The uncertainty in the position of an electro | vedclass

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Question

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.
Given: $\Delta x = 10 \ \mathring{A} = 10

Vedclass · Accepted Answer

$5.79 	imes 10^{4} \ 	ext{m s}^{-1}$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta v \geq \frac{h}{4 \pi m}$.
Given: $\Delta x = 10 \ \mathring{A} = 10 	imes 10^{-10} \ 	ext{m} = 10^{-9} \ 	ext{m}$.
Mass of electron $m = 9.11 	imes 10^{-31} \ 	ext{kg}$.
Planck's constant $h = 6.626 	imes 10^{-34} \ 	ext{J s}$.
Substituting the values:
$\Delta v = \frac{h}{4 \pi m \Delta x} = \frac{6.626 	imes 10^{-34}}{4 	imes 3.1416 	imes 9.11 	imes 10^{-31} 	imes 10^{-9}}$.
$\Delta v = \frac{6.626 	imes 10^{-34}}{114.53 	imes 10^{-40}} \approx 0.5785 	imes 10^{6} \ 	ext{m s}^{-1} = 5.79 	imes 10^{4} \ 	ext{m s}^{-1}$.

The uncertainty in the position of an electron is $10 \ \mathring{A}$. Find the uncertainty in its velocity.

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