Chemistry - Atomic Structure Question with Solution | TestHub

The Heisenberg uncertainty principle, $\Delta x\times \Delta p\ge \frac{h}{4\pi }$, where $\Delta x=$ Uncertainty in position, $\Delta p=$ Uncertainty in momentum and $h/4\pi =$ constant. As $\Delta x$ is same for electron and helium and $h/4\pi$ is a constant, therefore minimum uncertainty in the measurement of the momentum of the helium atom will be same as that of an electron which is $5.0\times 10^{-26}\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$.