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

Step 1 - First-Order Decay Equation For a first-order radioactive process, the ratio of initial to remaining activity is: $$\frac{A_0}{A_t} = e^{\lambda t} \implies \ln\left(\frac{A_0}{A_t}\right) = \lambda t$$ Step 2 - Calculate Decay Constant $\lambda$ Given: $A_0/A_t = 2.5$, $t = 7.0$ days. $$\lambda = \frac{\ln(2.5)}{t} = \frac{0.9163}{7.0\ \text{days}} = 0.1309\ \text{day}^{-1}$$ Step 3 - Calculate Half-Life $$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.1309\ \text{day}^{-1}} \approx \boxed{5.3\ \text{days}}$$ Step 4 - Evaluate Options - **(A) 10.58 days**: Double the correct value — arises from confusing factors in the formula. Incorrect. - **(B) 2.65 days**: Half the correct value. Incorrect. - **(C) 5.3 days**: Correct — matches the calculation of $t_{1/2} = \ln 2 / \lambda \approx 5.29$ days. **Correct.** - **(D) 4.2 days**: Incorrect distractor. $$\boxed{\text{Answer: C — 5.3 days}}$$