If the position of the electron was measured | vedclass

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Question

(A) According to the Heisenberg uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4 \pi}$.Given,$\Delta x = 0.002 \ nm = 2 	imes 10^{-3}

Vedclass · Accepted Answer

$2.637 	imes 10^{-23}$

Vedclass · Answer

(A) According to the Heisenberg uncertainty principle,$\Delta x \cdot \Delta p \ge \frac{h}{4 \pi}$.Given,$\Delta x = 0.002 \ nm = 2 	imes 10^{-3} 	imes 10^{-9} \ m = 2 	imes 10^{-12} \ m$.Using $h = 6.626 	imes 10^{-34} \ J \ s$ and $\pi = 3.14$:$\Delta p = \frac{h}{4 \pi \cdot \Delta x} = \frac{6.626 	imes 10^{-34}}{4 	imes 3.14 	imes 2 	imes 10^{-12}}$.$\Delta p = \frac{6.626 	imes 10^{-34}}{25.12 	imes 10^{-12}} \approx 0.2637 	imes 10^{-22} \ kg \ ms^{-1}$.$\Delta p = 2.637 	imes 10^{-23} \ kg \ ms^{-1}$.

If the position of the electron was measured with an accuracy of $\pm 0.002 \ nm$,the uncertainty in the momentum of it would be (in $kg \ ms^{-1}$) $(h=6.626 \times 10^{-34} \ J \ s)$.

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