If the uncertainty in momentum and uncertainty in the position of a particle are equal, then the uncertainty…

If the uncertainty in momentum and uncertainty in the position of a particle are equal, then the uncertainty in its velocity would be given by
  1. $\Delta v \geq \sqrt{\frac{h}{2 \pi}}$
  2. $\Delta v \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}}$
  3. $\Delta v \geq \sqrt{\frac{h}{\pi}}$
  4. $\Delta v \geq \frac{1}{m} \sqrt{\frac{h}{\pi}}$

Solution

$\begin{aligned} \text { } \Delta P=\Delta x & \text { (Given) } \\ \Delta x \cdot \Delta P \geq \frac{h}{4 \pi} & \text { (Uncertainty principle) } \\ \Delta P^2 & \geq \frac{h}{4 \pi} \\ (m \times \Delta v)^2 & \geq \frac{h}{4 \pi} \\ \Delta v & \geq \frac{1}{2 m} \sqrt{\frac{h}{\pi}} .\end{aligned}$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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