Question

The speed of a $1.0 g$ projectile is known to within $1 \times 10^{-6} ms ^{-1}$. Calculate the minimum uncertainty in its position.

Answer 1

SOLUTION : By Heisenberg's uncertainty principle $\Delta p \cdot \Delta x=\frac{h}{4 \pi}$

Given $\quad \Delta v=1 \times 10^{-6} m ; m=1 \cdot 0 g =10^{-3} kg .$

$\quad \Delta p=m \Delta v=10^{-3} \times 10^{-6}=10^{-9}$

Now, $\quad \Delta x=\frac{h}{4 \pi m \Delta v}=\frac{6 \cdot 626 \times 10^{-34}}{4 \times 3 \cdot 1416 \times 10^{-9}} m = 5 \cdot 2 7 \times 10^{-26} m$.

The uncertainty is completely negligible for all practical purposes concerning macroscopic objects.