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

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

Class 11 Chemistry Structure of Atom Uncertainty principle and Schrodinger wave equation

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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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A
$2.637 \times 10^{-23}$
B
$2.637 \times 10^{-24}$
C
$8.283 \times 10^{-23}$
D
$8.283 \times 10^{-24}$

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Uncertainty principle and Schrodinger wave equation

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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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