If 8 g of a radioactive isotope has a half-life of 10 h. The half-life of 2 g of the same substance — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Radioactive Decay Follows First-Order Kinetics Radioactive decay is a nuclear process governed by first-order kinetics. The half-life is: $$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$$ Step 2 - Half-Life Is an Intensive Property The formula $t_{1/2} = \ln 2 / \lambda$ contains no term for the initial mass or number of atoms $N_0$. Half-life depends solely on $\lambda$, the decay constant — an intrinsic property of the isotope. Therefore, half-life is an **intensive property**, completely independent of the amount of substance. Step 3 - Evaluate the Options - **(A) 2.5 h** — Incorrect. Assumes $t_{1/2} \propto m$ (linear reduction, which applies only to zero-order kinetics). - **(B) 5 h** — Incorrect. No physical or mathematical basis for this value. - **(C) 10 h** — Correct. Since half-life is independent of the starting mass, 2 g of the same isotope has the same half-life as 8 g. - **(D) 40 h** — Incorrect. Assumes $t_{1/2} \propto 1/m$ (inverse proportionality), which is also physically incorrect. $$\boxed{\text{Answer: C} \quad t_{1/2} = 10 \text{ h}}$$