Q.Calculate the uncertainty in position of an electron, if $\Delta v = 0.1\%$ and $v = 2.2\times10^{6}$ m s$^{-1}$.

Because a wave is inherently spread out rather than located at one precise point, the wave nature of matter imposes an unavoidable fundamental limit: it is impossible to simultaneously and precisely determine both the position and the momentum of a microscopic particle. Heisenberg expressed this quantitatively as

where $\Delta x$ and $\Delta p$ are the uncertainties in position and momentum respectively. Making one of these smaller (more precisely known) necessarily forces the other to grow larger - the product can never be pushed below $h/4\pi$. This limit is negligibly small compared to the momenta of everyday macroscopic objects, which is why classical mechanics works perfectly well for them, but it becomes physically significant for an electron: even pinning an electron's position in the first Bohr or …