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Medical Physics
Brain
The ear and hearing
Describe the basic structure of the human ear. The
structure should be limited to those features
affecting the physical operation of the ear.
State and explain how sound pressure variations in air are changed into
larger pressure variations in the cochlear fluid.
The eardrum is some fifteen times larger than
the oval window of the inner ear, giving an
amplification of about fifteen compared to a
case where the sound pressure interacted
with the oval window alone.
This can be dealt with in terms of the different areas of the eardrum and oval window,
together with the lever action of the ossicles. Although the concept of impedance
matching is not formally required, students should appreciate that, without
a mechanism for pressure transformation between media of different densities (air
and fluid), most sound would be reflected, rather than transmitted into the cochlear
fluid.
With a long enough lever, you can lift a
big rock with a small applied force on the
other end of the lever. The amplification of
force can be changed by shifting the pivot
point.
The ossicles can be thought of as a compound lever which achieves a
multiplication of force. This lever action is thought to achieve an amplification by a
factor of about three under optimum conditions, but can be adjusted by muscle
action to actually attenuate the sound signal for protection against loud sounds.
State the range of audible frequencies experienced by a
person with normal hearing.
i.e. 20Hz to 20kHz
State and explain that a change in observed loudness is
the response of the ear to a change in intensity.
Source
Intensity
Intensity
Level
# of Times
Greater Than TOH
Threshold of Hearing
(TOH)
1*10-12 W/m2
0 dB
100
Rustling Leaves
1*10-11 W/m2
10 dB
101
Whisper
1*10-10 W/m2
20 dB
102
Normal Conversation
1*10-6 W/m2
60 dB
106
Busy Street Traffic
1*10-5 W/m2
70 dB
107
Vacuum Cleaner
1*10-4 W/m2
80 dB
108
Large Orchestra
6.3*10-3 W/m2
98 dB
109.8
Ipod at Maximum
Level
1*10-2 W/m2
100 dB
1010
Front Rows of Rock
Concert
1*10-1 W/m2
110 dB
1011
Threshold of Pain
1*101 W/m2
130 dB
1013
Military Jet Takeoff
1*102 W/m2
140 dB
1014
Instant Perforation of
Eardrum
1*104 W/m2
160 dB
1016
State and explain that there is a logarithmic response of the
ear to intensity.
Intensity is the power of
the sound wave
reaching the eardrum. It
is measured in Wm-2.
Intensity level (dB)
1.Find the ratio of the intensities.
2.Express this as a power of 10.
3. Multiply this by 10 to give the
intensity level in dB.
Define intensity and also intensity level (IL).
Intensity is the power of
the sound wave
reaching the eardrum. It
is measured in Wm-2.
We define Loudness ≡ Intensity Level L
which is proportional to the {base-10}
logarithm of the intensity.
i.e. it is defined by the formula in words.
N.B. to add 2 sounds, you
must add the intensities, not
the dB.
Find the resulting intensity level when a
70dB and an 80db sound are added.
1. A mosquito's buzz is often rated with a decibel rating of 40 dB.
Normal conversation is often rated at 60 dB. How many times more
intense is normal conversation compared to a mosquito's buzz?
2. The table at the right represents the decibel level for several sound
sources. Use the table to make comparisons of the intensities of the following
sounds.
How many times more intense is the front row of a Smashin' Pumpkins
concert than ...
a. ... the 15th row of the same concert?
b. ... the average factory?
c. ... normal speech?
d. ... the library after school?
e. ... the sound which most humans can just barely hear?
3. On a good night, the front row of a concert would result in a 120 dB sound
level. An IPod produces 100 dB. How many IPods would be needed to
produce the same intensity as the front row of the concert?
1. Answer: C. 100 times
Normal conversation is 20 dB more intense. This 20 db difference
corresponds to a 2-Bel difference. This difference is equivalent to a
sound which is 102 more intense. Always raise 10 to a power which is
equivalent to the difference in "Bels."
