1. **Stating the problem:** Uranium decomposes at a rate proportional to the present amount. Initially, there are 55 grams, and after 18 years, 0.75% of the original amount has decomposed. We need to find: A) The half-life of uranium. B) The amount remaining after 650 years. 2. **Set up the exponential decay model:** The amount of uranium at time $t$ is given by $$ A(t) = A_0 e^{-kt} $$ where $A_0=55$ grams is the initial amount and $k$ is the decay constant. 3. **Find $k$ using the given data:** After 18 years, 0.75% decomposed, so 99.25% remains. Thus $$ A(18) = 0.9925 A_0 = 55 \times 0.9925 = 54.5875 $$ Substitute into the model: $$ 54.5875 = 55 e^{-18k} $$ Divide both sides by 55: $$ 0.9925 = e^{-18k} $$ Take the natural logarithm: $$ \ln(0.9925) = -18k $$ $$ k = - \frac{\ln(0.9925)}{18} $$ Calculate: $$ \ln(0.9925) \approx -0.007528 $$ So $$ k \approx -\frac{-0.007528}{18} = 0.00041822 $$ 4. **Find the half-life $T_{1/2}$:** The half-life is the time where the amount is half the original: $$ \frac{A_0}{2} = A_0 e^{-k T_{1/2}} $$ Simplify: $$ \frac{1}{2} = e^{-k T_{1/2}} $$ Take natural logarithm: $$ \ln\left(\frac{1}{2}\right) = -k T_{1/2} $$ $$ T_{1/2} = -\frac{\ln(1/2)}{k} = \frac{\ln 2}{k} $$ Calculate: $$ T_{1/2} = \frac{0.6931}{0.00041822} \approx 1657.4 \text{ years} $$ 5. **Find the amount remaining after 650 years:** $$ A(650) = 55 e^{-0.00041822 \times 650} $$ Calculate exponent: $$ -0.00041822 \times 650 = -0.271843 $$ Calculate: $$ A(650) = 55 e^{-0.271843} = 55 \times 0.76217 = 41.92 \text{ grams} $$ **Final answers:** A) Half-life of uranium is approximately $1657.4$ years. B) Amount remaining after 650 years is approximately $41.92$ grams.