AQA A Level Physics

Revision Notes

10.6.3 The PET Scanner

Positron Emission Tomography

  • Positron Emission Tomography (PET) is defined as:

A type of nuclear medical procedure that images tissues and organs by measuring the metabolic activity of the cells of body tissues

  • In PET scanning, a beta-plus emitting radioactive tracer is used in order to stimulate positron-electron annihilation to produce gamma photons
    • These are then detected using a ring of gamma cameras

Principles of PET Scanning

Before the scan

  • The patient is injected with a beta-plus emitting isotope, usually fluorine-18 (F-18)

During the scan

  • The part of the body being studied is surrounded by a ring of gamma cameras
  • The positrons from the F-18 nuclei annihilate with electrons in the patient
  • The annihilation of a positron and an electron produces two identical gamma photons travelling in opposite directions
  • The delay time between these two gamma-ray photons is used to determine the location of the annihilation due to the F-18 tracer
  • Photons that do not arrive within a nanosecond of each other are ignored since they cannot have come from the same point

After the scan

  • The signals from the gamma camera detectors are sent to a computer which builds up an image

A PET Scanner

Detecting Gamma Rays, downloadable AS & A Level Physics revision notes

Detecting gamma rays with a PET scanner

Annihilation

  • Annihilation can happen between any particle and antimatter counterpart, but in PET scanning, only the electron and positron interaction is considered
  • When a positron is emitted from a tracer in the body, it travels less than a millimetre before annihilating with an atomic electron
  • As with all collisions, mass, energy and momentum are always conserved

An Electron-Positron Annihilation

The Process of Annihilation, downloadable AS & A Level Physics revision notes

Annihilation of a positron and electron to form two gamma-ray photons

  • The gamma-ray photons produced each have an energy of 512 keV
    • This is the same each time as determined by the mass-energy equivalence of the positron-electron pair
  • The energy of each photon is given by

E space equals space h f space equals space m subscript e c squared

  • Where:
    • m subscript e = mass of the electron or positron (kg)
    • h = Planck's constant (J s)
    • f = frequency of the photon (Hz)
    • c = the speed of light in a vacuum (m s1)

Diagnosis Using PET Scanning

  • The signals produced by the photomultiplier tubes are used to produce an image
  • The γ rays travel in straight lines in opposite directions when formed from the annihilation of the positron and electron
    • 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 γ 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

Annihilation in the PET Scanner

PET Scanning Machine (1), downloadable AS & A Level Physics revision notesPET Scanning Machine (2), downloadable AS & A Level Physics revision notes

Annihilation of a positron and an electron is the basis of PET Scanning

  • Once the tracer is introduced to the body it has a short half-life, so, it begins emitting positrons open parentheses beta to the power of plus close parentheses 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

Worked example

Fluorine-18 decays by β+ emission. The positron emitted collides with an electron and annihilates producing two γ-rays.

(a)
Calculate the total energy released when a positron and an electron annihilate.
(b)
Show that each gamma ray has an energy of 512 keV.
(c)
Calculate the frequency of the gamma rays.
 

Answer:

Part (a)

Step 1: Write down the known quantities

  • Mass of an electron = mass of a positron, m subscript e = 9.11 × 10–31 kg
  • Total mass = mass of the electron and positron = 2 m subscript e
  • Speed of light, c = 3.00 × 108 m s–1

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

E space equals space m subscript e c squared

Step 3: Substitute in values and calculate energy E

E space equals space 2 cross times open parentheses 9.11 cross times 10 to the power of negative 31 end exponent close parentheses cross times open parentheses 3.00 cross times 10 to the power of 8 close parentheses squared space equals space 1.64 cross times 10 to the power of negative 13 end exponent space straight J

Part (b)

Step 1: Determine the energy of one photon

  • Planck's constant, h = 6.63 × 10−34 J s
  • Two photons are produced, so, the energy of one photon is equal to half of the total energy from part (a):

E space equals space fraction numerator 1.64 cross times 10 to the power of negative 13 end exponent over denominator 2 end fraction space equals space 8.2 cross times 10 to the power of negative 14 end exponent space straight J

  • Note: if you keep the exact value for energy in your calculator, you would have E = 8.199 × 10–14 J

Step 2: Convert the energy in J into eV:

  • 1 eV = 1.6 × 10−19 J

E space equals space fraction numerator 8.199 cross times 10 to the power of negative 14 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction space equals space 512 space 438 space eV space equals space 512 space keV

Part (c)

Step 1: Write out the equation for the energy of a photon

E space equals space h f

Step 2: Rearrange for frequency f and calculate

f space equals space E over h space equals space fraction numerator 8.2 space cross times space 10 to the power of negative 14 end exponent over denominator 6.63 space cross times space 10 to the power of negative 34 end exponent end fraction space equals space 1.2 space cross times space 10 to the power of 20 space Hz

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