If uncertainty in position Delta x equals uncertainty in momentum Delta p, then the uncertainty in velocity Delta v is:

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NEET Chemistry QUESTION #7350

Question 1

If uncertainty in position \(\Delta x\) equals uncertainty in momentum \(\Delta p\), then the uncertainty in velocity \(\Delta v\) is:

\(\sqrt{\dfrac{h}{\pi}}\)
\(\dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}\)✔️
\(\sqrt{\dfrac{h}{2\pi}}\)
\(\dfrac{1}{m}\sqrt{\dfrac{h}{\pi}}\)

Correct Answer Explanation

Heisenberg's Uncertainty Principle: \(\Delta x \cdot \Delta p \geq \dfrac{h}{4\pi}\)

Given \(\Delta x = \Delta p\), so \((\Delta p)^2 = \dfrac{h}{4\pi}\)

\(\Delta p = \sqrt{\dfrac{h}{4\pi}} = \dfrac{1}{2}\sqrt{\dfrac{h}{\pi}}\)

Since \(\Delta p = m \cdot \Delta v\):
\(\Delta v = \dfrac{\Delta p}{m} = \dfrac{1}{2m}\sqrt{\dfrac{h}{\pi}}\)

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