# When $$1/d^2$$ breaks down - part 2: area source

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.

Radiation is one of the important factors in evolution. It causes mutation, and some level of mutation is actually good for evolution

– David Grinspoon -