## Problem Restatement **Question:** The uncertainty in position of an electron is \( 3.25 \times 10^{-11} \) m. Find the minimum uncertainty in the electron's velocity. --- ## Step 1: Apply Heisenberg's Uncertainty Principle The Heisenberg Uncertainty Principle states: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where: - \( \Delta x \) = uncertainty in position - \( \Delta p \) = uncertainty in momentum - \( h \) = Planck’s constant (\(6.626 \times 10^{-34}\) Js) Since momentum \( p = m v \), we have \( \Delta p = m \cdot \Delta v \). The equation becomes: \[ \Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi} \] **Explanation:** We start by setting up the relationship between the uncertainties in position and velocity using Heisenberg’s principle. --- ## Step 2: Solve for Uncertainty in Velocity Rearrange for \( \Delta v \): \[ \Delta v \geq \frac{h}{4\pi m \Delta x} \] Plug in the values: - \( h = 6.626 \times 10^{-34} \) Js - \( m \) (electron mass) = \( 9.11 \times 10^{-31} \) kg - \( \Delta x = 3.25 \times 10^{-11} \) m \[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \cdot 9.11 \times 10^{-31} \cdot 3.25 \times 10^{-11}} \] Calculate the denominator: - \( 4\pi \approx 12.566 \) - \( 12.566 \times 9.11 \times 10^{-31} \times 3.25 \times 10^{-11} \approx 3.722 \times 10^{-40} \) Calculate \( \Delta v \): \[ \Delta v \geq \frac{6.626 \times 10^{-34}}{3.722 \times 10^{-40}} \approx 1.78 \times 10^{6} \text{ m/s} \] **Explanation:** After substituting all values and calculating, we find the minimum uncertainty in velocity. --- ## **Final Answer** \[ \boxed{1.78 \times 10^{6} \text{ m/s}} \] This is the minimum uncertainty in the electron's velocity.