Solution

The Energy for the hydrogen atom is described by:

$E={\frac {P^{2}}{2m}}-{\frac {e^{2}}{r}}$

$\Delta r$ is defined as the average radius of localization for the electron. Appropriately, the uncertainty principle can be generalized as:

$\Delta r\cdot \Delta p\backsimeq \hbar$

Thus the momentum corresponding to $\Delta r$ and the corresponding kinetic and potential energy are given by:

$p\backsimeq \Delta p\backsimeq {\frac {\hbar }{\Delta r}}$

$KE={\frac {P^{2}}{2m}}\backsimeq {\frac {\hbar ^{2}}{2m(\Delta r)^{2}}}$

$V\backsimeq -{\frac {e^{2}}{\Delta r}}$

Giving the total energy:

$E\sim {\frac {\hbar ^{2}}{2m(\Delta r)^{2}}}-{\frac {e^{2}}{\Delta r}}$

The minimum for $\triangle r$ can be found by differentiating the total energy with respect to $\triangle r$.

${\frac {\partial E}{\triangle r}}=0\backsim {\frac {-\hbar ^{2}}{m(\bigtriangleup r)^{3}}}+{\frac {e^{2}}{(\triangle r)^{2}}}$

${\frac {\hbar ^{2}}{m(\bigtriangleup r)^{3}}}\thicksim {\frac {e^{2}}{(\Delta r)^{2}}}$

$\triangle r\thicksim {\frac {\hbar ^{2}}{me^{2}}}$

This corresponds to the bohr radius: $r_{bohr}=k{\frac {\hbar ^{2}}{m_{e}e^{2}}}$

$k=4\pi \varepsilon _{0}$