The uncertainty in the position of a particle | vedclass

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Question

(B) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge \frac{h}{4\pi}$ Since $\Delta p = m 	imes \Delta v$,the equation b

Vedclass · Accepted Answer

$2.1 	imes 10^{-29}$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle: $\Delta x 	imes \Delta p \ge \frac{h}{4\pi}$ Since $\Delta p = m 	imes \Delta v$,the equation becomes: $\Delta x 	imes m 	imes \Delta v = \frac{h}{4\pi}$ Given: $m = 0.25 \, kg$,$\Delta x = 10^{-5} \, m$,$h = 6.626 	imes 10^{-34} \, J \cdot s$ $\Delta v = \frac{h}{4 	imes \pi 	imes \Delta x 	imes m}$ $\Delta v = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14159 	imes 10^{-5} 	imes 0.25}$ $\Delta v = \frac{6.626 	imes 10^{-34}}{3.14159 	imes 10^{-5}} \approx 2.1 	imes 10^{-29} \, m/s$

The uncertainty in the position of a particle of mass $0.25 \, kg$ is $10^{-5} \, m$. The uncertainty in its velocity is ...... $m/s$.

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