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

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$\frac{1}{2 m} \sqrt{\frac{h}{π}}$

$\sqrt{\frac{h}{2 π}}$

$\frac{1}{m} \sqrt{\frac{h}{π}}$

$\sqrt{\frac{h}{π}}$

answer is C.

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Detailed Solution

$Given, ∆ x = ∆ P ∆ x . ∆ P = \frac{h}{4 π} {(∆ P)}^{2} = \frac{h}{4 π} {(m ∆ V)}^{2} = \frac{h}{4 π} ∆ V = \sqrt{\frac{h}{4 {πm}^{2}}} = \frac{1}{2 m} \sqrt{\frac{h}{π}}$

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