• # 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 thrid Bohr orbit. Given the ionisation energy of hydrogen atom as 13.6 eV.

$114.25\overset{{}^\circ }{\mathop{A}}\,$ The energy of nth orbit of a hydrogen like atom is given as, ${{E}_{n}}=\frac{13.6{{Z}^{2}}}{{{n}^{2}}}$ Thus, for $\text{L}{{\text{i}}^{2+}}$ atom as $Z=3,$the electron energies for the first and third Bohr orbits are: For, $n=1,$      ${{E}_{1}}=-\frac{13.6\times {{(3)}^{2}}}{{{1}^{2}}}eV$                         $=-122.4\,\,eV$ 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, $E={{E}_{3}}-{{E}_{1}}$                         $=-136-(-122.4)=108.8\,eV$ Therefore, the radiation needed to cause this transition should have photons of this energy. $h\text{v=108}\text{.8}\,\,\text{eV}$ The wavelength of this radiation is, $\frac{hc}{\lambda }=10.8\,\,eV$ Or $\lambda =\frac{hc}{108.8\,eV}=\frac{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}{108.8\times 1.6\times {{10}^{-19}}}m$                         $=11425\overset{\text{o}}{\mathop{\text{A}}}\,$