1. **Problem Statement:** We are given the decay formula for polonium-210 (210 Po): $$A(t) = 0.7 \left(\frac{1}{2}\right)^{\frac{t}{138.4}}$$ where $A(t)$ is the amount of 210 Po (in mg) remaining after $t$ years, starting from 0.7 mg. 2. **Understanding the formula:** - The initial amount is 0.7 mg. - The base $\frac{1}{2}$ represents the half-life decay factor. - The exponent $\frac{t}{138.4}$ shows how many half-lives have passed, since the half-life of 210 Po is 138.4 years. 3. **Calculate the amount after $t=210$ years:** Substitute $t=210$ into the formula: $$A(210) = 0.7 \left(\frac{1}{2}\right)^{\frac{210}{138.4}}$$ 4. **Evaluate the exponent:** $$\frac{210}{138.4} \approx 1.517$$ 5. **Calculate the decay factor:** $$\left(\frac{1}{2}\right)^{1.517} = 2^{-1.517} = e^{-1.517 \ln 2}$$ Since $\ln 2 \approx 0.693$, then: $$e^{-1.517 \times 0.693} = e^{-1.051} \approx 0.349$$ 6. **Calculate the remaining amount:** $$A(210) = 0.7 \times 0.349 = 0.2443 \text{ mg}$$ **Answer:** After 210 years, approximately 0.244 mg of 210 Po remains in the sample.