• # question_answer A doubly ionised lithium atom is hydrogen like with atomic number Z = 3. Find the wavelength of the radiation required to excite the electron in $L{{i}^{2+}}$ from the first to the third Bohr orbit. Given the ionisation energy of hydrogen atom as 13.6 eV.

The energy of nth orbit of a hydrogen like atom is given as,  ${{E}_{n}}=\frac{13.6{{Z}^{2}}}{{{n}^{2}}}$ Thus, for $L{{i}^{2+}}$ atom, as Z = 3, the electron energies for the first and third Bohr orbit are: For n = 1, ${{E}_{1}}=-\frac{13.6\times {{(3)}^{2}}}{{{1}^{2}}}ev$             $=-122.4eV$ For        n = 3, ${{E}_{3}}=-\frac{13.6\times {{(3)}^{2}}}{{{(3)}^{2}}}eV=-13.6\,eV$ Thus, the energy required to transfer an electron from ${{E}_{1}}$ level to ${{E}_{3}}$ level is,             $E={{E}_{3}}-{{E}_{1}}$             $=-13.6-(-122.4)=108.8eV$ Therefore, the radiation needed to cause this transition should have photons of this energy.             $hv=108.8\,eV$ The wavelength of this radiation is, $\frac{hc}{\lambda }=108.8\,eV$                       $[\because v=c/\lambda ]$             or $\lambda =\frac{hc}{108.8\,eV}=\frac{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}{108.8\times 1.6\times {{10}^{-19}}}m$ $=114.25\overset{{}^\circ }{\mathop{A}}\,$
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