In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005%. Certainty to which the position of the electron can be located is $(h = 6.6 \times 10^{- 34} k g m^{2} s^{- 1},$ mass of an electron, $m_{e} = 9.1 \times 10^{- 31} k g)$

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$1.92 \times 10^{- 3} m$

$3.84 \times 10^{- 3} m$

$1.52 \times 10^{- 3} m$

$5.10 \times 10^{- 3} m$

answer is A.

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Detailed Solution

According to the uncertainty principle $Δ p Δ x \geq (h / 2 π)$ i.e. $(m Δ υ) Δ x \geq (h / 4 π)$

Hence, $Δ x \geq \frac{h}{4 π} \frac{1}{m (Δ υ)} = \frac{(6.6 \times 10^{- 34} k g m^{2} s^{- 1}}{4 (22 / 7) (9.1 \times 10^{- 31} k g) {(600 m s^{- 1}) (0.005 / 100)}} = 1.92 \times 10^{- 3} m$

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