A drug undergoes first-order elimination with a half-life of 4 hours. What percentage of the drug will remain in the body after 12 hours of administration?

Explanation

## First-Order Elimination and Half-Life Calculation **Key Point:** In first-order kinetics, a constant **fraction** (not amount) of drug is eliminated per unit time. The half-life (t₁/₂) is the time required for plasma concentration to fall to 50% of its initial value. ### Calculation Method **Number of half-lives in 12 hours:** $$n = \frac{\text{Total time}}{t_{1/2}} = \frac{12 \text{ hours}}{4 \text{ hours}} = 3 \text{ half-lives}$$ **Remaining drug after each half-life:** - After 1st half-life (4 hrs): 50% remains - After 2nd half-life (8 hrs): 25% remains - After 3rd half-life (12 hrs): 12.5% remains ### General Formula $$C_t = C_0 \times \left(\frac{1}{2}\right)^n$$ Where: - $C_t$ = concentration at time t - $C_0$ = initial concentration - $n$ = number of half-lives **Mnemonic:** **HALF-LIFE HALVING** — Each half-life period cuts the remaining drug in half. 3 half-lives = 1/2 × 1/2 × 1/2 = 1/8 = 12.5%. **High-Yield:** This is a fundamental concept tested in every NEET PG exam. Always count the number of half-lives first, then apply the halving rule. **Clinical Pearl:** Understanding half-life is critical for dosing intervals. A drug with t₁/₂ = 4 hours should ideally be dosed every 4–8 hours to maintain therapeutic levels and avoid accumulation. [cite:KD Tripathi 8e Ch 5]