Which statement correctly defines the effective half-life in internal exposure?

• A. The time for activity to be eliminated by biological processes only
• B. The combination of biological and radiological half-lives
• C. The time for activity to be eliminated by radiological decay
• D. The time for activity to be eliminated by chemical processes

Answer: (B)

Effective half-life in internal exposure reflects how fast activity decreases due to both radioactive decay and biological elimination. When a radionuclide is inside the body, it decays physically at its own rate and is cleared biologically by metabolic and excretory processes. The overall disappearance rate is the sum of these two processes, so the combined effect produces a shorter half-life than either process alone. Mathematically, Te = (Tphys × Tbi) / (Tphys + Tbi), or Te = ln(2) / (ln(2)/Tphys + ln(2)/Tbi). This is why the statement that defines effective half-life as the combination of biological and radiological half-lives is the correct one. If you consider examples, a short physical half-life or a fast biological clearance will both pull Te down, illustrating why the mixed, not single, mechanism governs the internal case. The other options describe only one mechanism or misattribute the process, so they don’t capture the full picture for internal exposure.