1. **State the problem:** We have two radioactive samples with given amounts over time. Sample 1 starts at 800 mg and decreases to 400 mg after 1 month. Sample 2 starts at 1000 mg and decreases to 800 mg after 1 month. We need to determine if Sample 2 decreases more rapidly than Sample 1. 2. **Formula and concept:** Radioactive decay is modeled by exponential decay: $$A(t) = A_0 e^{-kt}$$ where $A_0$ is the initial amount, $k$ is the decay constant, and $t$ is time. 3. **Find decay constant $k$ for Sample 1:** $$400 = 800 e^{-k \times 1}$$ Divide both sides by 800: $$\frac{400}{800} = e^{-k}$$ $$\frac{1}{2} = e^{-k}$$ Taking natural logarithm: $$\ln\left(\frac{1}{2}\right) = -k$$ $$-\ln 2 = -k$$ $$k = \ln 2 \approx 0.693$$ 4. **Find decay constant $k$ for Sample 2:** $$800 = 1000 e^{-k \times 1}$$ Divide both sides by 1000: $$\frac{800}{1000} = e^{-k}$$ $$0.8 = e^{-k}$$ Taking natural logarithm: $$\ln(0.8) = -k$$ $$k = -\ln(0.8) \approx -(-0.2231) = 0.2231$$ 5. **Compare decay constants:** Sample 1 decay constant $k \approx 0.693$. Sample 2 decay constant $k \approx 0.2231$. 6. **Interpretation:** A larger decay constant means faster decay. Since Sample 1 has a larger $k$, it decays faster than Sample 2. 7. **Conclusion:** The claim that Sample 2 decreases more rapidly is incorrect. Sample 1 decreases more rapidly. **Final answer:** Sample 1 decays faster because its decay constant is larger ($k \approx 0.693$) compared to Sample 2 ($k \approx 0.2231$).