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The uncertainity in momentum of an electron is 1 × 10^–5 kg.m/s. The uncertainity in its position will be (h = 6.62 × 10^–34 kg.m/s)

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$1.05 \times 10^{- 28} m$

$1.05 \times 10^{- 26} m$

$5.27 \times 10^{- 30} m$

$5.27 \times 10^{- 28} m$

answer is C.

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Detailed Solution

$∆ P = 1 \times 10^{- 5} kg . m / s ∆ x = \frac{h}{4 π ∆ P} = \frac{6.62 \times 10^{- 34}}{4 \times 3.14 \times 1 \times 10^{- 5}} = 5.27 \times 10^{- 30} m$

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