2 a. 10 X more intense
b. 102 X more intense
c. 105 X more intense
d. 107 X more intense
e. 1011 X more intense
3. Answer: 100 IPods
Since 120 db is 102 times or 100 times more intense than 100 dB. It is
necessary to wear 100 IPods to produce the same sound level.
IB question
Answer
State the approximate magnitude of the intensity level at
which discomfort is experienced by a person with
normal hearing.
Describe the effects on hearing of short-term and long-term
exposure to noise.
How do we diagnose hearing
problems?
Air Conduction assesses the function of both the conduction (outer and
middle ear) and sensorineural (cochlea and auditory nerve) components of
the ear.
Bone Conduction (BC) assesses the function of the cochlea and auditory
nerve only.
Analyse and give a simple interpretation of graphs
where IL is plotted against the logarithm of
frequency for normal and for defective hearing.
air conduction test
] - left, bone conduction
bone conduction test
[ - right, bone conduction Conductive loss
Sensorineural loss
Hyperlink
X = left ear air
conduction
O = right ear air
conduction
Conductive loss happens when there is a problem conducting sound
waves through the outer ear, eardrum or middle ear (ossicles). This can be
corrrected by surgery or a hearing aid. This is tested with an air conduction
test.
Sensorineural loss occurs when there is damage to the inner ear (cochlea) or
to the nerve pathways from the inner ear to the brain. Sensorineural hearing
loss cannot be medically or surgically corrected. It is a permanent loss. This is
diagnosed with a bone conduction test.
Conductive hearing loss
If there is a problem in the external or middle ear (conductive hearing
loss) the AC threshold will be less than the BC threshold because
the person will hear better by bone than air conduction
Sensorineural hearing loss
If there is a problem with the cochlea or the auditory nerve, the AC and
BC thresholds will be the same (See diagram below).
Mixed hearing loss
Mixed hearing loss is a reduction in hearing of both AC and BC, but
they are not the same. (See diagram below).
The bone conduction test is normal, but
there are losses in the air conduction
test. Therefore the problem is in the
outer/middle ear, not the cochlea.
How do you diagnose a middle ear problem?
How do you diagnose a inner ear problem?
How do you diagnose a mixed problem?
You should consider which test to use and how the audiograms would look.
Tsokos
Page 698 Q’s 1-9.
Medical imaging
X-ray scan
Define the terms attenuation
coefficient and half-value thickness.
The intensity I is the amount of energy per unit area
in the beam.
X is the distance travelled through the material.
The attenuation coefficient describes the extent to
which the intensity of an energy beam is reduced as
it passes through a specific material.
When used in the context of X-rays or Gamma-rays,
it is represented using the symbol μ, and measured
in cm-1.
Lead has a high attenuation coefficient.
The half-value thickness is the thickness
of the material that reduces the intensity
of the X-rays by 50%
Derive the relation between
attenuation coefficient and half-value thickness.
Staring with
As the intensity drops to 50%
Taking logs
Solve problems using the equation
I = I0 e− μx .
Describe X-ray detection, recording
and display techniques.
Students should be aware of photographic film, enhancement, electronic
detection and display.
Photographic plates are sensitive (the efficiency of X-Ray film to absorb x-ray
photons is only » 1%) to X-rays, they provide a means of recording the image, but
require a lot of exposure (to the patient), so intensifying screens were devised
(scintilator and photomultiplier tube). They allow a lower dose to the patient, because
the screens take the X-ray information and intensify it so that it can be recorded on
film positioned next to the intensifying screen.
Scintillators
Some materials such as sodium iodide (NaI) can "convert" an X-ray photon to a
visible photon; an electronic detector can be built by adding a photomultiplier tube.
These detectors are called “scintilators", filmscreens or “scintilation counters". The
main advantage of using these is that an adequate image can be obtained while
subjecting the patient to a much lower dose of X-rays.
Scintillators
Some materials such as sodium iodide (NaI) can "convert" an X-ray photon to
a visible photon; an electronic detector can be built by adding a
photomultiplier tube. These detectors are called “scintilators", filmscreens or
“scintilation counters". The main advantage of using these is that an
adequate image can be obtained while subjecting the patient to a much lower
dose of X-rays.
