An electron is located within an interval of 0.187 nm in the north-south direction. What is the minimum uncertainty Av in the electron's velocity in that direction? The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing > Av= x10 TOOLS m/s
An electron is located within an interval of 0.187 nm in the north-south direction. What is the minimum uncertainty Av in the electron's velocity in that direction? The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing > Av= x10 TOOLS m/s
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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:
\[
h \approx 6.626 \times 10^{-34} \text{ J}\cdot\text{s}
\]
\[
m \approx 9.109 \times 10^{-31} \text{ kg}
\]
\[
\Delta x = 0.187 \times 10^{-9} \text{ m}
\]
\[
\Delta v \ge \frac{6.626 \times 10^{-34}}{4\pi \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe1657792-05f2-4f41-a2f4-a113dcf6b483%2F1d14fc4c-83be-42ba-ac23-07374e87f870%2Fg7j9xdn_processed.png&w=3840&q=75)
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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