Question

How can you determine the size of hydrogen atom using uncertainty principle potential energy of an electron $V=\frac{-e}{4\Pi E_{0}a}$, were 'a' is the distance of electron from the nucleus.

Answer 1

$\Delta x\times \Delta V=\frac{h}{4\Pi m}{V}_n=\frac{2.8\times {10}^6}{n}\Delta x\times \frac{2.8\times {10}^6}{n}=\frac{6.626\times {10}^{-34}}{4\Pi x}\Delta x\times V=\frac{h}{4\Pi m}\Delta x=\frac{h}{2\Pi \sqrt{\frac{m^2{V}^2}{2}}\times 2}\Delta x=\frac{h}{2\Pi \sqrt{mV}}\to \left(1\right)Ke=eV=e.\frac{e^2}{4\Pi E9}\to \left(11\right)\Delta x=\frac{h}{2\Pi \sqrt{m\frac{e^2}{4\Pi E9}}}\Delta x=\frac{h}{\sqrt{4m^2\frac{{me}^2}{4\Pi E9}}}$