations, an x-ray is taken.
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Lesson 46 questions – X-Ray Imaging
Name……………………………….
Class……………
(
/26)…..%………
SOME
1.
Full-body CT scans produce detailed 3-D information about a patient and can
identify cancers at an early stage in their development.
(a)
Describe how a CT scan image is produced, referring to the physics principles
involved.
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[7]
(b)
State and explain two reasons why full-body CT scans are not offered for
regular checking of healthy patients.
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[3]
[Total 10 marks]
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2.
Describe the use of a contrast medium, such as barium, in the imaging of internal
body structures. Your answer should include
•
how an image of an internal body structure is produced from an X-ray
beam
•
an explanation of the use of a contrast medium
•
examples of the types of structure that can be imaged by this process.
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[Total 8 marks]
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3.
The figure below shows a cross-section of a radiographic detector which uses fllm
and intensifying screens. Describe how an image of an internal body structure may
be produced using X-ray film. Within your answer you should include details of the
use and advantages of an intensifying screen.
X-ray photons
plastic coating
intensifying screen
film
intensifying screen
metal backing
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[Total 8 marks]
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Lesson 47 notes – Radioactive tracers and the Gamma
Camera
Objectives
(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);
Outcomes
Be able to describe the use of medical tracers like technetium-99m to diagnose the function of
organs;
Be able to describe the main components of a gamma camera;
Be able to describe the principles of positron emission tomography (PET);
Use of gamma radiation in diagnosis - Tracer techniques
A source of gamma radiation is injected or ingested and moves around the body
collecting in the region of interest.
The gamma rays given off are detected outside the body and the location is determined.
Technetium-99m
Technetium-99m (m means metastable since it has a comparatively long half life
compared to a lot of other gamma emitters).
It is used as a radioactive tracer.
It is well suited to the role because it emits readily detectable 140 keV gamma rays, and
its half-life is 6.01 hours (meaning that about 94% of it decays to technetium-99 in
24 hours).
Technetium-99 decays almost entirely by beta decay.
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Gamma Cameras
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PET Scans
While a CT scan provides anatomical detail (size and location of the tumour, mass,
etc.), a PET scan provides metabolic detail (cellular activity of the tumour, mass, etc.).
Combined PET/CT is more accurate than PET and CT alone.
This is a whole-body PET acquisition, showing
abnormality in the region of the stomach.
Normal physiological isotope uptake is seen in
the brain, renal collection systems and
bladder.
Radionuclides used are typically isotopes with short half-lives.
Examples are: carbon-11 (~20 min), nitrogen-13 (~10 min), oxygen-15 (~2 min), and
fluorine-18 (~110 min).
These radionuclides are incorporated into compounds normally used by the body such
as glucose, water or ammonia
These labelled compounds are known as radiotracers.
PET technology can be used to trace the biologic pathway of any compound.
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How PET Scans work
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Lesson 48 notes – Radioactive tracers and the Gamma
Camera
Objectives
(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);
Outcomes
Be able to describe what MRI scans are used for and the advantages and
disadvantages of them.
Be able to describe the principles of MRI.
Be able to describe the main components of an MRI scanner.
MRI Basics
The Angular momentum of an electron makes a magnet
(The strength of which is given by “magnetic moment, μ”.)
The Nuclei of atoms have a spin and associated magnetic moments just like electrons,
except that the magnetic moments are 1000-10000 times smaller. Each different type
of atom has different moment, or “nuclear fingerprint”.
By looking at nucleus spin flip transitions, we can identify different types of atoms but
this nuclear spin spectroscopy (“nuclear magnetic resonance”, NMR) will also work in a
solid, including living tissue.
Can use NMR to measure distribution of H atoms in human body, and chemical
environment = “magnetic resonance imaging, MRI”.
Larmor Frequency, Resonance and Relaxation time
A spinning disc of charge has angular momentum.
Produces magnetic dipole moment, µ.
