The minimum uncertainty in the speed of an el | vedclass

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Question

(A) According to Heisenberg's uncertainty principle:$\Delta x 	imes \Delta p_{x} \geq \frac{h}{4 \pi}$Given $\Delta x = 2 a_{0} = 2 	imes 52.9 	ime

Vedclass · Accepted Answer

$548$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle:$\Delta x 	imes \Delta p_{x} \geq \frac{h}{4 \pi}$Given $\Delta x = 2 a_{0} = 2 	imes 52.9 	imes 10^{-12} \ m = 105.8 	imes 10^{-12} \ m$.For minimum uncertainty,$\Delta x 	imes m \Delta v = \frac{h}{4 \pi}$.$\Delta v = \frac{h}{4 \pi 	imes m 	imes \Delta x}$Substituting the values:$\Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14159 	imes 9.1 	imes 10^{-31} 	imes 105.8 	imes 10^{-12}}$$\Delta v \approx 548273 \ m \ s^{-1} = 548.273 \ km \ s^{-1}$.Rounding to the nearest integer,the value is $548 \ km \ s^{-1}$.

The minimum uncertainty in the speed of an electron in a one-dimensional region of length $2 a_{0}$ (where $a_{0} = \text{Bohr radius} = 52.9 \ pm$) is $km \ s^{-1}$. (Given: Mass of electron $= 9.1 \times 10^{-31} \ kg$,Planck's constant $h = 6.63 \times 10^{-34} \ J \ s$)

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