A radioactive sample has an initial activity of 28 dpm. Half hour later, the activity is 14 dpm. How — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Determine the Half-Life Activity drops from 28 dpm to 14 dpm (exactly half) in 30 minutes. $$\text{One half-life elapsed} \implies t_{1/2} = 30\ \text{min}$$ Step 2 - Calculate Decay Constant $$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.7}{30}\ \text{min}^{-1}$$ Step 3 - Find Initial Number of Atoms Using $A_0 = \lambda N_0$: $$N_0 = \frac{A_0}{\lambda} = \frac{28\ \text{dpm}}{\dfrac{0.7}{30}\ \text{min}^{-1}}$$ $$N_0 = \frac{28 \times 30}{0.7} = \frac{840}{0.7} = \boxed{1200\ \text{atoms}}$$ Step 4 - Evaluate Options - **(A) 1200**: Correct — matches our calculation. **Correct.** - **(B) 200**: Would give $A_0 = \frac{0.7}{30} \times 200 \approx 4.67$ dpm, not 28. Incorrect. - **(C) 600**: Would give $A_0 = \frac{0.7}{30} \times 600 = 14$ dpm (half the actual value). Incorrect. - **(D) 300**: Would give $A_0 = \frac{0.7}{30} \times 300 = 7$ dpm. Incorrect. $$\boxed{\text{Answer: A — 1200 atoms}}$$