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(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$ Since $\Delta p = m \cdot \Delta v$,we have $\Delt

Vedclass · Accepted Answer

$5.79 	imes 10^{-4} \ m$

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(C) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}$ Since $\Delta p = m \cdot \Delta v$,we have $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \pi}$ Substituting the values: $\Delta x \cdot (9.1 	imes 10^{-31} \ kg) \cdot (0.1 \ m/s) = \frac{6.626 	imes 10^{-34} \ J \cdot s}{4 	imes 3.14159}$ $\Delta x \cdot (9.1 	imes 10^{-32} \ kg \cdot m/s) = 5.274 	imes 10^{-35} \ kg \cdot m^2/s$ $\Delta x = \frac{5.274 	imes 10^{-35}}{9.1 	imes 10^{-32}} \approx 5.79 	imes 10^{-4} \ m$

If the uncertainty in velocity of electron $(\Delta v)$ is $0.1 \ m/s$,the uncertainty in its position $(\Delta x)$ is (given: $m_e = 9.1 \times 10^{-31} \ kg$)

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