 Solution

For Coulomb binding, the virial theorem gives a typical kinetic energy of the order of the binding energy. Taking $p_{\mathrm{typ}}\simeq \sqrt{2m_e|E_1|}$ and using the Heisenberg uncertainty relation estimate $r_{\mathrm{typ}}\sim \hslash /p_{\mathrm{typ}}$ givesThe Bohr model gives $r_1=a_0/Z=a_0/2=2.65\times 10^{-11}\mathrm{m}$ for a heavy helium nucleus, agreeing with the characteristic scale. Equivalently, minimizing the localization estimate $E(r)\sim {\hslash }^2/(2m_e{r}^2)-Z{e}^2/(4\pi {ϵ}_{0}r)$ gives that same scale.

This is an order-of-magnitude radius estimate, not an exact deduction of the cloud's expected value from the inequality alone. Numerical coefficients depend on which spread is used; $\Delta x\Delta p\ge \hslash /2$ does not assert $⟨r⟩=\hslash /p$. For the exact hydrogenic $1s$ state, $⟨r⟩=3a_0/(2Z)=3.97\times 10^{-11}\mathrm{m}$ at $Z=2$, one and a half times the Bohr orbital radius. An electron cloud has a radial distribution, whereas the Bohr model assigns a definite radius.