Given: The mass of an electron is 9.1 times 1 | vedclass

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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$.Given: $\Delta x = 0.1 \ \mathring{A} = 0.1 	im

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$5.79 	imes 10^6 \ m \cdot s^{-1}$

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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$.Given: $\Delta x = 0.1 \ \mathring{A} = 0.1 	imes 10^{-10} \ m$,$m = 9.1 	imes 10^{-31} \ kg$,$h = 6.62 	imes 10^{-34} \ J \cdot s$.Substituting the values:$\Delta v = \frac{h}{4 \pi m \Delta x} = \frac{6.62 	imes 10^{-34}}{4 	imes 3.1416 	imes 9.1 	imes 10^{-31} 	imes 0.1 	imes 10^{-10}}$.$\Delta v = \frac{6.62 	imes 10^{-34}}{114.354 	imes 10^{-41}} \approx 5.79 	imes 10^6 \ m \cdot s^{-1}$.

Given: The mass of an electron is $9.1 \times 10^{-31} \ kg$,Planck's constant is $6.62 \times 10^{-34} \ J \cdot s$. The uncertainty involved in the measurement of velocity within a distance of $0.1 \ \mathring{A}$ is:

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