What is the uncertainty product $\Delta v \cdot \Delta x$ for a particle of mass $1 \ mg$? What does this imply?
(N/A) According to Heisenberg's uncertainty principle,$\Delta x \cdot \Delta v \ge \frac{h}{4 \pi m}$.
Given mass $m = 1 \ mg = 10^{-6} \ kg$.
$\Delta x \cdot \Delta v \ge \frac{6.626 \times 10^{-34} \ J \cdot s}{4 \times 3.14 \times 10^{-6} \ kg} \approx 5.27 \times 10^{-29} \ m^2 \ s^{-1}$.
This value is extremely small and negligible,implying that for macroscopic objects like $1 \ mg$ or heavier,the uncertainty is not observable and classical mechanics is applicable.
Given mass $m = 1 \ mg = 10^{-6} \ kg$.
$\Delta x \cdot \Delta v \ge \frac{6.626 \times 10^{-34} \ J \cdot s}{4 \times 3.14 \times 10^{-6} \ kg} \approx 5.27 \times 10^{-29} \ m^2 \ s^{-1}$.
This value is extremely small and negligible,implying that for macroscopic objects like $1 \ mg$ or heavier,the uncertainty is not observable and classical mechanics is applicable.
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