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(C) Given: $\Delta p = 1 	imes 10^{-3} \, g \, cm \, sec^{-1}$,$h = 6.626 	imes 10^{-27} \, erg \, sec$.According to Heisenberg's uncertainty princi

Vedclass · Accepted Answer

$0.527 	imes 10^{-24} \, cm$

Vedclass · Answer

(C) Given: $\Delta p = 1 	imes 10^{-3} \, g \, cm \, sec^{-1}$,$h = 6.626 	imes 10^{-27} \, erg \, sec$.According to Heisenberg's uncertainty principle:$\Delta x \cdot \Delta p \geqslant \frac{h}{4\pi}$$\Delta x \geqslant \frac{h}{4 \cdot \pi \cdot \Delta p}$Substituting the values:$\Delta x \geqslant \frac{6.626 	imes 10^{-27}}{4 	imes 3.1416 	imes 10^{-3}}$$\Delta x \geqslant \frac{6.626}{12.5664} 	imes 10^{-24} \, cm$$\Delta x \geqslant 0.527 	imes 10^{-24} \, cm$.

Calculate the uncertainty in the position of a particle when the uncertainty in momentum is $1 \times 10^{-3} \, g \, cm \, sec^{-1}$.

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