Uncertainty in position is twice the uncertainty in momentum. Uncertainty in velocity is

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$\frac{1}{2 m} \sqrt{\frac{h}{π}}$

$\frac{1}{2 m} \sqrt{h}$

$\sqrt{\frac{h}{π}}$

$\frac{h}{4 π}$

answer is C.

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Detailed Solution

$\begin{array}{l} Given, Δx = 2 Δp \\ Δx . Δp = \frac{h}{2} = \frac{h}{4 π} \\ \Rightarrow 2 Δp . Δp = \frac{h}{2} \\ 2 {(mΔV)}^{2} = \frac{h}{2}; {(ΔV)}^{2} = \frac{h}{4 m^{2}} \\ \Rightarrow ΔV = \frac{\sqrt{h}}{2 m} . \end{array}$

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