Photostimulable phosphors (PSPs) (Electronic detection)
An common method is the use of photostimulated luminescence (PSL), pioneered
by Fuji in the 1980s. In modern hospitals a photostimulable phosphor plate (PSP
plate) is used in place of the photographic plate. After the plate is x-rayed, excited
electrons in the phosphor material remain "trapped" in "colour centres" in the crystal
lattice until stimulated by a laser beam passed over the plate surface. The light
given off during laser stimulation is collected by a photomultiplier tube and the
resulting signal is converted into a digital image by computer technology, which
gives this process its common name, computed radiography (also referred to as
digital radiography). The PSP plate can be reused, and existing X-ray equipment
requires no modification to use them.
Image plates have the following advantages.
1. dynamic range larger than 5 orders of magnitude in x-ray dose,
2. lower limit of useful dose compared to the x-ray film,
3. reusability,
4. no wet chemical processing,
5. images are available in digital form,
6. image processing and pattern recognition possible, and
7. simple data storage on optical or digital media.
Disadvantages of image plates compared to the conventional x-ray film are
1. a poorer spatial resolution due to light scattering at the storage phosphor
grains during the readout process.
How X Rays Work
X Rays (continued)
X-rays: x-radiography
How do we obtain a clear x-ray
image?
X rays are scattered or absorbed as they pass through the body.
If the energy (tube voltage) of the x rays is too high, then all the x –
rays get through (penetration) and there is no contrast on the image.
If the x-ray energy is too low, then no x-rays reach the x-ray plate.
The choice of tube voltage (energy) depends upon the type of tissue
and the thickness of the tissue.
X ray contrast
• Structures in the body like bones are very dense and contain
elements such as calcium that have a high atomic number. This
makes bone absorb a high proportion of the x-rays. Soft tissues like
fat and muscle allow more x-rays to pass though. The body casts an
x-ray shadow onto the film. Where the x-rays have passed though
bone, the film is less exposed so it looks white; where they have not
passed though anything the film is exposed and turns black; and
where the x-rays have passed through soft tissues the film has
different levels of grey.
• In order to make some parts of the body show up better, contrast
media with a high atomic number can be used. This can be a
'barium meal', where the patient drinks a liquid containing barium
(atomic number 56) which makes the digestive tract show up clearly
on x-rays, or the patient can have an injection of iodine (atomic
number 53) which makes the blood vessels stand out (this is called
angiography).
Explain standard X-ray imaging
techniques used in medicine.
Students should appreciate the causes of loss of sharpness and of contrast in
X-ray imaging. They should be familiar with techniques for improving
sharpness and contrast.
The general relationship
between scattering angle
and size of the object is,
Angle = (object size)-1
This will cause the edge of
objects to lose sharpness
X-ray Sharpness
A quantitative measure of the loss of edge detail which is due to geometric
properties of the object and imaging system and not due to image noise or
X-ray scatter. It is usually expressed as the width of the band of changing
density or brightness arising from a sudden change in the intensity of the
radiation incident on the film or fluorescent screen. From this definition it
can be understood that unsharpness and resolution are different concepts.
It is possible for an edge to be "spread" by one of many factors, and at the
same time for two such edges to be resolved in the image. The factors
which contribute to the total image unsharpness include geometric
unsharpness, movement unsharpness, absorption unsharpness, image
receptor unsharpness, and parallax unsharpness. The various unsharpness
factors all contribute to the observed unsharpness of structures in an
image. However, the quantitative manner in which the factors combine is in
general complicated and is not completely understood. It is known from
observation that the total unsharpness is not the direct sum of the
contributing factors. In general, it appears that the total image unsharpness
is dominated by the unsharpness of the largest individual factor.
X-ray Sharpness
• Unsharpness is introduced by
1.A wide source (parallax)
2.A moving patient
IB Questions
Questions 7,11, 19b,c from review pack.
(b) An X-ray beam that consists of photons of the same energy is used to
image a possible bone fracture in the leg of a patient. At this photon energy
attenuation coefficient of bone = 0.62 cm–1
attenuation coefficient of tissue = 0.12 cm–1.
