Measuring Radiation Dosage — Question 8

Question 8

Physical decay and biological removal combine into an effective half-life

In a one-compartment teaching model, a retained radionuclide is removed independently by physical decay and biological clearance. Its physical half-life is 8.00 d and clearance half-life is 4.00 d. The dose rate in a fixed target is proportional to retained activity and starts at .300 mGy/d.

1. Derive and calculate the effective half-life.

2. Find the integrated dose from t=0 to infinity under the stated constant-deposition model.

Question 8 - Solution

Physical decay and biological removal combine into an effective half-life

(a) Method and calculation

$$\dot N=-(\lambda_p+\lambda_b)N\Rightarrow\boxed{\lambda_{\mathrm{eff}}=\lambda_p+\lambda_b}.$$$$\boxed{T_{\mathrm{eff}}^{-1}=T_p^{-1}+T_b^{-1},\quad T_{\mathrm{eff}}=8/3=2.667\,\mathrm d}.$$

(b) Interpretation and further calculation

$$D_{\infty}=\int_0^\infty\dot D_0e^{-\lambda_{\mathrm{eff}}t}dt=\dot D_0/\lambda_{\mathrm{eff}}.$$$$\boxed{D_{\infty}=.300(8/3)/\ln2=1.154\,\mathrm{mGy}}.$$

Physical check

Clearance removes undecayed nuclei from the modeled compartment; it is not itself radioactive decay. Real internal dosimetry may require multiple compartments and time-dependent deposition.