------------------When a proton is in a magnetic field it experiences a torque and so it precesses with a
certain frequency called the Larmor Frequency.
The Larmor frequency depends on the magnetic moment, µ,and the magnetic flux
density, B but is in the range of radio frequencies.
If a radio wave at the Larmor frequency is absorbed by a proton it will flip on it axis –
changing energy state.
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Over time it will flip back to the lower energy state and as it does so will emit a radio
wave that can be detected. The time it takes to flip back is called the relaxation time.
Magnetic resonance imaging. (MRI)
Detects the density of H atoms throughout body. There is more H than anything else,
and its magnetic moment is the biggest of common stuff. Different tissues have
different molecules = different number of H atoms.
H atoms--tiny magnets
MRI Scanner
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How to look at details in a particular location
The magnetic field is made different across body weak at one end and strong at the
other. This uses the magnetic field dependence of resonance.
1.0 T
1.1 T
BLE
B BRE
x
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Advantages and disadvantages
Advantages
No ionising radiation
Image quality is excellent
Can tell the difference between different tissues
Radio waves pass through all materials (except metal) easily
No side effects known
Disadvantages
No metal can be scanned and so no pins etc allowed in
machine
Machines very expensive
Machines sensitive to radio waves in the machine
Scans take a long time
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Lesson 48 questions – MRI
Name……………………………….
Class……………
(
/6)…..%………
ALL
1.
Discuss briefly the advantages and disadvantages of scanning using MRI
techniques.
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[Total 6 marks]
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Lesson 49 notes – Non-invasive techniques
Objectives
(h) describe the need for non-invasive techniques in diagnosis (HSW 6a);
(i) explain what is meant by the Doppler effect;
Outcomes
Be able to describe the need for non-invasive techniques in diagnosis (HSW 6a);
Be able to explain what is meant by the Doppler effect;
Endoscopes
An endoscope is a bundle of fibre optic cables that can be used to see inside a patient
without having to cut them open! Light is shone down the cables that reflects off the part
of the body that is being examined and back up to the eyepiece by total internal
reflection. Each fibre optic cable is one pixel of the image – so the more there are the
better the image quality.
They can be used to examine lungs, intestines, stomach etc.
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With a small incision keyhole surgery is also made possible with the use of this
technology.
Ultrasound and the Doppler effect
Ultrasound is any frequency above 20000Hz. It can also be used for non-invasive
imaging and is especially useful where ionising radiation may be particularly dangerous.
Ultrasound uses the Doppler effect to help image moving blood flow so an explanation
of the Doppler effect follows:
Moving sound produces high frequency in font and low frequency behind.
Remember c=fλ
If a wave of frequency f and a speed c moves with a speed v then since T=1/f, in this
time it has moved a distance v/f.
The wave is being caught up with its source and so:
λ’=(c-v)/f
f’=c/λ’=cf/(c-v)
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Lesson 50 notes – Ultrasound
Objectives
(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
I = (Z2-Z1)2
Ir
(Z2+Z1)2
(h) describe the importance of impedance matching;
(i) explain why a gel is required for effective ultrasound imaging techniques.
Outcomes
Be able to describe the properties of ultrasound;
Be able to describe the piezoelectric effect;
Be able to explain why a gel is required for effective ultrasound imaging techniques.
Be able to explain how ultrasound transducers emit and receive high-frequency sound;
Be able to describe the principles of ultrasound scanning;
Be able to describe the difference between A-scan and B-scan;
Be able to calculate the acoustic impedance using the equation Z = ρc;
Be able to describe the importance of impedance matching;
Be able to calculate the fraction of reflected intensity using the equation
I = (Z2-Z1)2
Ir
(Z2+Z1)2
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Ultrasound
Ultrasound is a sound that has a higher frequency than our ears can hear.
Other animals have different ranges.
Uses of Ultrasound in Medicine
•Ultrasound is used for examining soft tissue inside the body.
•Parts of the body that may be examined include muscles and unborn babies.
