Answer:
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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