The mass of an electron is 9.11 times 10^{-31 | vedclass

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(B) According to Heisenberg's uncertainty principle:$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$Since $\Delta p = m \cdot \Delta v$,the formula beco

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

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(B) According to Heisenberg's uncertainty principle:$\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$Since $\Delta p = m \cdot \Delta v$,the formula becomes:$\Delta x \cdot m \cdot \Delta v \geq \frac{h}{4 \pi}$$\Delta v \geq \frac{h}{4 \pi \cdot m \cdot \Delta x}$Given:$\Delta x = 0.1 \ \mathring{A} = 0.1 	imes 10^{-10} \ m = 10^{-11} \ m$$m = 9.11 	imes 10^{-31} \ kg$$h = 6.626 	imes 10^{-34} \ J \ s$$\pi = 3.14$Substituting the values:$\Delta v = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14 	imes 9.11 	imes 10^{-31} 	imes 10^{-11}}$$\Delta v = \frac{6.626 	imes 10^{-34}}{114.46 	imes 10^{-42}}$$\Delta v \approx 0.05789 	imes 10^8 \ m \ s^{-1} = 5.79 	imes 10^6 \ m \ s^{-1}$

The mass of an electron is $9.11 \times 10^{-31} \ kg$,and the Planck constant is $6.626 \times 10^{-34} \ J \ s$. The uncertainty involved in the measurement of velocity within a distance of $0.1 \ \mathring{A}$ is:

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