•Blood flow can also be monitored using ultrasound.
A Scan
1-dimensional, distance only.
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B Scan – creates a picture
Piezo crystals
Piezo crystals create electricity when they are
squashed but if they are given an electrical
impulse they will vibrate and create sounds that
can be amplified and used in ultra sonograms.
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•From previous lesson we had:
•f’=c/λ’=cf/(c-v)
•So if it appears to go twice the distance in the same time we get:
•f’=c/λ’=cf/(c-2v)
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Acoustic impedance
At a boundary, part of the initial intensity is transmitted, and part is reflected.
It is important to try and match the impedance by using gel to get as much ultrasound to
be transmitted instead of being reflected straight off the surface.
•The acoustic impedance is given by:
•Z = v ( - density; v-speed of sound in material)
•It is used to find the fraction of the intensity refracted at a boundary of different
materials.
•Ir = (Z2-Z1)2
I0 (Z2+Z1)2
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Lesson 50 questions – Ultrasound
Name……………………………….
Class……………
(
/44)…..%………
SOME
1.
The quality of ultrasound images in increasing at a phenomenal pace, thanks to
advances in computerised imaging techniques. The computer technology is
sophisticated enough to monitor and display tiny ultrasound signals from a patient.
The ratio of reflected intensity to incident intensity for ultrasound reflected at a
boundary is related to the acoustic impedance Z1 of the medium on one side of the
boundary and the acoustic impedance Z2 of the medium on the other side of the
boundary by the following equation.
reflected intensity (Z 2  Z1 ) 2

incident intensity (Z 2  Z1 ) 2
(a)
State two factors that determine the value of the acoustic impedance.
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[2]
(b)
An ultrasound investigation was used to identify a small volume of substance
in a patient. It is suspected that this substance is either blood or muscle.
During the ultrasound investigation, an ultrasound pulse of frequency of
3.5 × 106 Hz passed through soft tissue and then into the small volume of
unidentified substance. A pulse of ultrasound reflected from the front surface
of the volume was detected 26.5 μs later. The ratio of the reflected intensity to
the incident intensity, for the ultrasound pulse reflected at this boundary was
found to be 4.42 × 10–4. The table below shows data for the acoustic
impedances of various materials found in a human body.
medium
air
blood
water
brain tissue
soft tissue
bone
muscle
© science-spark.co.uk 2010
acoustic impedance Z/ kg m–2 s–1
4.29 × 102
1.59 × 106
1.50 × 106
1.58 × 106
1.63 × 106
7.78 × 106
1.70 × 106
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(i)
Use appropriate data from the table above to identify the unknown
medium. You must show your reasoning.
medium = .....................................................
[4]
(ii)
Calculate the depth at which the ultrasound pulse was reflected if the
speed of ultrasound in soft tissue is 1.54 km s–1.
depth = .................................................. cm
[2]
(iii)
Calculate the wavelength of the ultrasound in the soft tissue.
wavelength = ............................................m
[2]
[Total 10 marks]
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MOST
2.
Describe the principles of the production of a short pulse of ultrasound using a
piezoelectric transducer.
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[Total 5 marks]
3.
The diagram below shows a trace on a cathode-ray oscilloscope (CRO) of an
ultrasound reflection from the front edge and rear edge of a foetal head.
20 s
The CRO timebase is set to 20 s cm–1. The speed of ultrasound in the foetal head
is 1.5 × 103 m s–1.
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(i)
Calculate the size of the foetal head.
size = ................................................... cm
[4]
(ii)
State and explain what would be seen on the CRO screen if gel had not been
applied between the ultrasound transducer and the skin of the mother.
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........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
........................................................................................................................
[3]
[Total 7 marks]
4.
Explain how ultrasound is produced using a piezoelectric crystal such as quartz.
.................................................................................................................................
.................................................................................................................................
.................................................................................................................................
[Total 2 marks]
© science-spark.co.uk 2010
Page 43 of 49
5.
