The uncertainty in the location of a circulat | vedclass

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(B) Given that the uncertainty in position $\Delta x$ is equal to the de Broglie wavelength $\lambda$: $\Delta x = \lambda = \frac{h}{mv}$ According t

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$8$

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(B) Given that the uncertainty in position $\Delta x$ is equal to the de Broglie wavelength $\lambda$: $\Delta x = \lambda = \frac{h}{mv}$ According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}$ Substituting $\Delta x = \frac{h}{mv}$ into the uncertainty equation: $(\frac{h}{mv}) \cdot m \Delta v \geq \frac{h}{4\pi}$ Simplifying the expression: $\frac{\Delta v}{v} \geq \frac{1}{4\pi}$ The percentage error in velocity is given by: $\frac{\Delta v}{v} 	imes 100 \geq \frac{100}{4\pi} \approx \frac{100}{12.56} \approx 7.96 \% \approx 8 \%$

The uncertainty in the location of a circulating electron is equal to its de Broglie wavelength. The minimum percent error in its measurement of velocity under these circumstances will be approximately:

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