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The uncertainity in the momentum of a particle is 3.31 × 10^–2 kgms^–1. The uncertainity in its position is (in metres)

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1.59 × 10^–33

0.4 × 10^–20

3.3 × 10^–24

0.33 × 10^–30

answer is A.

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Detailed Solution

$ΔX = \frac{h}{4 πΔP} = \frac{6.625 \times 10^{- 34}}{4 \times 3.14 \times 3.31 \times 10^{- 2}} = 1.59 \times 10^{- 33}$

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