 Solution

Take the hydrogen ground-state binding-energy magnitude as $E_B\simeq 13.6\mathrm{eV}=2.18\times 10^{-18}\mathrm{J}$, and use $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$ as its characteristic radius. A definite volume convention isIt givesA quantum atom has no sharp outer surface. If instead its characteristic volume is taken as $a_0^3$, the estimate is $1.47\times 10^{13}\mathrm{J}{\mathrm{m}}^{-3}$ and the ratio is about $4.2\times 10^2$. Both conventions place the two densities roughly three orders of magnitude apart at order-of-magnitude accuracy. The comparison uses the magnitude of the binding energy, not its negative sign relative to ionization.