# Step 1: Apply Heisenberg Uncertainty Principle The Heisenberg Uncertainty Principle states: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] Given: - Uncertainty in momentum, \( \Delta p = 1.5 \times 10^{-11} \) kg·m/s - Planck’s constant, \( h = 6.626 \times 10^{-34} \) J·s Rearrange to solve for minimum uncertainty in position (\( \Delta x \)): \[ \Delta x \geq \frac{h}{4\pi \Delta p} \] ### Explanation This equation relates the uncertainties in position and momentum, setting a lower limit on how precisely we can know both. --- # Step 2: Plug in the Values \[ \Delta x \geq \frac{6.626 \times 10^{-34}}{4\pi \times 1.5 \times 10^{-11}} \] Calculate the denominator: \[ 4\pi \times 1.5 \times 10^{-11} \approx 18.85 \times 10^{-11} = 1.885 \times 10^{-10} \] Now substitute: \[ \Delta x \geq \frac{6.626 \times 10^{-34}}{1.885 \times 10^{-10}} \] \[ \Delta x \geq 3.516 \times 10^{-24} \text{ m} \] ### Explanation By substituting the given values, we find the minimum uncertainty in the position. --- # Final Answer **The minimum uncertainty in the position of the electron is:** \[ \boxed{3.52 \times 10^{-24} \text{ m}} \] (None of the provided options exactly match, but this is the correct calculation.)