In Heisenberg's uncertainty equation $\Delta x \times \Delta p \ge \frac{h}{4\pi}$,$\Delta p$ stands for:

The effect of the uncertainty principle is significant only for the motion of microscopic particles and is negligible for macroscopic particles. Justify the statement with the help of a suitable example.

Given below are two statements $:$
Statement $(I):$ It is impossible to specify simultaneously with arbitrary precision,both the linear momentum and the position of a particle.
Statement $(II) :$ If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron,then the uncertainty in the measurement of velocity is $\geq \sqrt{\frac{h}{4\pi}} \times \frac{1}{m}$ which simplifies to $\geq \frac{1}{2m} \sqrt{\frac{h}{\pi}}$. In the light of the above statements,choose the correct answer from the options given below $:$

The correct form of the $Schrodinger$ wave equation for an electron,as derived by $Erwin$ $Schrodinger$,is:

Which one of the following conditions is incorrect for a well-behaved wave function $(\psi)$?

The position of both an electron and a helium atom is known within $1.0 \, nm$ and the momentum of the electron is known within $50 \times 10^{-26} \, kg \, m \, s^{-1}$. The minimum uncertainty in the measurement of the momentum of the helium atom is