Given the following data:
Mass of electron: 9.11 × 10⁻³¹ kg
Planck's constant: 6.626 × 10⁻³⁴ Js
In light of th

Question

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:

Accepted Answer

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 ($\Delta x$) = $0.1 	ext{ \AA} = 0.1 	imes 10^{-10} 	ext{ m} = 10^{-11} 	ext{ m}$ .

2.  **Substitute Values:**
    *   $h = 6.626 	imes 10^{-34} 	ext{ Js}$
    *   $m = 9.11 	imes 10^{-31} 	ext{ kg}$
    *   $\pi \approx 3.14159$

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

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