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

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5.27 × 10^–30 m

1.05 × 10^–28m

1.05 × 10^–26 m

5.27 × 10^–28 m

answer is C.

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Detailed Solution

$∆ x . ∆ p = \frac{h}{4 π} 1 \times 10^{- 5} \times ∆ p = \frac{6.626 \times 10^{- 34}}{4 \times 3.14} ∆ p = 5.27 \times 10^{- 30} m$

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