Home / CIE AS & A Level Physics 9702: Topic 24: Medical physics- Unit : 24.3 PET scanning Study Notes

CIE AS & A Level Physics 9702: Topic 24: Medical physics- Unit : 24.3 PET scanning Study Notes

RADIOACTIVE TRACERS

  • A radioactive tracer is defined as:
    A radioactive substance that can be absorbed by tissue in order to study the structure and function of organs in the body
  •  Radioactive isotopes, such as technetium-99m or fluorine-18, are suitable for this purpose because:
    •  They both bind to organic molecules, such as glucose or water, which are readily available in the body
    •  They both emit gamma $(\gamma)$ radiation and decay into stable isotopes
    • Technetium-99m has a short half-life of 6 hours (it is a short-lived form of Technetium-99)
    • Fluorine-18 has an even shorter half-life of 110 minutes, so the patient is exposed to radiation for a shorter time

Using Tracers in PET Scanning

  •  Positron Emission Tomography (PET) is:
                     A type of nuclear medical procedure that images tissues and organs by measuring the metabolic activity of the cells of body tissues
  • A common tracer used in PET scanning is a glucose molecule with radioactive fluorine attached called fluorodeoxyglucose
    • The fluorine nuclei undergoes $\boldsymbol{\beta}^{+}$decay – emitting a positron ( $\beta^{+}$particle)
  • The radioactive tracer is injected or swallowed into the patient and flows around the body
  • Once the tissues and organs have absorbed the tracer, then they appear on the screen as a bright area for a diagnosis
    •  This allows doctors to determine the progress of a disease and how effective any treatments have been
  • Tracers are used not only for the diagnosis of cancer but also for the heart and detecting areas of decreased blood flow and brain injuries, including Alzheimer’s and dementia

Worked example: Positron emission of fluorine

(2) Write a nuclear decay for the decay of fluorine-18 by $\beta^{+}$emission.

Answer/Explanation

Step 1:

Work out what will be on the reactants and products side
Reactant:
Fluorine $={ }_9^{18} F$
Products:
Beta-plus $\beta+($ positron $)={ }_1^0 \beta$
Oxygen $={ }_8^{18} 0$
Gamma-ray $=\gamma$

Step 2:
Write the nuclear equation

$
{ }_9^{18} F \rightarrow{ }_8^{18} O+{ }_1^0 \beta+\gamma
$

 ANNIHILATION IN PET SCANNING

The Process of Annihilation

  • When a positron is emitted from a tracer in the body, it travels less than a millimetre before it collides with an electron
  • The position and the electron will annihilate, and their mass becomes pure energy in the form of two gamma rays which move apart in opposite directions
  • Annihilation doesn’t just happen with electrons and positrons, annihilation is defined as:
    When a particle meets its equivalent antiparticle they are both destroyed and their mass is converted into energy
  •  As with all collisions, the mass, energy and momentum are conserved
  • Positron Emission Tomography (PET) Scanning
  •  Once the tracer is introduced to the body it has a short half-life, so, it begins emitting positrons $\left(\beta^{+}\right)$immediately
    • This allows for a short exposure time to the radiation
    • A short half-life does mean the patient needs to be scanned quickly and not all hospitals have access to expensive PET scanners
  • In PET scanning:
    • Positrons are emitted by the decay of the tracer
    • They travel a small distance and annihilate when they interact with electrons in the tissue
    • This annihilation produces a pair of gamma-ray photons which travel in opposite directions

Detecting Gamma-Rays from PET Scanning

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

Creating an Image from PET Scanning

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

Calculating Energy of Gamma-Ray Photons

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

$
E=h f=m_e \mathbf{c}^2
$

  •  The momentum $p$ of the photon is given by

$
p=\frac{E}{c}
$

  •  Where:
    • $\mathrm{m}_{\mathrm{e}}=$ mass of the electron or positron $(\mathrm{kg})$
    • $\mathrm{h}=$ Planck’s constant $(\mathrm{J}$ s)
    • $f=$ frequency of the photon $(\mathrm{Hz})$
    • $\mathrm{c}=$ the speed of light in a vacuum $\left(\mathrm{m} \mathrm{s}^{-1}\right)$

Worked example: Calculating energy of gamma-ray photons

Fluorine-18 decays by $\beta^{+}$emission. The positron emitted collides with an electron and annihilates producing two $y$-rays.
a) Calculate the energy released when a positron and an electron annihilate.
b) Calculate the frequency of the $y$-rays emitted.
c) Calculate the momentum of one of the $y$-rays.
Mass of an electron $=$ mass of a positron $=9.11 \times 10^{-31} \mathrm{~kg}$.

Answer/Explanation

Part (a)
Step 1: $\quad$ Write down the known quantities
Total mass is equal to the mass of the electron and positron:

$
m=2 \times\left(9.11 \times 10^{-31}\right)=1.822 \times 10^{-30} \mathrm{~kg}
$

Step 2:
Write out the equation for mass-energy equivalence

$
E=m_e c^2
$

Step 3:
Substitute in values and calculate energy $\mathrm{E}$

$
E=2 \times\left(9.11 \times 10^{-31}\right) \times\left(3.0 \times 10^8\right)^2=1.6 \times 10^{-13} \mathrm{~J}
$

Part (b)

Step 1: $\quad$ Write down the known quantities
Two photons are produced, so, the energy of one photon is equal to half of the total energy:

$
E=\frac{1.6 \times 10^{-13}}{2}=0.8 \times 10^{-13} \mathrm{~J}
$

Step 2:
Write out the equation for energy of a photon

$
E=h f
$

Step 3:
Rearrange for frequency $f$, and calculate

$
\mathrm{f}=\frac{E}{h}=\frac{0.8 \times 10^{-13}}{6.63 \times 10^{-34}}=1.2 \times 10^{20} \mathrm{~Hz}
$

Part (c)
Step 1: Write out the equation for momentum of a photon

$
\mathrm{p}=\frac{E}{c}
$

Step 2: Substitute in values and calculate momentum $p$

$
\mathrm{p}=\frac{E}{c}=\frac{0.8 \times 10^{-13}}{3.0 \times 10^8}=2.7 \times 10^{-22} \mathrm{~N} \mathrm{~s}
$

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