Quartz is a compound of silicon and oxygen. Each silicon atom is attached to four
oxygen atoms. See Fig. 1 below. Each oxygen atom carries a negative charge. The
silicon atom carries a positive charge.
–
oxygen ion
silicon ion
–
–
–
–
–
–
+
+
–
–
E-field
–
–
–
+
Fig. 1
(i)
+
+
+
+
+
+
Fig. 2
On Fig. 2, draw possible positions for the negatively-charged oxygen ions
when an electric field is applied in the direction shown. The central silicon ion
and one oxygen ion have been drawn in for you.
[1]
(ii)
Use your answer to (i) to explain why a single crystal of quartz is piezoelectric.
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........................................................................................................................
[1]
[Total 2 marks]
© science-spark.co.uk 2010
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SOME
6.
(a)
Acoustic impedance Z is the product of the density  of a medium and the
speed of ultrasound v.
The fraction f of ultrasound reflected at a boundary between two media of
acoustic impedances Z1 and Z2 is given by the equation
f=
(Z 2  Z1 ) 2
(Z 2  Z1 ) 2
medium
density / kg m–3
ultrasound velocity v / m s–1
air
1.299
330
skin
1075
1590
coupling medium
1090
1540
bone
1750
4080
Fig. 1
(i)
Use the data in Fig. 1 to find the fraction f of ultrasound reflected at an
air-skin boundary.
f = ...........................
[2]
© science-spark.co.uk 2010
Page 45 of 49
MOST
(ii)
Hence explain the need for a coupling medium in ultrasound imaging.
...............................................................................................................
...............................................................................................................
...............................................................................................................
...............................................................................................................
[2]
(b)
Fig. 2 is a CRO display showing the reflected ultrasound signal from the front
edge F and the rear edge R of a bone. The time-base setting is 1.0 × 10–5 s
cm–1.
F
R
1 cm
1 cm
1.0 × 10–5 s
Fig. 2
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Page 46 of 49
Using appropriate data from Fig. 1 and Fig. 2, calculate the thickness of the
bone.
thickness = ........................................... cm
[4]
[Total 8 marks]
7.
The ratio of reflected intensity to incident intensity for ultrasound reflected at a
boundary is related to the acoustic impedance Z1 of the medium on one side of the
boundary and the acoustic impedance Z2 of the medium on the other side of the
boundary by the following equation.
reflected intensity (Z 2  Z1 ) 2

incident intensity (Z 2  Z1 ) 2
(a)
State the two factors that determine the value of the acoustic impedance.
........................................................................................................................
........................................................................................................................
[2]
© science-spark.co.uk 2010
Page 47 of 49
(b)
An ultrasound investigation was used to identify a small volume of substance
in a patient. It is suspected that this substance is either blood or muscle.
During the ultrasound investigation, an ultrasound pulse of frequency
3.5 × 106 Hz passed through soft tissue and then into the small volume of
unidentified substance. A pulse of ultrasound reflected from the front surface
of the volume was detected 26.5 μs later. The ratio of the reflected intensity to
incident intensity for the ultrasound pulse reflected at this boundary was found
to be 4.42 × 10–4. The table below shows data for the acoustic impedances of
various materials found in a human body.
(i)
medium
acoustic impedance Z /kg m–2 s–1
air
4.29 × 102
blood
1.59 × 106
water
1.50 × 106
brain tissue
1.58 × 106
soft tissue
1.63 × 106
bone
7.78 × 106
muscle
1.70 × 106
Use appropriate data from the table to identify the unknown medium.
You must show your reasoning.
medium = ............................
[4]
© science-spark.co.uk 2010
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(ii)
Calculate the depth at which the ultrasound pulse was reflected if the
speed of ultrasound in soft tissue is 1.54 km s–1.
depth = ...................... cm
[2]
(iii)
Calculate the wavelength of the ultrasound in the soft tissue.
wavelength = ............................... m
[2]
[Total 10 marks]
© science-spark.co.uk 2010
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