If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is (a) (1)/(m)√((h)/(π

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Q. If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is

(a)

\dfrac{1}{m}\sqrt{\dfrac{h}{\pi}}

(b)

\sqrt{\dfrac{h}{\pi}}

(c)

\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}

(d)

\dfrac{h}{4\pi}

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Step 1. At the minimum-uncertainty limit, $\Delta x\cdot\Delta p=\dfrac{h}{4\pi}$ . Given that $\Delta x=\Delta p$ (call this common value $x$ ), substitute:

x\cdot x=\frac{h}{4\pi}\qquad\Rightarrow\qquad x^2=\frac{h}{4\pi}\qquad\Rightarrow\qquad x=\sqrt{\frac{h}{4\pi}}=\frac12\sqrt{\frac{h}{\pi}}

Step 2. Since $\Delta p=m\,\Delta v$ , and $\Delta p=x$ from Step 1:

$\Delta v=\frac{x}{m}=\frac{1}{2m}\sqrt{\frac{h}{\pi}}$ …

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Q.If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is

Concept understanding — Heisenberg's Uncertainty Principle

Q.If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is

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