CIE A Level Physics

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First teaching 2020

Last exams 2024

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24.2.3 Detecting Gamma-Rays from PET Scanning

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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 (γ-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

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

Detecting gamma rays with a PET scanner

Creating an Image from PET Scanning

  • The γ 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 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

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 = hf = mec2

  • The momentum p of the photon is given by

Calculating Energy of Gamma-Ray Photons equation 1

  • Where:
    • me = 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)

Worked example

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

(a) Calculate the energy released when a positron and an electron annihilate.

(b) Calculate the frequency of the γ-rays emitted.

(c) Calculate the momentum of one of the γ-rays.

Part (a)

Step 1: Write down the known quantities

    • Mass of an electron = mass of a positron, me = 9.11 × 10–31 kg
    • Total mass is equal to the mass of the electron and positron = 2me

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

E = mec2

Step 3: Substitute in values and calculate energy E

E = 2 × (9.11 × 10-31) × (3.0 × 108)2 = 1.6 × 10–13 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):

Calculating Energy of Gamma-Ray Photons Worked Example equation 1

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

E = hf

Step 3: Rearrange for frequency f, and calculate

Calculating Energy of Gamma-Ray Photons Worked Example equation 2

Part (c)

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

Calculating Energy of Gamma-Ray Photons Worked Example equation 3

Step 2: Substitute in values and calculate momentum, p

Calculating Energy of Gamma-Ray Photons Worked Example equation 4

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