In passing through the leg, the X-rays effectively encounter a thickness of
tissue equal to 14 cm and thickness of bone equal to 8.0 cm.
Use the above data to explain why X-rays of this energy are suitable for
imaging a possible leg fracture.
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . [4]
(b) intensity after passing through bone = I0e-0.62x8.0
= 7x10-3I0
intensity after passing through tissue =0.19I0
reduction by bone much greater than by tissue so good contrast
between bone and tissue;
Outline the principles of computed
tomography (CT).
A CT scan (sometimes called computed axial tomography, or a CAT
scan) uses x-rays.
In a CT scan the patient lies on a table and is moved though a
doughnut-shaped machine. It creates images that are slices through
the patient.
It does this by moving the x-ray tube and detector in a circle taking xray images of the slice from all angles around the body.
A computer then processes these images to produce a cross
sectional image (a picture of a slice through the body).
CT scans are useful as they can show a range of very different tissue
types clearly: lung tissue, bone, soft tissue and blood vessels.
CT Scan
X-rays: Computed tomography image (CT
scan)
Second
metatarsal bone
(the bone that
David Beckham and
Wayne Rooney broke!)
X-rays: Computed tomography image (CT
scan)
Ultrasound
Ultrasound scan
Ultrasound
Ultrasound imaging: What does it look
like?
Ultrasound imaging: development of a
pregnancy
24 weeks
8 weeks gestation (out of a 40 week pregnancy)
18 weeks
Describe the principles of the generation and the
detection of ultrasound using piezoelectric
crystals.
When the ultrasound beam is
generated, an electrical signal
is sent to the crystal, and
converted to mechanical
vibrations.
Naturally-occurring crystals
Cane sugar, quartz, topaz, bone,
enamel, dentine and tendons.
When the reflected signal is
received, the mechanical
vibrations are converted back
to electrical signals and sent to
the computer for processing
into an image.
Define acoustic impedance (Z) as the product of
the density (ρ) of a substance and the speed of
sound (c) in that substance.
Z = ρc
The acoustic impedance of the eardrum, for instance, corresponds
well with that of the auditory canal, guaranteeing maximum
efficiency of energy transfer, but it does not correspond well with
air. The pinna may be described as an impedance matching device
between the air and the auditory canal. Likewise, the flared end of
a trumpet results in less energy being lost by being reflected back
down the tube of the instrument.
Solve problems involving acoustic
impedance.
Students should understand the use of a gel on the surface of the skin.
Z = ρc
Z1
Z2
Gel is generally necessary because the
acoustic impedance mismatch between air
and the body is large. Without gel, nearly all
of the energy is reflected and very little is
transmitted into the body.
Outline the differences between
A-scans and B-scans.
B-scan
A-scan
An A-scan, is routine type of diagnostic test used in
ophthalmology. The A-scan provides data on the
length of the eye. (i.e. is 1 dimensional)
A B-scan, is a diagnostic test used in ophthalmology
to produce a two-dimensional, cross-sectional view
of the eye.
Ultrasound imaging: A-scan How does it
work?
• An ultrasound element acts like a bat.
• Emit ultrasound and detect echoes
• Map out boundary of object
Ultrasound imaging: B-scan How does it
work?
• Now put many elements together to make a
probe and create an image
How do the scans work?
A-mode: A-mode is the simplest type of ultrasound. A single
transducer scans a line through the body with the echoes
plotted on screen as a function of depth.
B-mode: In B-mode ultrasound, a linear array of transducers
simultaneously scans a plane through the body that can be
viewed as a two-dimensional image on screen.
M-mode: M stands for motion. In m-mode a rapid sequence
of B-mode scans whose images follow each other in
sequence on screen enables doctors to see and measure
range of motion, as the organ boundaries that produce
reflections move relative to the probe.
Identify factors that affect the
choice of diagnostic frequency.
Students should appreciate that attenuation and resolution are dependent on
frequency.
High frequency gives you high
resolution but high attenuation i.e.
limited depth penetration.
