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The uncertainity in the momentum of a particle is $3.31 \times 10^{- 2}$ $kgm s^{- 1}$ . The uncertainity in its position is (in metres)

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$0.4 \times 10^{- 20}$

$3.3 \times 10^{- 24}$

$0.33 \times 10^{- 30}$

$1.59 \times 10^{- 33}$

answer is A.

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Detailed Solution

\large \Delta x = \frac{h}{{4\pi }} \times \frac{1}{{\Delta P}} ; \Delta x = \frac{{6.625 \times 1{0^{ - 34}}}}{{4 \times 3.14}} \times \frac{1}{{3.31 \times 1{0^{ - 2}}}}\

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