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In an atom, an electron is moving with a speed of 500 m/s with an accuracy of 0.006%. certainty with which the position of the electron can be located is $(h = 6.6 5 10^{- 34} {kg m}^{2} s^{- 1} {, mass of electron, m}_{e} = 9.1 \times 10^{- 31} kg)$

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${6.10 × 10}^{-3} m$

${1.62 × 10}^{-4} m$

${3.89 × 10}^{-3} m$

${1.92 × 10}^{-3} m$

answer is D.

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Detailed Solution

$Δ v = \frac{0.006}{100} \times 500 m s^{- 1}$

$= 3 \times 10^{- 2} m s^{- 1}$

$Δ x \times m Δ v = \frac{h}{4 π}$

$∴ Δ x = \frac{h}{4 π m Δ v}$

$= \frac{6.6 \times 10^{- 34} k t g m^{2} s^{- 1}}{4 \times 3.14 \times 9.1 \times 10^{31} k g \times 3 \times 10^{2} m s^{- 1}}$

$= 1.92 \times 10^{- 3} m$

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