The uncertainty in momentum of an electron is 1 x 10^-5kg - m/s. The uncertainty in its position will be (h = 6.62x 10^-34 kg - m²/s)

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

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

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

$5 .25 \times 10^{- 28} m$

answer is C.

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Detailed Solution

$According to Δx \times Δp = \frac{h}{4 π}$
$Δx = \frac{h}{Δp \times 4 π} = \frac{6 .62 \times 10^{- 34}}{1 \times 10^{- 5} \times 4 \times 3 .14} = 5 .27 \times 10^{- 30} m$

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