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**Problem Statement:**

An electron is located within an interval of 0.187 nm in the north-south direction. What is the minimum uncertainty (\(\Delta v\)) in the electron's velocity in that direction?

**Given Information:**

The Heisenberg uncertainty relation is given in different forms in different textbooks. Use the form employing \(\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\).

**Approach:**

1. Start with the Heisenberg Uncertainty Principle equation:
   \[
   \Delta x \cdot \Delta p \ge \frac{h}{4\pi}
   \]
   where \(\Delta x\) is the uncertainty in position and \(\Delta p\) is the uncertainty in momentum.

2. Convert the uncertainty in position to meters:
   \[
   \Delta x = 0.187 \text{ nm} = 0.187 \times 10^{-9} \text{ m}
   \]

3. The uncertainty in momentum (\(\Delta p\)) is related to the uncertainty in velocity (\(\Delta v\)) by the equation:
   \[
   \Delta p = m \cdot \Delta v
   \]
   where \(m\) is the mass of the electron (approximately \(9.109 \times 10^{-31} \text{ kg}\)).

4. Substitute \(\Delta p\) into the Heisenberg equation:
   \[
   \Delta x \cdot (m \cdot \Delta v) \ge \frac{h}{4\pi}
   \]

5. Solve for \(\Delta v\):
   \[
   \Delta v \ge \frac{h}{4\pi \cdot \Delta x \cdot m}
   \]

**Calculation:**

Plug in the values:
   \[
   h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s}
   \]
   \[
   m \approx 9.109 \times 10^{-31} \text{ kg}
   \]
   \[
   \Delta x = 0.187 \times 10^{-9} \text{ m}
   \]

   \[
   \Delta v \ge \frac{6.626 \times 10^{-34}}{4\pi \
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Transcribed Image Text:**Problem Statement:** An electron is located within an interval of 0.187 nm in the north-south direction. What is the minimum uncertainty (\(\Delta v\)) in the electron's velocity in that direction? **Given Information:** The Heisenberg uncertainty relation is given in different forms in different textbooks. Use the form employing \(\Delta x \cdot \Delta p \ge \frac{h}{4\pi}\). **Approach:** 1. Start with the Heisenberg Uncertainty Principle equation: \[ \Delta x \cdot \Delta p \ge \frac{h}{4\pi} \] where \(\Delta x\) is the uncertainty in position and \(\Delta p\) is the uncertainty in momentum. 2. Convert the uncertainty in position to meters: \[ \Delta x = 0.187 \text{ nm} = 0.187 \times 10^{-9} \text{ m} \] 3. The uncertainty in momentum (\(\Delta p\)) is related to the uncertainty in velocity (\(\Delta v\)) by the equation: \[ \Delta p = m \cdot \Delta v \] where \(m\) is the mass of the electron (approximately \(9.109 \times 10^{-31} \text{ kg}\)). 4. Substitute \(\Delta p\) into the Heisenberg equation: \[ \Delta x \cdot (m \cdot \Delta v) \ge \frac{h}{4\pi} \] 5. Solve for \(\Delta v\): \[ \Delta v \ge \frac{h}{4\pi \cdot \Delta x \cdot m} \] **Calculation:** Plug in the values: \[ h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \] \[ m \approx 9.109 \times 10^{-31} \text{ kg} \] \[ \Delta x = 0.187 \times 10^{-9} \text{ m} \] \[ \Delta v \ge \frac{6.626 \times 10^{-34}}{4\pi \
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