The uncertainity in momentum of an electron is $1 \times 10^{- 5} k g . m / s$ . The uncertainity in its position will be ( $h$ = $6.62 \times 10^{- 34} k g . m / s$ )

1.05 X 10^-28 m

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1.05 X 10^-26 m

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5.27 X 10^-30 m

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5.27 X 10^-28 m

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Solution

The correct option is C
5.27 X 10^-30 m

$Δ x = \frac{h}{4 π Δ p}$

= $\frac{6.62 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 5}}$

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

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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)

The correct option is C 5.27 X 10-30 m Δx = h4πΔp = 6.62 × 10−344 × 3.14 × 10−5 = 5.27 × 10−30 m

The uncertainity in momentum of an electron is $1 \times 10^{- 5} k g . m / s$ . The uncertainity in its position will be ( $h$ = $6.62 \times 10^{- 34} k g . m / s$ )

The correct option is C
5.27 X 10^-30 m

$Δ x = \frac{h}{4 π Δ p}$

= $\frac{6.62 \times 10^{- 34}}{4 \times 3.14 \times 10^{- 5}}$

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