The uncertainty in momentum of an electron is | vedclass

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Question

(A) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4 \pi}$.Given: $\Delta p = 1 	imes 10^{-5} \ kg \ m/s$ an

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$5.28 	imes 10^{-30} \ m$

Vedclass · Answer

(A) According to Heisenberg's uncertainty principle,$\Delta x 	imes \Delta p \geq \frac{h}{4 \pi}$.Given: $\Delta p = 1 	imes 10^{-5} \ kg \ m/s$ and $h = 6.63 	imes 10^{-34} \ Js$.Rearranging for uncertainty in position: $\Delta x = \frac{h}{4 \pi 	imes \Delta p}$.Substituting the values: $\Delta x = \frac{6.63 	imes 10^{-34}}{4 	imes 3.14159 	imes 10^{-5}}$.$\Delta x = \frac{6.63 	imes 10^{-34}}{12.566 	imes 10^{-5}} \approx 5.28 	imes 10^{-30} \ m$.

The uncertainty in momentum of an electron is $1 \times 10^{-5} \ kg \ m/s$. The uncertainty in its position will be $(h = 6.63 \times 10^{-34} \ Js)$.

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