The accumulated radiation dose when moving up to a source
Published: Nov 27, 2023
Source: Dr Chris Robbins, Grallator
The accumulated dose when moving up to a source
Dr Chris Robbins, Grallator
Consider the situation where you move up to a radioactive source, perform some operations for a period of time, and then move away. What
total dose will be received?
Let's introduce a few definitions and assumptions.
\(\qquad R_r \qquad \) is the reference dose rate measured at a reference distance \(x_r\) from the source.
\(\qquad T_m \qquad \) is the time it takes to move up to the source.
\(\qquad T_w \qquad \) is the working time spent stationary near the source.
\(\qquad x_s \qquad \) is the starting distance away from the source (m).
\(\qquad x_w \qquad \) is the working distance from the source. (m)
It will be assumed that:
\(\qquad\) \(x_r = 1 \) m.
\(\qquad\) \(x_w < x_s\).
\(\qquad\) You move up to the source at constant speed.
\(\qquad\) You move away from the source at the same constant speed.
The easy bit is the dose received at a fixed distance near the source while working.
The dose rate at a distance \(x\) from the source is obtained from the reference dose rate at a distance \(x_r\) by
\[ R(x)=R_r \left (\frac{x_r}{x} \right )^2. \]
With the assumption \(x_r = 1 \) this simplifies to
\[ R(x)=\frac{R_r}{{x}^2}. \]
The total dose at this fixed distance is the simply given by \[ D = RT_w = \frac{R_r T_w}{{x_w}^2} \] (which is simply the dose received per unit time multiplied by the time).
The harder bit is the dose received when moving as \(x\) is constantly changing. The first thing to do is to find an expression that links the distance from the source to time. At time \(t = 0, x = x_s\), and at time \(t = T_m, x = x_w\). The [constant] speed of movement is
\[ s=\frac{x_s-x_w}{T_m} \]
and at any point in time \(0 \leq t \leq T_m, x\) is given by
\[ x=x_s-st \]
The dose received, \(\delta D\) in a small amount of time \(\delta t\) at time \(t\) is given in a similar way to that found while working by
multiplying dose rate by time as
\[ \delta D=R \cdot \delta t = \frac{R_r \cdot \delta t}{{x}^2}=\frac{R_r \cdot \delta t}{\left ( x_s-st \right )^2} \]
The total dose is found by letting \(\delta t \rightarrow 0\) and integrating.
\[ D = \int_{0}^{T_m} \frac{R_r}{\left ( x_s-st \right )^2}dt = R_r \int_{0}^{T_m} \frac{1}{\left ( x_s-st \right )^2}dt \]
To solve this use the substitution \(u=x_s-st\). This gives \[ \begin{align} \frac{du}{dt}&=-s \Rightarrow \frac{-1}{s}du=dt \\ \\ u(t=0) &= x_s \\ \\ u(t=T_m) &= x_s - sT_m = x_w \\ \\ \end{align} \]
So that
\[ \begin{align} D &= \frac{-R_r}{s} \int_{x_s}^{x_w} \frac{1}{u ^2}du \\ \\ &=\frac{R_r}{s} {\left [ \frac{1}{u} \right]_{x_s}^{x_w}} = \frac{R_r T_m}{x_s-x_w} {\left [ \frac{1}{u} \right]_{x_s}^{x_w}} \\ \\ &=\frac{R_r T_m}{x_s-x_w} {\left ( \frac{1}{x_w} - \frac{1}{x_s} \right )} \\ \\ &=\frac{R_r T_m}{x_s-x_w} {\left ( \frac{x_s-x_w}{x_s x_w} \right )} \\ \\ &=\frac{R_r T_m}{x_s x_w} \end{align} \]
We can now answer the original question:
Consider the situation where you move up to a radioactive source, perform some operations for a period of time, and then move away. What
total dose will be received?
\[ \text{Total dose} = \underbrace {\frac{R_r T_m}{x_s x_w} }_{\text{moving to}} + \underbrace {\frac{R_r T_w}{{x_w}^2} }_{\text{working}} + \underbrace {\frac{R_r T_m}{x_s x_w} }_{\text{moving away}} \]
As a last note, if \(x_r \neq 1\) then all the doses need scaling by \(x_r^2\) to give
\[ \text{Total dose} = \underbrace {\frac{R_r T_m x_r^2}{x_s x_w} }_{\text{moving to}} + \underbrace {\frac{R_r T_w x_r^2}{{x_w}^2} }_{\text{working}} + \underbrace {\frac{R_r T_m x_r^2}{x_s x_w} }_{\text{moving away}} \]
Or, tidying things up
\[ \begin{align} \text{Total } &=\frac{R_r x_r^2}{x_w} \left ( \underbrace {\frac{T_m}{x_s} }_{\text{moving to}} + \underbrace {\frac{T_w}{x_w} }_{\text{working}} + \underbrace {\frac{T_m}{x_s} }_{\text{moving away}} \right ) \\ \\ &=\frac{R_r x_r^2}{x_w} \left ( \frac{2 T_m}{x_s} + \frac{T_w}{x_w} \right ) \end{align} \]
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