The minimum uncertainty in the speed of an electron in an one dimensional region of length $2 a_{0}$ (Where $a_{0} =$ Bohr radius 52.9 pm) is __________ $k m s^{- 1} .$ (Given: Mass of electron = $9.1 \times 10^{- 31}$ kg, Planck’s constant $h = 6.63 \times 10^{- 34}$ Js)

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answer is 548.

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Detailed Solution

Heisenberg’s uncertainty principle $Δ x \times Δ p_{x} \underline{>} \frac{h}{4 π} \Rightarrow 2 a_{0} \times m Δ v_{x} = \frac{h}{4 π} (minimum)$

$\Rightarrow Δ v_{x} = \frac{h}{4 π} \times \frac{1}{2 a_{0}} \times \frac{1}{m} = \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 2 \times 52.9 \times 10^{- 12} \times 9.1 \times 10^{- 31}}$

$= 548273 m s^{- 1} = 548.273 k m s^{- 1} = 548 k m s^{- 1}$

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