# Problem Restatement **Question:** The uncertainty in position of an electron is \(1.5 \times 10^{-10} \, \text{m}\). Find the minimum uncertainty in the electron’s velocity. --- ## Step 1: Write Down the Relevant Equation We use **Heisenberg's Uncertainty Principle**: \[ \Delta x \cdot \Delta p \geq \frac{h}{4\pi} \] where: - \(\Delta x\) = uncertainty in position \(= 1.5 \times 10^{-10} \, \text{m}\) - \(\Delta p\) = uncertainty in momentum \(= m \Delta v\) - \(h\) = Planck's constant \(= 6.626 \times 10^{-34} \, \text{Js}\) So, \[ \Delta x \cdot m \Delta v \geq \frac{h}{4\pi} \] \[ \Delta v \geq \frac{h}{4\pi m \Delta x} \] **Explanation:** We express the uncertainty in velocity (\(\Delta v\)) using the uncertainty principle and the definition of momentum (\(p = mv\)). Plug in the values in the next step. --- ## Step 2: Plug in Values and Calculate Given: - \(h = 6.626 \times 10^{-34} \, \text{Js}\) - \(m\) (electron mass) \(= 9.1 \times 10^{-31} \, \text{kg}\) - \(\Delta x = 1.5 \times 10^{-10} \, \text{m}\) \[ \Delta v \geq \frac{6.626 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 1.5 \times 10^{-10}} \] Calculate the denominator: - \(4\pi \approx 12.566\) - \(9.1 \times 1.5 = 13.65\) - \(10^{-31} \times 10^{-10} = 10^{-41}\) - So, \(12.566 \times 13.65 = 171.544\) - Denominator: \(171.544 \times 10^{-41} = 1.71544 \times 10^{-39}\) Now, plug in the numbers: \[ \Delta v \geq \frac{6.626 \times 10^{-34}}{1.71544 \times 10^{-39}} = 3.86 \times 10^{5} \, \text{m/s} \] **Explanation:** Plug all known values into the formula and carefully handle the powers of ten. The result is the minimum uncertainty in the electron's velocity. --- # Final Answer \[ \boxed{3.86 \times 10^5 \; \text{m/s}} \] is the minimum uncertainty in the electron's velocity.