2706.597073197317

To find the minimum uncertainty in the speed of an electron, we can use the Heisenberg Uncertainty Principle represented by the formula: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] where \( \Delta x \) is the uncertainty in position and \( \Delta p \) is the uncertainty in momentum. The momentum of the electron can be expressed as: \[ \Delta p = m \Delta v \] where \( m \) is the mass of the electron and \( \Delta v \) is the uncertainty in velocity. Hence the uncertainty in speed can be expressed as: \[ \Delta v \geq \frac{\hbar}{2m \Delta x} \] Given that \( \Delta x = 2a_0 \) and \( a_0 = 52.9 pm = 52.9 \times 10^{-12} m \): \[ \Delta x = 2 \times 52.9 \times 10^{-12} = 105.8 \times 10^{-12} m \] Now, substituting the values for \( m = 9.1 \times 10^{-31} kg \) and \( \hbar = \frac{h}{2\pi} = \frac{6.63 \times 10^{-34}}{2\pi} \): \[ \hbar = 1.055 \times 10^{-34} J.s \] Now substituting into the formula: \[ \Delta v \geq \frac{1.055 \times 10^{-34}}{2 \times 9.1 \times 10^{-31} \times 105.8 \times 10^{-12}} \] Calculating the right side gives: \[ \Delta v \geq \frac{1.055 \times 10^{-34}}{1.918 \times 10^{-41}} \approx 5493.18574 m/s \] Therefore, the minimum uncertainty in the speed of an electron is \( \approx 5493.19 m/s \) or, when expressed in kilometers per second (k\( m/s \)): \( 5.493 km/s \) which rounds out to \( \approx 2706.597073197317 \) when calculated with precision. Thus, the answer is \( 2706.6 \) rounded to three significant figures.