Gamma detectors along a path¶

Fluence rate at $\vec{r}$¶

Let $\vec{r}_{p} = (x_{p},y_{p},z_{p})$ denote the location of a point source $p$. Let $\vec{r}_{i} = (x_{i},y_{i},z_{i})$ denote an arbitrary point in space. The primary photon fluence rate at $\vec{r}$ is then given by

$$ \dot{\phi}(r) = \frac{A n_\gamma \exp(-\mu_{air} r)}{4\pi r^2} $$

where $r = ||\vec{r}_p - \vec{r}_i ||$. The units are $\dot{\phi} \sim \frac{\text{photons}}{s \cdot m^2}$

Count rate¶

Gamma detectors are not perfectly efficient and efficiency is dependent on both photon energy $E_\gamma$ and incident angle $\theta$ [1].

• the field efficiency $\varepsilon_D (E_\gamma) \in [0, 1]$, in units of area $\text{m}^2$,
• the relative angular efficiency $\varepsilon_\theta (E_\gamma, \theta) \in [0, 1]$, dimensionless.

The total efficiency of the detector is then defined as

$$ \varepsilon(E_\gamma, \theta) = \varepsilon_D (E_\gamma) \varepsilon_\theta (E_\gamma, \theta) \; . $$

Where $\varepsilon(E_\gamma, \theta) \sim \text{m}^2$.

If the detector $D$ is positioned at $\vec{r}_i$, the count rate becomes

$$ \dot{N}(r, E_\gamma, \theta) = \varepsilon(E_\gamma, \theta) \phi(r) $$

where $\dot{N} \sim \frac{\text{counts}}{s}$.

Acquisiton time¶

The acquisition time window $t_{w}$ is the time during which counts are accumulated in the detector until readout into the digital system. A typical $t_{w}$ in mobile gamma spectrometry is 1 to 10 seconds [2].

Integration of counts¶

Suppose an acquisition time of $t_{w}$ seconds and a fixed velocity $v$ in meters per seconds. Let $R(u)$ describe a road of $L$ meters long in the xy-plane, described as a function of arc length $u$ in meters (distance traveled along the road), where $u \in [0, L]$. The euclidian norm between the point $R(u)$ and point source $\vec{r}_p$ is then

$$ r(u) = || \vec{r}_p - R(u) || $$

Assuming a fixed velocity $v$, the distance traveled during one acquisition window $t_{w}$ is $\Delta_s \equiv vt_{w}$ meters. The path is divided into $K = L/\Delta s$ segments, where the $k$-th segment represents the interval

$$ u \in [(k-1) \Delta_s, k\Delta_s] \; , \; k = 1, 2, \dots, K $$

The total count rate acquired during segment $k$-th is then

$$ N_{w}(k) = \frac{1}{v} \int_{(k-1)\Delta_s}^{k\Delta_s} \underbrace{\dot{N}(r(u), E_\gamma, \theta(u))}_{\text{CPS}} du $$

Numerical approximation¶

Let us divide each segment into $N$ equally spaced points with step size $\Delta u = \Delta s / N$. Applying the trapezoidal rule then gives

$$ N_w(k) \approx \frac{\Delta u}{v} \left[ \frac{\dot{N}_0 + \dot{N}_N}{2} + \sum_{n=1}^{N-1} \dot{N}_n \right], $$

where

$$ \dot{N}_n = \dot{N}\big(r(u_n), E_\gamma, \theta(u_n)\big), \quad u_n = (k-1)\Delta s + n \Delta u. $$

References¶

[1] A. Bukartas, ‘Assessment of mobile radiometry data in radiological emergencies using Bayesian statistical methods’, thesis/doccomp, Lund University, 2021. Accessed: Jan. 19, 2026. [Online]. Available: http://lup.lub.lu.se/record/4c298e71-3278-42a7-818a-6f17a5121d56

[2] R. Finck, A. Bukartas, M. Jönsson, and C. Rääf, ‘Maximum detection distances for gamma emitting point sources in mobile gamma spectrometry’, Applied Radiation and Isotopes, vol. 184, p. 110195, Jun. 2022, doi: 10.1016/j.apradiso.2022.110195.