An electron is located within an interval of 0.211 nm in the north-south direction. What is the minimum uncertainty Au in the electron's velocity in that direction? The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing > Av= m/s
An electron is located within an interval of 0.211 nm in the north-south direction. What is the minimum uncertainty Au in the electron's velocity in that direction? The Heisenberg uncertainty relation is given different forms in different textbooks. Use the form employing > Av= m/s
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Question
![### Understanding the Heisenberg Uncertainty Principle
**Problem Statement:**
An electron is located within an interval of 0.211 nm in the north-south direction. What is the minimum uncertainty (∆v) in the electron's velocity in that direction?
**Concept Explanation:**
The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that states that it is impossible to simultaneously know both the position and the velocity of a particle with absolute certainty. This principle is expressed in different forms in various textbooks.
**Calculation Requirement:**
Use the form of the Heisenberg uncertainty relation employing:
\[
\Delta x \cdot \Delta p \geq \frac{h}{4\pi}
\]
Where:
- ∆x is the uncertainty in position (0.211 nm, as given)
- ∆p is the uncertainty in momentum
- h is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
**Solution:**
To find the uncertainty in velocity (∆v), use the relationship:
\[
\Delta p = m \cdot \Delta v
\]
This leads to:
\[
\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}
\]
Therefore,
\[
\Delta v \geq \frac{h}{4\pi \cdot m \cdot \Delta x}
\]
**Box for Solution:**
\[
\Delta v = \text{(Insert calculated result here)} \, \text{m/s}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76884ef0-0050-4a4d-a83f-5c6d84fc3fcd%2Feaa922e4-7f3a-4ec9-9eb1-fec2766cd176%2Fhlx7ctf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Understanding the Heisenberg Uncertainty Principle
**Problem Statement:**
An electron is located within an interval of 0.211 nm in the north-south direction. What is the minimum uncertainty (∆v) in the electron's velocity in that direction?
**Concept Explanation:**
The Heisenberg Uncertainty Principle is a fundamental theory in quantum mechanics that states that it is impossible to simultaneously know both the position and the velocity of a particle with absolute certainty. This principle is expressed in different forms in various textbooks.
**Calculation Requirement:**
Use the form of the Heisenberg uncertainty relation employing:
\[
\Delta x \cdot \Delta p \geq \frac{h}{4\pi}
\]
Where:
- ∆x is the uncertainty in position (0.211 nm, as given)
- ∆p is the uncertainty in momentum
- h is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\))
**Solution:**
To find the uncertainty in velocity (∆v), use the relationship:
\[
\Delta p = m \cdot \Delta v
\]
This leads to:
\[
\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4\pi}
\]
Therefore,
\[
\Delta v \geq \frac{h}{4\pi \cdot m \cdot \Delta x}
\]
**Box for Solution:**
\[
\Delta v = \text{(Insert calculated result here)} \, \text{m/s}
\]
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