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(C) According to Heisenberg's uncertainty principle,the relation is $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m 	imes \Delta v

Vedclass · Accepted Answer

$5.8 	imes 10^{-5} \ m^2 \ s^{-1}$

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(C) According to Heisenberg's uncertainty principle,the relation is $\Delta x 	imes \Delta p \geq \frac{h}{4\pi}$.Since $\Delta p = m 	imes \Delta v$,the equation becomes $\Delta x 	imes \Delta v 	imes m = \frac{h}{4\pi}$.Therefore,the product of uncertainties in position and velocity is $\Delta x 	imes \Delta v = \frac{h}{4 \pi 	imes m}$.Given: $h = 6.63 	imes 10^{-34} \ kg \ m^2 \ s^{-1}$,$m = 9.1 	imes 10^{-31} \ kg$,and $\pi = 3.14$.Substituting the values: $\Delta x 	imes \Delta v = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14 	imes 9.1 	imes 10^{-31}}$.$\Delta x 	imes \Delta v = \frac{6.63 	imes 10^{-34}}{114.296 	imes 10^{-31}} \approx 0.058 	imes 10^{-3} \ m^2 \ s^{-1} = 5.8 	imes 10^{-5} \ m^2 \ s^{-1}$.

According to Heisenberg's uncertainty principle,the product of uncertainties in position and velocity for an electron of mass $9.1 \times 10^{-31} \ kg$ is

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