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.

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 \[\lambda =\frac{hc}{108.8\,eV}=\frac{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}{108.8\times 1.6\times {{10}^{-19}}}m\] \[=114.25\overset{{}^\circ }{\mathop{A}}\,\]