INGENIA

NMD-06

Well-counter geometry

ε = ½ (1 − h/√(h²+R²)). On-axis geometric efficiency of a circular well.

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CountingWell counter

Governing equation

ε=12(1−h/h2+R2)\varepsilon=\tfrac12\left(1-h/\sqrt{h^{2}+R^{2}}\right)

h∈[0.1,10]h \in \left[0.1, 10\right]

where

h
Source height (cm)
R
Well radius (cm)
\varepsilon
Geometric efficiency (—)

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.

NMD-06 · gauge
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①

Governing law — red

ε=12(1−h/h2+R2)\varepsilon=\tfrac12\left(1-h/\sqrt{h^{2}+R^{2}}\right)

h∈[0.1,10]h \in \left[0.1, 10\right]

h=0.1h = 0.1ε=0.467\varepsilon = 0.467

②

Domain of the variable — blue

hmin⁡=0.1,hmax⁡=10h_{\min} = 0.1,\quad h_{\max} = 10

h∈[0.1,10]h \in \left[0.1, 10\right]

h=0.1h = 0.1h=0.1h = 0.1

③

Response curve — green

ε=12(1−h/h2+R2)\varepsilon=\tfrac12\left(1-h/\sqrt{h^{2}+R^{2}}\right)

h∈[0.1,10]h \in \left[0.1, 10\right]

h=0.1h = 0.1ε=0.467\varepsilon = 0.467

④

Rate of change — amber

dεdh\frac{\mathrm{d}\varepsilon}{\mathrm{d}h}

h∈[0.1,10]h \in \left[0.1, 10\right]

h=0.162h = 0.162ε′=−0.327\varepsilon^{\prime} = -0.327

Lecture brief

Historical brief

MIRD schema, in-vivo decay, uptake fractions and PET coincidence turned unsealed sources into organ dose. The sheets compute activity, residence and well-counter geometry. This sheet (NMD-06 — Well-counter geometry) is the form associated with Well counter. Working symbols: hh, RR →\rightarrow ε\varepsilon. A NaI well almost 4π-wraps a small tube. Solid angle Ω/4π is the geometric ε.

Purpose

Purpose: compute ε\varepsilon from hh, RR in Nuclear medicine via ε=12(1−h/h2+R2)\varepsilon=\tfrac12\left(1-h/\sqrt{h^{2}+R^{2}}\right) ε = ½ (1 − h/√(h²+R²)). On-axis geometric efficiency of a circular well. Use it when a real nuclear medicine question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given h=2.000 cmh = 2.000\,\mathrm{cm}, R=1.500 cmR = 1.500\,\mathrm{cm}, the governing relation ε=12(1−h/h2+R2)\varepsilon=\tfrac12\left(1-h/\sqrt{h^{2}+R^{2}}\right) yields ε=0.1000 —\varepsilon = 0.1000\,\mathrm{—}. A crystal cup and a point source on axis. Move a slider: the numbers are this situation, not a canned story.

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Inputs

Outputs

  • Geometric efficiency \varepsilon0.1000 —
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Narration of this film

A crystal cup and a point source on axis.

A NaI well almost 4π-wraps a small tube. Solid angle Ω/4π is the geometric ε.

Reading speed

Watch on YouTube