# When \( 1/d^2 \) breaks down - part 2: area source

**Published:** Mar 22, 2024

**Source:** Dr Chris Robbins, Grallator

#### When \( 1/d^2 \) breaks down - part 2: area source

##### Dr Chris Robbins, Grallator

A previous note has looked at the variation of source with distance for a line source (**part 1: line source**). This
note considers the same problem for an area source and might, for example, represent
the radiation field due to a radioactive liquid spill.

Consider a radioactive area source of radius \(R\) and activity of \(S_A\) per unit area. We seek
to evaluate the intensity at a point \(D\), a distance \(d\) along the vertical axis, as shown below.

Let \(dA\) be an infinitely small area element centred a distance \(r\) from the centre. The contribution to the total intensity at \(D\) will be given by the source from the area element modified by \(1/l^2\), the distance to the detector. \[ dS=\frac{S_A \space dA}{l^2} \] Using polar coordinates the area element is given by \(dA=r dr d\theta\), and the total intensity at \(D\) is giving by integrating all the contributions: \[ S=\int^{2\pi}_0 \int^R_0 \frac{S_A r \space dr d\theta}{l^2} \]

The integral over \(\theta\) is trivial. Changing the order of integration, evaluating the now inner integral, and using \(l^2 = d^2 + r^2\) gives \[ S=2\pi S_A \int^R_0 \frac{r \space dr }{d^2+r^2} \] To solve this the substitution \(u= {\left( d^2 + r^2 \right)}^\frac{1}{2}\) so that \[ \frac{du}{dr}=\frac{r}{{\left( d^2 + r^2 \right)}^\frac{1}{2}} \Rightarrow r \space dr = u \space du \Rightarrow \frac{r \space dr }{d^2+r^2}=\frac{u \space du }{u^2}= \frac{1 }{u}du \] to obtain \[ S=2\pi S_A \int^{{\left( d^2 + r^2 \right)}^\frac{1}{2}}_{d} \frac{1 }{u}du = 2\pi S_A \mathrm{ln} \left( \sqrt{1+\frac{R^2}{d^2}} \right) \]

Now for a sanity check! We know that are large distances the intensity of radiation should fall as \(1/d^2\). Can we obtain this from the above?

For \(d \gg R\) (a long distance from the area source compared with its size) a binomial expansion of the square root gives the following approximation to first order \[ \sqrt{1+\frac{R^2}{d^2}} \approx 1+ \frac{1}{2} \frac{R^2}{d^2} \] Now using a Maclaurin expansion for the log function to first order, \[ \mathrm{ln}(1+x) \approx x \Rightarrow \mathrm{ln} \left( \sqrt{1+\frac{R^2}{d^2}} \right) \approx \mathrm{ln} \left( 1+ \frac{1}{2} \frac{R^2}{d^2} \right) \approx \frac{1}{2} \frac{R^2}{d^2} \] so that \[ S = 2\pi S_A \mathrm{ln} \left( \sqrt{1+\frac{R^2}{d^2}} \right) \approx \frac{\pi R^2 S_A}{d^2} \] This expression gives the expected result; the total source strength is the source per unit area multiplied by the area, and its intensity falls as \(1/d^2\) . As with the line source, things break down close to the source and care must be taken when defining a reference point from which to use a \(1/d^2\) approximation.

To look at this we'll use: \[\quad R = 1 \quad \mathrm{arbitrary \space distance \space units} \] \[\quad S_A = 1 \quad \mathrm{arbitrary \space activity \space units} \] and \(d\) will range from 1 to 30 distance units. As with the line source, a reference measurement will be taken at 1 distance unit and 20 distance units.

The results of a \(1/d^2\) calculation using the above two reference points is shown in the plot below.

As with the line source the following is concluded

- Taking a reference measurement very near the source results in a \(1/d^2\) model under-predicting intensity at larger distances.
- Taking a reference measurement at a larger distance from the source results in a \(1/d^2\) model over-predicting intensity at smaller distances.