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,

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

where Z = atomic number of the atom.
The ionization energy of this atom is equal in magnitude to energy of ground state$=E_{\mathrm{\infty }}=13.6{\mathrm{Z}}^2$

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