Answer
$ \Delta p_x =1.0\times10^{-24}kg \cdot m/s$.
Work Step by Step
The Heisenberg uncertainty principle limits our knowledge of its momentum.
$$\Delta x \Delta p_x \geq \frac{\hbar}{2}$$
Find the minimum uncertainty in momentum.
$$ \Delta p_x =\frac{\hbar}{2 \Delta x}
=\frac{\hbar}{2 (0.529\times10^{-10}m)} $$
$$ \Delta p_x =1.0\times10^{-24}kg \cdot m/s$$
The magnitude of the momentum of the electron in the n=1 orbit is the same.
We see that the momentum’s magnitude is of the same order as the uncertainty in the momentum. The uncertainty principle plays a large role in atomic structure, and cannot be ignored.
