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(C) According to Heisenberg's uncertainty principle:$ \Delta x \cdot \Delta p \geq \frac{h}{4 \pi} $Given that uncertainty in position $( \Delta x )$

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

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(C) According to Heisenberg's uncertainty principle:$ \Delta x \cdot \Delta p \geq \frac{h}{4 \pi} $Given that uncertainty in position $( \Delta x )$ and momentum $( \Delta p )$ are equal,i.e.,$ \Delta x = \Delta p $.Substituting this into the equation:$ (\Delta p)^{2} = \frac{h}{4 \pi} $Since $ \Delta p = m \cdot \Delta v $:$ (m \cdot \Delta v)^{2} = \frac{h}{4 \pi} $$ m^{2} \cdot (\Delta v)^{2} = \frac{h}{4 \pi} $$ (\Delta v)^{2} = \frac{h}{4 \pi m^{2}} $Taking the square root on both sides:$ \Delta v = \sqrt{\frac{h}{4 \pi m^{2}}} $$ \Delta v = \frac{1}{2m} \sqrt{\frac{h}{\pi}} $

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

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