# Step 1: Apply the Uncertainty Principle When an electron passes through a hole of diameter \( d = 13.7\,\mu\text{m} = 13.7 \times 10^{-6}\,\text{m} \), its position in the \( x \)-direction is known within about \( \Delta x \approx d \). By Heisenberg’s uncertainty principle: \[ \Delta x \, \Delta p_x \geq \frac{\hbar}{2} \] Given: - \( \Delta x = 13.7 \times 10^{-6}\,\text{m} \) - \( \hbar = 1.055 \times 10^{-34}\,\text{Js} \) So, \[ \Delta p_x \geq \frac{\hbar}{2 \Delta x} \] \[ \Delta p_x \geq \frac{1.055 \times 10^{-34}}{2 \times 13.7 \times 10^{-6}} \] \[ \Delta p_x \geq \frac{1.055 \times 10^{-34}}{2.74 \times 10^{-5}} \] \[ \Delta p_x \geq 3.85 \times 10^{-30}\,\text{kg m/s} \] **Explanation**: - The width of the hole sets the uncertainty in position (\( \Delta x \)). - The uncertainty principle gives us the minimum uncertainty in momentum (\( \Delta p_x \)). --- # Step 2: Convert Momentum Uncertainty to Velocity Range The electron's momentum in \( x \)-direction: \( p_x = m v_x \). So, \[ \Delta p_x = m \Delta v_x \] The mass of the electron \( m = 9.11 \times 10^{-31}\,\text{kg} \). \[ \Delta v_x = \frac{\Delta p_x}{m} \] \[ \Delta v_x = \frac{3.85 \times 10^{-30}}{9.11 \times 10^{-31}} \] \[ \Delta v_x \approx 4.23\,\text{m/s} \] Because the electron initially had \( v_x = \), after passing through the hole, its \( v_x \) is likely in the range: \[ -2.12\,\text{m/s} \leq v_x \leq +2.12\,\text{m/s} \] This range is \( \pm \Delta v_x/2 \), but commonly, the full uncertainty is taken as the range \( (-\Delta v_x/2, +\Delta v_x/2) \) or just \( \pm \Delta v_x \). Here, using the full calculated uncertainty as the likely spread: \[ -4.23\,\text{m/s} \leq v_x \leq +4.23\,\text{m/s} \] **Explanation**: - The range for \( v_x \) comes from the uncertainty in momentum, divided by the electron mass. - After passing through the hole, the electron’s \( v_x \) could be positive or negative within this range. --- # Final Answer **After passing through the hole, the electron’s \( v_x \) is likely in the range:** \[ -4.23\,\text{m/s} \leq v_x \leq +4.23\,\text{m/s} \] **This range is set by the uncertainty principle due to the position constraint of the hole.**