Measuring Radiation Dosage — Question 7

Question 7

Source decay limits the integrated dose in a fixed geometry

In a fixed source-detector arrangement, dose rate initially is $\dot D_0=.800\,\mathrm{mGy/h}$ and follows the source decay with half-life 2.00 h. Geometry and absorption remain fixed.

1. Derive dose accumulated in time T and its infinite-time limit.

2. Calculate the dose over 6.00 h and the overestimate from assuming constant initial dose rate.

Question 7 - Solution

Source decay limits the integrated dose in a fixed geometry

(a) Method and calculation

$$D(T)=\int_0^T\dot D_0e^{-\lambda t}dt=\boxed{\frac{\dot D_0}{\lambda}(1-e^{-\lambda T})}.$$$$\boxed{D(\infty)=\dot D_0/\lambda=.800(2.00)/\ln2=2.308\,\mathrm{mGy}}.$$

(b) Interpretation and further calculation

$$\boxed{D(6)=D(\infty)(1-1/8)=2.020\,\mathrm{mGy}}.$$$$\boxed{D_{\mathrm{constant}}=.800(6)=4.800\,\mathrm{mGy},\quad D_{\mathrm{constant}}/D(6)=2.377}.$$

Physical check

The infinite-time total is finite because the integrated number of parent decays is finite. Time units for the rate and decay constant must agree.