If the uncertainty in velocity of a particle of mass 0.1 kg is 6.626 × 108 m.s–1, the uncertainty in its distance (in m) ish=6.626×10−34Jss

∆x × m∆v =h4π ∆x =h4π× m∆v ∆x =6.625 ×10-344π × 0.1× 6.626 × 108 ∆x =10-420.4π

If the uncertainty in velocity of a particle of mass 0.1 kg is 6.626 × 108 m.s–1, the uncertainty in its distance (in m) ish=6.626×10−34Jss

∆x × m∆v =h4π ∆x =h4π× m∆v ∆x =6.625 ×10-344π × 0.1× 6.626 × 108 ∆x =10-420.4π

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If the uncertainty in velocity of a particle of mass 0.1 kg is 6.626 × 10⁸ m.s^–1, the uncertainty in its distance (in m) is
$(h = 6 .626 \times 10^{- 34} {Js}_{s})$

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$\frac{1}{40 π} \times 10^{- 42}$

$\frac{1}{2 π} \times 10^{- 40}$

$\frac{1}{4 π} \times 10^{- 40}$

$\frac{1}{0 .4 π} \times 10^{- 42}$

answer is D.

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$∆ x \times m ∆ v = \frac{h}{4 π} ∆ x = \frac{h}{4 π \times m ∆ v} ∆ x = \frac{6.625 \times 10^{- 34}}{4 π \times 0.1 \times 6.626 \times 10^{8}} ∆ x = \frac{10^{- 42}}{0.4 π}$

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If the uncertainty in velocity of a particle of mass 0.1 kg is 6.626 × 108 m.s–1, the uncertainty in its distance (in m) ish=6.626×10−34Jss

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If the uncertainty in velocity of a particle of mass 0.1 kg is 6.626 × 10⁸ m.s^–1, the uncertainty in its distance (in m) is
$(h = 6 .626 \times 10^{- 34} {Js}_{s})$