Download I2 Medical imaging

I2 Medical imaging
X-rays
• X-rays pass through human (or dog) flesh very
easy, but do not pass through bone as easily.
Absorption
• Light passing through a solid is absorbed if the
photon energy is enough to increase the
energy of an electron so it can jump from a
low energy level to a higher one.
Absorption
• Since light photons have an energy about the
same as an electron transition (1 eV approx)
most are absorbed.
X-ray photons
• X-ray photons have much more energy (>120
eV)
• They can knock electrons out of the atom in 2
ways
1. Photoelectric effect
• X-ray photon is completely absorbed – giving
all of its energy to the KE of the electron
(minus the work function of course)
2. Compton scattering
• The photon gives some of its energy to the
electron and continues with less energy and
so longer wavelength (E = hf = hc/λ)
2. Compton Scattering
Attenuation
Attenuation
• Reduction in X-ray intensity with thickness
Number of photons x 1015
100
# interaction is proportional to the # photons,
so the curve is exponential decay
50
3
3
Thickness/cm
6
Attenuation coefficient
I=
-μx
I0e
• where I0 = the original intensity
• x = distance through absorber
• μ = attenuation coefficient
Half-value thickness x½
• Thickness required to reduce the intensity to
half its original intensity
I=
-μx½
I0/2 = I0e
• Taking logs
x½ = ln2/μ = 0.693/μ
• Half value thickness depends on the energy of
the X-rays and the substance
Example
• A parallel beam of X-rays of intensity 0.2
kW.m-2 is passed through 5 mm of a material
of half-value thickness 2 mm. Calculate the
intensity of the beam.
HL Physics, C. Hamper, Pearson 2009
Example
• A parallel beam of X-rays of intensity 0.2 kW.m-2 is passed
through 5 mm of a material of half-value thickness 2 mm.
Calculate the intensity of the beam.
• First calculate the attenuation
coefficient μ
• μ = 0.693/x½ = 0.693/2 = 0.35 mm-1
• I = I0e-μx = 0.2e-0.35x5 = 0.035 kW.m-2
HL Physics, C. Hamper, Pearson 2009
Taking an X-ray picture
Taking an X-ray picture
• http://www.youtube.com/watch?v=GG2XYSp
nyus
• Have to choose X-rays of the right
intensity/energy
• X-rays are dangerous (ionising) – need to keep
exposure time and intensity to a minimum
• Simplest way is to place broken part on
photographic film (which is enclosed in a light
proof box
Black areas are
where lots of
photons hit the
photographic
film and
“exposed” it
Few X-ray photons
travelled through
and hit paper
Intensifying screen
• A screen of fluorescent material which gives
out visible light when X-rays hit it is put on
either side of the photographic film. This
means X-rays of lower intensity can be used.
Barium Meal
• To help see soft-tissue such as the gut, it can
be filled with a material with is opaque to Xrays like barium sulphate. Drinking a “barium
meal”
Digital images
• A array of photosensitive diodes works in a
similar way to a CCD. They gain charge when
exposed to X-rays. This produces a p.d. which
can be converted to a digital signal.
Digital images
• The digital image can be enhanced and
coloured electronically – as in airport security.
Tomography
• For parts deep in the body a “slicing”
technique called tomography can be used
Tomography
Computer Tomography (CT scan)
Computer Tomography (CT scan)
• X-ray source and a circular array of sensors are
rotated around the patient
Computer Tomography (CT scan)
• By moving along the length of the body and
using computers to analyse the digital signals
a complete 3-D image can be built.
Let’s try a worksheet
Ultrasound
Ultrasound
• Higher frequency than a healthy human can
hear (>20 kHz).
• Ultrasounds are reflected off different layers in
the body and a picture is built up.
Piezoelectric effect
• When a quartz crystal is stretched, a p.d. is
produced (due to polarisation of the
molecules in the crystal)
• If a p.d. is applied, the same happens, the
quartz is deformed.
Piezoelectric effect
Piezoelectric effect
Piezoelectric effect
Piezoelectric effect
Ultrasound production
• An alternating p.d. of frequency >20 kHz is
applied to a quartz crystal causing it to vibrate
Ultrasound detection
• A sound wave causes a crystal to vibrate,
producing an alternating p.d.
Reflections
• Because the reflections are analysed, the
transmitter and receiver are in the same
place. The same crystal can be used as
transmitter and receiver if pulsed signals are
used that are short enough so the reflected
wave returns before a new pulse is produced.
Example
• If the pulse length is 10-6 s and the speed of
sound in body tissue is 1500 m.s-1 then the
minimum distance the wave can travel before
the pulse has finished transmitting is 1500 x
10-6 = 1.5 mm. So it could be reflected off
something at a depth of 0.75 mm. This is fine!
Frequency?
• Need to use a short wavelength to avoid the
wave spreading out from diffraction. Smallest
object a doctor might be interested in is
around a few mm, so wavelength needs to be
a bit less. If λ = 1mm and velocity = 1500 m.s-1
then f = 1500/10-3 = 1.5 MHz.
Frequency
• Higher frequencies would give higher
resolution but are absorbed easier (higher
attenuation). Operator adjusts frequency and
pulse length to get the best image for organs
of different depth and size.
Acoustic impedance
• When ultrasound is incident on a boundary
between two media (materials), part of the
wavefront is reflected and part refracted.
• http://www.falstad.com/ripple/
Acoustic impedance - Z
• The percentage reflected depends upon the
relative acoustic impedance of the two media.
• Acoustic impedance (Z) = ρc
where
ρ = density of medium
c = the velocity of the ultrasound
Unit of Z = kg.m-2.s-1
Acoustic impedance
• The greater the difference in acoustic
impedance, the greater the % reflection.
Acoustic impedance
• The greater the difference in acoustic
impedance, the greater the % reflection. The
difference between air and skin is great. To
prevent all of the ultrasound being reflected
by the skin, gel is used to fill the gap between
the transmitter and the skin.
A - scans
• A plot of strength of reflected beam against
time
Signal strength
Time
A - scans
probe
• The depth and
thickness of the
organ can be
deduced from
the times of the
reflected pulse
organ
tissue
gel
A - scans
probe
organ
• Notice the 2nd
pulse is smaller
due to
attenuation
tissue
gel
A - scans
probe
• The last pulse is
large as most of
the ultrasound
is reflected by
the last
boundary
organ
tissue
gel
A - scans
probe
• To gain more
information the
probe can be moved
up and down to
reveal the size, shape
and changes in
thickness of the
organ.
organ
tissue
gel
A - scans
probe
• To gain more
information the
probe can be moved
up and down to
reveal the size, shape
and changes in
thickness of the
organ.
organ
tissue
gel
B - scans
• An A scan gives information but not an image.
B – scans converts the signal into a dot whose
brightness corresponds to the strength of the
signal. By sweeping the probe across the
organ an image can be produced.
3-D ultrasound
• An array of probes moved around the patient
can be used to construct a 3D image using a
computer.
Questions!