Find the ratio of ionization energy of hydrogen-like lithium atom and Bohr's hydrogen atom. 

Energy of an electron in ground state of Bohr's hydrogen-like atom is given by,

         $\mathrm{E}=-\frac{13.6{\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{eV}$

where Z = atomic number of the atom.

The ionization energy of this atom is equal in magnitude to energy of ground state$={\mathrm{E}}_{\mathrm{\infty }}=13.6{\mathrm{Z}}^2$

$\Rightarrow \frac{{\left({\mathrm{E}}_{\mathrm{\infty }}\right)}_{\mathrm{Li}}}{{\left({\mathrm{E}}_{\mathrm{\infty }}\right)}_{\mathrm{H}}}=\frac{{\left({\mathrm{Z}}_{\mathrm{Li}}\right)}^2}{{\left({\mathrm{Z}}_{\mathrm{H}}\right)}^2}={\left(\frac{3}{1}\right)}^2=9$