What is the minimum uncertainty in a helium atoms velocity if the position is known within 1.1 A

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What is the minimum uncertainty in a helium atoms velocity if the position is known within 1.1 A
**Problem Statement:**

"What is the minimum uncertainty in a helium atom's velocity (\(\Delta v_{\text{min}}\)) if the position is known within 1.1 Å?"

**User Input:**

\[
\Delta v_{\text{min}} = 5.27 \times 10^5 \, \text{m/s} 
\]

**Feedback:**

- Incorrect  

**Explanation:**

This problem is addressing the concept of uncertainty as described by Heisenberg's Uncertainty Principle. The aim is to calculate the minimum uncertainty in the velocity of a helium atom when its position is known to an accuracy of 1.1 Ångströms. The user provided an answer, 5.27 x 10^5 m/s, which is marked as incorrect.

In order to find the correct answer, one would typically use the Heisenberg Uncertainty Principle formula:

\[
\Delta x \cdot \Delta p \geq \frac{\hbar}{2}
\]

Where:
- \(\Delta x\) is the uncertainty in position (1.1 Å)
- \(\Delta p\) is the uncertainty in momentum
- \(\hbar\) is the reduced Planck's constant \(\left(\frac{h}{2\pi}\right)\)

Since momentum (\(p\)) is the product of mass (\(m\)) and velocity (\(v\)), the formula becomes:

\[
\Delta x \cdot m\Delta v \geq \frac{\hbar}{2}
\]

Rearrange to solve for \(\Delta v\):

\[
\Delta v \geq \frac{\hbar}{2m\Delta x}
\]

To calculate \(\Delta v_{\text{min}}\), you'd need to know the mass of the helium atom and use the value for \(\hbar\).
Transcribed Image Text:**Problem Statement:** "What is the minimum uncertainty in a helium atom's velocity (\(\Delta v_{\text{min}}\)) if the position is known within 1.1 Å?" **User Input:** \[ \Delta v_{\text{min}} = 5.27 \times 10^5 \, \text{m/s} \] **Feedback:** - Incorrect **Explanation:** This problem is addressing the concept of uncertainty as described by Heisenberg's Uncertainty Principle. The aim is to calculate the minimum uncertainty in the velocity of a helium atom when its position is known to an accuracy of 1.1 Ångströms. The user provided an answer, 5.27 x 10^5 m/s, which is marked as incorrect. In order to find the correct answer, one would typically use the Heisenberg Uncertainty Principle formula: \[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \] Where: - \(\Delta x\) is the uncertainty in position (1.1 Å) - \(\Delta p\) is the uncertainty in momentum - \(\hbar\) is the reduced Planck's constant \(\left(\frac{h}{2\pi}\right)\) Since momentum (\(p\)) is the product of mass (\(m\)) and velocity (\(v\)), the formula becomes: \[ \Delta x \cdot m\Delta v \geq \frac{\hbar}{2} \] Rearrange to solve for \(\Delta v\): \[ \Delta v \geq \frac{\hbar}{2m\Delta x} \] To calculate \(\Delta v_{\text{min}}\), you'd need to know the mass of the helium atom and use the value for \(\hbar\).
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