$$r_n = \left(\frac{h^2}{4\pi^2me^2}\right)\times\frac{n^2}{Z} $$

Would I be able to calculate the radius of sodium ion ($\ce{Na+}$) with the help of the above Bohr atomic model formula?

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    $\begingroup$ This formula is for single electron systems, and $\ce{Na+}$ has 10 electrons. You could do $\ce{Na^10+}$, though. $\endgroup$ – Karsten Theis Feb 21 '19 at 17:55
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    $\begingroup$ The atom is the chemical equivalent of the three body gravity problem in physics. There is no exact solution only numerical ones when you have more than one electron. $\endgroup$ – MaxW Feb 21 '19 at 18:15
  • $\begingroup$ To follow-up on @MaxW's point, I believe the system is also chaotic, so even having a nice, closed form solution is kind of useless because very small changes to initial conditions can lead to wildly different behavior of the system. $\endgroup$ – Zhe Feb 21 '19 at 20:12

The hydrogen atom wavefunctions can be useful for multielectron atoms as a means of looking up their size (by means of a parameterization). The hydrogen atom wavefunctions form one possible basis for the definition of effective nuclear charges, see for instance here, where Hartree–Fock orbitals are used to compute effective charges:

$$ Z_{eff} = \frac{<r>_H}{<r>_Z}$$

Here $<r>_H$ and $<r>_Z$ are the mean hydrogenic and Hartree-Fock radii (for nuclear charge Z).

The mean hydrogenic radius for ground state hydrogen is related to the Bohr radius $a_0$ as $$<r>_H = \frac{3}{2}\frac{a_o}{Z}$$

Expressions for $<r>_H$ are available for other orbitals, allowing $<r>_Z$ to be computed from the tabulated values of $ Z_{eff}$ as

$$ <r>_Z = \frac{<r>_H}{Z_{eff}}$$

So using the hydrogen wavefunctions in this way is a little more complicated than using the equation for the Bohr model, but just the s-orbitals can already be useful to give you an idea about the size of atoms.

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