Detecting Gamma-Rays from PET Scanning

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

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

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

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)

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.

Answer:

Part (a)

Step 1: Write down the known quantities

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

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

Step 3: Substitute in values and calculate energy E

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

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):

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

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

$E = h f$

Step 3: Rearrange for frequency f, and calculate

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

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

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

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

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