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Uncertainty in the position of an electron (mass = 9.1 × 10^–31 kg) moving with a velocity 300ms^–1accurate upto 0.001% will be
$(h = 6 .63 \times 10^{- 34} Js)$

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$3.84 \times 10^{- 2} m$

$19.2 \times 10^{- 2} m$

$5.76 \times 10^{- 2} m$

$1.92 \times 10^{- 2} m$

answer is D.

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Detailed Solution

$∆ V = 300 \times \frac{0.001}{100} = 0.003 = 3 \times 10^{- 3} m / s ∆ x = \frac{h}{4 πm ∆ V} = \frac{6.625 \times 10^{- 34} J}{4 \times 3.14 \times 9.1 \times 10^{- 31} kg \times 3 \times 10^{- 3}} = 1.92 \times 10^{- 2} m$

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