Low frequencies are less
absorbed (attenuated),
therefore can penetrate
deeply. But have lower
resolution.
Tsokos
Page 710 Q’s 1-8.
IB Q 2 from the review pack.
NMR and lasers
Hyperlink to video
MRI Scan
Hyperlink
Outline the basic principles of nuclear
magnetic resonance (NMR) imaging.
Students need only give a simple qualitative description of the principle,
including the use of a non-uniform magnetic field in conjunction with the
large uniform field.
Magnetic Field
When a person is lying in the magnetic
field of the MRI scanner the nuclei of the
hydrogen atoms in their body line up,
like compass needles in the Earth's
magnetic field, either pointing in the
direction of the field or opposite to it.
The hydrogen nuclei (protons) don’t stay
still though, but move like a spinning top
around the direction of the magnetic
field.
How MRI works
1.The body is mainly composed of water molecules which
each contain two hydrogen nuclei or protons. When a
person goes inside the powerful magnetic field of the
scanner these protons align with the direction of the field.
2. A second radio frequency electromagnetic field is then
briefly turned on causing the protons to absorb some of
its energy. When this field is turned off the protons
release this energy at a radio frequency which can be
detected by the scanner.
3. The position of protons in the body can be determined by
applying additional magnetic fields during the scan which
allows an image of the body to be built up. These are
created by turning gradients coils on and off which
creates the knocking sounds heard during an MR scan.
Proton nuclear magnetic resonance (NMR) detects the presence of hydrogen nuclei
(protons) by subjecting them to a large magnetic field to align the nuclear spins, then
exciting the spins with properly tuned radio frequency (RF) radiation, and then
detecting weak radio frequency radiation from them as they “relax" from this
magnetic interaction. The frequency of this proton "signal" is proportional to the
magnetic field to which they are subjected during this relaxation process. In the
medical application known as Magnetic Resonance Imaging (MRI), an image of a
cross-section of tissue can be made by producing a well-calibrated magnetic field
gradient across the tissue so that a certain value of magnetic field can be associated
with a given location in the tissue. Since the proton signal frequency is proportional
to that magnetic field, a given proton signal frequency can be assigned to a location
in the tissue. This provides the information to map the tissue in terms of the protons
present there. Since the proton density varies with the type of tissue, a certain
amount of contrast is achieved to image the organs and other tissue variations in the
subject tissue.
Since the MRI uses proton NMR, it images the concentration of protons. Many of
those protons are the protons in water, so MRI is particularly well suited for the
imaging of soft tissue,
It is estimated that about 80% of the body's atoms are hydrogen atoms, so most
parts of the body have an abundance of sources for the hydrogen NMR signals
which make up the magnetic resonance image.
Describe examples of the use of lasers
in clinical diagnosis and therapy.
Applications such as the use in pulse oximetry and in endoscopes should be
discussed. Students should be familiar with the use of a laser as a scalpel
and as a coagulator.
Infrared: Pulse oximetry
Heart rate:81 bpm
Blood oxygenation: 99%
Pulse oximetry
Pulse oximetry is a simple non-invasive method of monitoring the percentage of
haemoglobin (Hb) which is saturated with oxygen. The pulse oximeter consists
of a probe attached to the patient's finger or ear lobe which is linked to a
computerised unit.
The principle of pulse oximetry is based on the red and infrared light absorption
characteristics of oxygenated and deoxygenated haemoglobin. Oxygenated
haemoglobin absorbs more infrared light and allows more red light to pass
through. Deoxygenated (or reduced) haemoglobin absorbs more red light and
allows more infrared light to pass through.
Visible: Endoscopy
Visible: Endoscopy
Visible: Endoscopy
This is the
endoscope coming
out of the oesophagus
Parasitic
Worm!
The stomach wall has
relapsed back
into the oesophagus.
This is a hernia.
Radiation in medicine
State the meanings of the terms exposure,
absorbed dose, quality factor (relative
biological effectiveness) and dose
equivalent as used in radiation dosimetry.
