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The minimum uncertainty in the speed of an electron in an one dimensional region of length $2 a_{0}$ $(W h e r e a_{0} = B o h r r a d i u s 52.9 p m)$ is ___________ $k m s^{- 1}$
(Given mass of electron $= 9.1 \times 10^{- 31} k g, P l a n c k' s c o n s \tan t h = 6.63 \times 10^{- 34} J s$

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answer is 548.

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Detailed Solution

Heisenberg’ uncertainty principle

$Δ x \times Δ p_{x} \geq \frac{h}{4 π}$

$\Rightarrow 2 a_{0} \times m Δ v_{x} = \frac{h}{4 π} (\min i m u m)$

$\Rightarrow Δ v_{x} = \frac{h}{4 π} \times \frac{1}{2 a_{0}} \times \frac{1}{m}$

$= \frac{6.63 \times 10^{- 34}}{4 \times 3.14 \times 2 \times 52.9 \times 10^{- 12} \times 9.1 \times 10^{- 31}}$

$= 548273 m s^{- 1}$

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

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

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