Detecting Gamma-Rays from PET Scanning (Cambridge International A-Level Physics)

Calculating Energy of Gamma-Ray Photons

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

$E=hf=m_{e}c^{2}$

• The momentum of a photon is given by

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

• Where:

  • m_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 s^–1)

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.

Answer:

Part (a)

Step 1: Write down the known quantities

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

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

$E=m_{e}c^{2}$

Step 3: Substitute in values and calculate energy

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

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):

$E=\frac{1.6\times10^{-13}}{2}=0.8\times10^{-13}J$

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

$E=hf$

Step 3: Rearrange for frequency and calculate

$f=\frac{E}{h}=\frac{0.8\times10^{-13}}{6.63\times10^{-34}}=1.2\times10^{20}\mathrm{Hz}$

Part (c)

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

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

Step 2: Substitute in values and calculate momentum

$p=\frac{E}{c}=\frac{0.8\times10^{-13}}{3.0\times10^{8}}=2.7\times10^{-22}Ns$

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:

  • A crystal scintillator
  • A photomultiplier

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
• Therefore, they need to be amplified and converted to an electrical signal by a photomultiplier tube

PET scanner

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 is due to the conservation of 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 formed at 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