Students should be able to discuss the significance of these quantities in
radiation dosimetry.
Exposure
X = Q/m
Coulombs per Kg
Intensity of Radiation (Exposure)
The roentgen (R) is a measure of radiation intensity of x-rays or gamma rays. It is
formally defined as the radiation intensity required to produce and ionization charge
of 0.000258 coulombs per kilogram of air. It is one of the standard units for
radiation dosimetry, but is not applicable to alpha, beta, or other particle emission
and does not accurately predict the tissue effects of gamma rays of extremely high
energies. The roentgen has mainly been used for calibration of x-ray machines.
Absorbed dose (J/kg)
The charge deposited per Kg of air does not take into account the type of tissue
and therefore the energy deposition.
The absorbed dose is equal to the energy deposited per unit mass of the
medium, and so has the unit J/kg, which is given the special name Gray (Gy).
Note that the absorbed dose is not a good indicator of the likely biological
effect.
Radiation type Quality factor
The dose equivalent is a measure of the
radiation dose to tissue where an attempt has been
made to allow for the different relative biological
effects of different types of ionising radiation.
Equivalent dose is therefore a less fundamental
quantity than radiation absorbed dose, but is more
biologically significant. Equivalent dose has units of
sieverts.
X -rays
1
gamma
1
beta
1
alpha
20
slow neutrons
5
fast neutrons
10
protons
5
Discuss the precautions taken in situations
involving different types of radiation.
Students should consider shielding, distance and time-of-exposure factors.
They should be familiar with the film badge.
There is a light-proof packet of
photographic film inside the badge.
The more radiation this absorbs, the
darker it becomes when it is
developed. To get an accurate
measure of the dose received, the
badge contains different materials
that the radiation must penetrate to
reach the film. These may include
aluminium, copper, lead-tin alloy and
plastic. There is also an open area at
the centre of the badge.
TOK: They should appreciate that current practice is determined from a gradual
increase in available data.
Penetrative power
Alpha Particles
(2n, 2p)
Beta Particles
(e-or+)
Photons (hv)
(x or gamma rays)
Paper
Concrete
Minimizing Radiation
Exposure
Basic Concepts
• Time
• Distance
• Shielding
Minimizing Exposure - Time
• Minimize the
amount of time
spent near
sources of
radiation.
Minimize Exposure by
Maximizing Distance
As the distance from a radioactive
source doubles, the exposure rate
decreases by a factor of four.
Minimizing Exposure By
Using Shielding
Lead shielding
around radiation
sources
The type of shielding required
depends on the type of
radiation present.
Surry Power Station
Loss of Life Expectancy
Activity or Behavior
LLE (DAYS)__________________
Recreational swimming
40
Being 15 percent overweight
900
Smoking 20 cigarettes per day
1,600
Using pesticides at home
12
Being exposed to radon in a home
35
Living within 10 miles of a nuclear power plant
0.4
Riding a bicycle
6
Driving a car
200
Skydiving
25
Consuming alcohol (U.S. average)
230
Discuss the concept of balanced risk.
Aim 8, Int, TOK: Students should appreciate that codes of practice have been
developed for conduct involving the use of radiations.
Alara principle
Distinguish between physical half-life,
biological half-life and effective half-life.
Students should be able to calculate the effective half-life from the physical halflife and the biological half-life.
Radiation Dose
Dose or radiation dose is a generic term for a
measure of radiation exposure. In radiation
protection, dose is expressed in millirem.
External Dose
After
X-Ray
Machine
Image
(film)
Subject is not radioactive but
has been exposed to a
radiation dose (single chest x
ray = 5-10 mrem).
Contamination
Contamination is the presence of a radioactive
material in any place where it is not desired,
and especially in any place where
its presence could be harmful.
Yuck!
Solve problems involving radiation
dosimetry.
1.
This question is about radiation used in medicine.
(a)
Define the terms exposure and absorbed dose.
(2)
Exposure: .........................................................................................................
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Absorbed dose:
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(b)
Explain, with reference to α and γ radiation, the distinction between absorbed dose and dose
equivalent.
