Measuring Radiation Dosage — Question 5

Question 5

A moving detector accumulates an integral of the inverse-square dose rate

An ideal point source gives dose rate $\dot D(r)=\dot D_0(r_0/r)^2$ in a small detector. It moves radially outward with $r(t)=r_0+vt$, with $r_0=1.00\,\mathrm m$, $v=.100\,\mathrm{m/min}$ and $\dot D_0=12.0\,\mu\mathrm{Gy/min}$. Ignore attenuation and source decay.

1. Integrate dose over a travel time T.

2. Evaluate dose for T=10.0 min and compare it with remaining at $r_0$ for that interval.

Question 5 - Solution

A moving detector accumulates an integral of the inverse-square dose rate

(a) Method and calculation

$$D(T)=\dot D_0r_0^2\int_0^T\frac{dt}{(r_0+vt)^2}.$$$$\boxed{D(T)=\frac{\dot D_0r_0}{v}\left(1-\frac{r_0}{r_0+vT}\right)}.$$

(b) Interpretation and further calculation

$$\boxed{D(10)=\frac{12.0(1.00)}{.100}(1-1/2)=60.0\,\mu\mathrm{Gy}}.$$$$\boxed{D_{\mathrm{fixed}}=\dot D_0T=120\,\mu\mathrm{Gy},\quad D/D_{\mathrm{fixed}}=.500}.$$

Physical check

The point-source model applies when detector and source dimensions are small relative to their separation. The time average of an inverse-square rate is not generally its value at the mean distance.