Given the following data: Mass of electron: 9.11 × 10⁻³¹ kg Planck's constant: 6.626 × 10⁻³⁴ Js In light of the data given above, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å will be:

According to the Heisenberg Uncertainty Principle, the product of the uncertainty in position ($\Delta x$) and the uncertainty in momentum ($\Delta p$) is given by: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ Since $\Delta p = m \cdot \Delta v$, the equation becomes: $\Delta x \cdot (m \cdot \Delta v) \ge \frac{h}{4\pi} \implies \Delta v \ge \frac{h}{4\pi \cdot m \cdot \Delta x}$

1. Convert Units:

• Distance ($\Delta x$) = $0.1 \text{ \AA} = 0.1 \times 10^{-10} \text{ m} = 10^{-11} \text{ m}$ .

2. Substitute Values: $h = 6.626 \times 10^{-34} \text{ Js}$ $m = 9.11 \times 10^{-31} \text{ kg}$

• $\pi \approx 3.14159$

3. Calculation: $\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14159 \times (9.11 \times 10^{-31}) \times 10^{-11}}$ $\Delta v = \frac{6.626 \times 10^{-34}}{114.48 \times 10^{-42}}$ $\Delta v = 0.05788 \times 10^8 \text{ ms}^{-1}$ $\Delta v = 5.79 \times 10^6 \text{ ms}^{-1}$