IMG-19

PET coincidence rate

R = 2 ε² A e^{−μd}. True coincidences through a pair of detectors.

Reading speed

PETPET coincidence

Governing equation

R=2\varepsilon^{2}A e^{-\mu d}

$A \in \left[10, 800\right]$

where

A: Activity (MBq)
\varepsilon: Detector efficiency (—)
\mu: Attenuation μ (1/cm)
d: Path through patient (cm)
R: True coincidence rate (kcps)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

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①

Governing law — red

R=2\varepsilon^{2}A e^{-\mu d}

$A \in \left[10, 800\right]$

$A = 10$ $R = 0.041$

Along the curve

②

Domain of the variable — blue

A_{\min} = 10,\quad A_{\max} = 800

$A \in \left[10, 800\right]$

$A = 10$ $A = 10$

Along the curve

③

Response curve — green

R=2\varepsilon^{2}A e^{-\mu d}

$A \in \left[10, 800\right]$

$A = 10$ $R = 0.041$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}R}{\mathrm{d}A}

$A \in \left[10, 800\right]$

$A = 14.97$ $R^{\prime} = 0.004$

Along the curve

Lecture brief

Historical brief

Larmor precession, CT Beer projections, SNR and Nyquist sampling are why MRI and CT images exist as numbers. The lab computes frequency, dose and resolution limits. This sheet (IMG-19 — PET coincidence rate) is the form associated with PET coincidence. Working symbols: $A$ , $\varepsilon$ , $\mu$ , $d$ $\rightarrow$ $R$ . A 511 keV pair. Randoms grow as singles² × 2τ. Scatter fraction is the other contaminant.

Purpose

Purpose: compute

R

from

A

\varepsilon

\mu

d

in Medical imaging via

R=2\varepsilon^{2}A e^{-\mu d}

R = 2 ε² A e^{−μd}. True coincidences through a pair of detectors. Use it when a real medical imaging question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

A = 200.000\,\mathrm{MBq}

\varepsilon = 0.150\,\mathrm{—}

\mu = 0.096\,\mathrm{1/cm}

d = 25.000\,\mathrm{cm}

, the governing relation

R=2\varepsilon^{2}A e^{-\mu d}

yields

R = 0.82\,\mathrm{kcps}

. Two opposed crystals and a line of response. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Activity A200 MBqDetector efficiency \varepsilon0.15 —Attenuation μ \mu0.096 1/cmPath through patient d25 cm

Outputs

True coincidence rate R0.82 kcps

Reading speed

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Narration of this film

Two opposed crystals and a line of response.

A 511 keV pair. Randoms grow as singles² × 2τ. Scatter fraction is the other contaminant.

Reading speed

Watch on YouTube