The activity of a certain preparation decreases 2.5 times after 7.0 days. Find its half-life. — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

From $A = A_0 e^{-\lambda t}$, we have $A_0/A = e^{\lambda t} = 2.5$. Thus $\lambda \times 7.0 = \ln(2.5) \approx 0.916$. This gives $\lambda = 0.1308 \text{ day}^{-1}$. Half-life $t_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.1308} \approx 5.3 \text{ days}$. Therefore, correct answer is C.