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

Vedclass · Accepted Answer

$2.1 	imes 10^{-28} \, m/s$

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(A) According to Heisenberg's uncertainty principle: $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$.Since $\Delta p = m \cdot \Delta v$,the formula becomes $\Delta x \cdot m \cdot \Delta v = \frac{h}{4\pi}$.Given: $m = 25 \, g = 0.025 \, kg = 25 	imes 10^{-3} \, kg$,$\Delta x = 10^{-5} \, m$,$h = 6.6 	imes 10^{-34} \, J \cdot s$.Substituting the values: $\Delta v = \frac{h}{4 \pi \cdot \Delta x \cdot m}$.$\Delta v = \frac{6.6 	imes 10^{-34}}{4 	imes 3.14 	imes 10^{-5} 	imes 25 	imes 10^{-3}}$.$\Delta v = \frac{6.6 	imes 10^{-34}}{314 	imes 10^{-6}} \approx 2.1 	imes 10^{-28} \, m/s$.

$A$ particle of mass $25 \, g$ has an uncertainty in position of $10^{-5} \, m$. The uncertainty in velocity is ... (Planck's constant $h = 6.6 \times 10^{-34} \, J \cdot s$).

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