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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 × 10^–28 m

1.05 × 10^–26 m

5.27 × 10^–30 m

5.27 × 10^–28 m

answer is C.

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Detailed Solution

Δx.ΔP = h/4π

Δx = h/4π. 1/ΔP

$\large \ = \frac{{6.625 \times 1{0^{ - 34}}}}{{4 \times 3.14}} \times \frac{1}{{{{10}^{ - 5}}}} = 5.27 \times 1{0^{ - 30}}m\$

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