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(3)
(c)
Explain why, when using radioactive tracer elements in the treatment of cancer, it is better to use
radioactive isotopes that have a long physical half-life and a short biological half-life.
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(2)
(Total 7 marks)
1.
(a)exposure:total ionised charge produced in unit mass of air
by a particular radiation;
absorbed dose: energy absorbed per unit mass;
2
(b)
dose equivalent is the amount of energy absorbed;
but a quality factor is introduced to describe the effects of different
types of radiation;
α is absorbed more that γ radiation and so has a much higher Q factor; 3
Do not look for this precise wording but look for the understanding.
(c)
if the biological half-life is long then the tracer can do a lot of damage
to healthy cells;
with a short biological half-life and long physical half-life the tracer
will have a high activity during the time it is in the body;
2
Again do not look for this precise wording but look for the understanding.
[7]
2.
This question is about dosimetry.
(a)
Describe what is meant by the term relative biological effectiveness (quality
factor).
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.....................................................................................................................................(2)
The whole body of a person of mass 70 kg is exposed to monochromatic Xrays of energy 200 keV. As a result of this exposure, the person receives a dose
equivalent of 500 μSv in 2.0 minutes.
(b)
Deduce that the person absorbs about 1010 X-ray photons per second.
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(Total 6 marks)
2.(a)
a factor that compares (the effectiveness) of different types of radiation
to that of X-rays;
some detail e.g. because different radiations (of the same intensity)
produce different amounts of ionization / cause different amounts of damage;
2
Second marking point can be implied in first point.
(b)
D = 5 × 10–4 J kg–1 (Q = 1);
E = m × D = 70 × 5 × 10–4 J;
1 photon has energy 200 × 103 × 1.6 × 10–19 J;
number of photons in 1 second = » 1010;
4
[6]
Outline the basis of radiation
therapy for cancer.
Radiation Therapy
Used for treating cancer. Why does it work?
Photo by Karen Sheehan
Image courtesy of
External Beam
Brachytherapy (implants)
X-rays: Radiotherapy
This should include the differential effects on normal and malignant cells, as
well as a description of the types of sources available.
X-rays: Radiotherapy
Cancer occurs when cells divide too
quickly. The dividing cells become a
tumour which can damage
surrounding tissue or spread to the
rest of the body
X-rays: Radiotherapy
• X-rays or other radiation can damage
the DNA in cells and kill them
• This is why radiation can be
dangerous
• But cells which are dividing rapidly
are more likely to be killed
• So we use x-rays to kill the rapidlydividing cancer cells
• We must still ensure that healthy
tissue is undamaged
X-rays: Radiotherapy
A linear accelerator generates
x-rays. It rotates around the
body, irradiating the tumour
from all directions
X-rays: Radiotherapy
X-ray CT scan of chest shows lungs, heart and tumour (red)
X-rays: Radiotherapy
A medical physicist decides which
angles to shine x-rays from
to destroy tumour and minimise
damage to other tissue
X-rays: Radiotherapy
The treatment plan lists the directions the x-rays will come from and calculates the
radiation dose to the tumour (in purple) and rest of body (grey)
Solve problems involving the choice of
radio-isotope suitable for a particular diagnostic or
therapeutic application.
Students should be familiar with a variety of techniques. Where reference is
made to a specific technique, sufficient description will be given for
the student to be able to answer any questions on that technique.
Things to consider;
Type of radiation. How far do you want it to travel in the body?
Half life. How long do you want the substance to be radioactive for?
Solve problems involving particular
diagnostic applications.
For example, assessment of total blood volume. Where reference is made to a
specific technique, sufficient description will be given for the student
to be able to answer any questions on that technique.
Nuclear Medicine
Diagnostic Procedures
• Radioactive injection
• Short half-life
radionuclide
• Pictures taken with
special gamma camera
• Many different studies:
Thyroid
Lung
Cardiac
White Blood Cell
Photo by Karen Sheehan
Bone Scans
Image courtesy of
Tsokos
Page 717 Q’s 1-9.
IB review pack Q’s 4,9,17,20,25,30.