What is the minimum uncertainty in a helium atom's velocity ( Delta $v \mathrm{~min}$ ) if the position is known within $1.4 \operatorname{dot} \mathrm{A}$.

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∻Solution∻ ‖ 1 ⋮ **Identify the given data**: The position uncertainty $\Delta x$ is given as $1.4 \, ext{Å}$ (angstroms). ‖ ‖ 2 ⋮ **Recall Heisenberg's Uncertainty Principle**: The principle states that $\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$, where $\Delta p$ is the uncertainty in momentum and $\hbar$ is the reduced Planck's constant ($\hbar = \frac{h}{2\pi}$). ‖ ‖ 3 ⋮ **Express momentum uncertainty in terms of velocity**: Since momentum $p = m \cdot v$, the uncertainty in momentum $\Delta p = m \cdot \Delta v$. Therefore, $\Delta x \cdot m \cdot \Delta v \geq \frac{\hbar}{2}$. ‖ ‖ 4 ⋮ **Solve for velocity uncertainty**: Rearrange the equation to find $\Delta v$: $$\Delta v \geq \frac{\hbar}{2m \Delta x}$$ For a helium atom, the mass $m$ is approximately $4 imes 1.66 imes 10^{-27} \, ext{kg}$. ‖ ‖ 5 ⋮ **Substitute the values**: $$\Delta v \geq \frac{1.054 imes 10^{-34} \, ext{Js}}{2 imes 4 imes 1.66 imes 10^{-27} \, ext{kg} imes 1.4 imes 10^{-10} \, ext{m}}$$ Calculate the result to find the minimum uncertainty in velocity. ‖ ∻Answer∻ ⚹ The minimum uncertainty in the helium atom's velocity is approximately $1.4 imes 10^3 \, ext{m/s}$. ⚹ ∻Key Concept∻ ⚹ Heisenberg's Uncertainty Principle ⚹ ∻Explanation∻ ⚹ The principle states that the product of the uncertainties in position and momentum of a particle is always greater than or equal to $\frac{\hbar}{2}$. By knowing the position uncertainty, we can calculate the minimum uncertainty in velocity. ⚹◊What is the relationship between the uncertainty in velocity $(\Delta v)$ and the uncertainty in position $(\Delta x)$ for a particle in quantum mechanics?⍭ Generate me a similar question◊

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