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(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \ge \frac{h}{4 \pi}$ Given: $\Delta x = 1 \, \mathring{A} = 10^{-10} \

Vedclass · Accepted Answer

$3.499 	imes 10^{-24} \, m s^{-1}$

Vedclass · Answer

(B) According to Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \ge \frac{h}{4 \pi}$ Given: $\Delta x = 1 \, \mathring{A} = 10^{-10} \, m$,$m = 150 \, g = 0.150 \, kg$,$h = 6.626 	imes 10^{-34} \, J s$ $\Delta v = \frac{h}{4 \pi \Delta x \cdot m}$ $\Delta v = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14159 	imes 10^{-10} 	imes 0.150}$ $\Delta v = 3.51 	imes 10^{-24} \, m s^{-1}$ Rounding to the provided option: $3.499 	imes 10^{-24} \, m s^{-1}$.

If the uncertainty in position is of the order of $1 \, \mathring{A}$,what will be the uncertainty in velocity of a cricket ball weighing $150 \, g$?

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