When the uncertainty in the position of a moving particle is $0$,then the uncertainty in momentum $(p)$ is equal to:

The uncertainty in momentum of an electron is $1 \times 10^{-5} \ kg \ m/s$. The uncertainty in its position will be $(h = 6.63 \times 10^{-34} \ Js)$.

The uncertainty principle gave the concept of:

The uncertainties in the velocities of two particles,$A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times that of the mass of $A$. What is the ratio of uncertainties $\frac{\Delta x_A}{\Delta x_B}$ in their positions?

The uncertainty in the position of an electron $(mass = 9.1 \times 10^{-28} \ g)$ moving with a velocity of $3.0 \times 10^4 \ cm \ s^{-1}$ accurate up to $0.001\%$ will be ................. $cm$ (Use $\frac{h}{4\pi}$ in the uncertainty expression,where $h = 6.626 \times 10^{-27} \ erg \ s$)

What is the nature of the solution to the $Schrodinger$ wave equation for multi-electron atoms? How is it handled?