1. **State the problem:** We are given the radioactive decay function for Strontium-90 as $$A(t) = A_0 e^{-0.0244t}$$ where $A_0$ is the initial amount and $A(t)$ is the amount at time $t$ (in years). The initial amount is 500 grams. 2. **Decay rate:** The decay rate is the coefficient of $t$ in the exponent, which is 0.0244 per year. 3. **Calculate amount after 10 years:** Use the formula $$A(10) = 500 e^{-0.0244 \times 10}$$ Calculate the exponent: $$-0.0244 \times 10 = -0.244$$ So, $$A(10) = 500 e^{-0.244}$$ Using approximate value $e^{-0.244} \approx 0.7835$, $$A(10) \approx 500 \times 0.7835 = 391.75$$ grams. 4. **Find time when 400 grams remain:** Set $$A(t) = 400$$ So, $$400 = 500 e^{-0.0244 t}$$ Divide both sides by 500: $$\frac{400}{500} = e^{-0.0244 t}$$ Simplify fraction: $$\frac{\cancel{400}}{\cancel{500}} = \frac{4}{5} = 0.8 = e^{-0.0244 t}$$ Take natural logarithm on both sides: $$\ln(0.8) = -0.0244 t$$ Solve for $t$: $$t = \frac{\ln(0.8)}{-0.0244}$$ Calculate: $$\ln(0.8) \approx -0.2231$$ So, $$t = \frac{-0.2231}{-0.0244} \approx 9.14$$ years. 5. **Calculate half-life:** The half-life $t_{1/2}$ satisfies $$\frac{1}{2} A_0 = A_0 e^{-0.0244 t_{1/2}}$$ Divide both sides by $A_0$: $$\frac{1}{2} = e^{-0.0244 t_{1/2}}$$ Take natural logarithm: $$\ln\left(\frac{1}{2}\right) = -0.0244 t_{1/2}$$ Solve for $t_{1/2}$: $$t_{1/2} = \frac{\ln(\frac{1}{2})}{-0.0244}$$ Calculate: $$\ln(\frac{1}{2}) = \ln(0.5) \approx -0.6931$$ So, $$t_{1/2} = \frac{-0.6931}{-0.0244} \approx 28.4$$ years. **Final answers:** - Decay rate: 0.0244 per year - Amount after 10 years: approximately 391.75 grams - Time when 400 grams remain: approximately 9.14 years - Half-life: approximately 28.4 years