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.

Answer:

\[114.25\overset{{}^\circ }{\mathop{A}}\,\] 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 \[\lambda =\frac{hc}{108.8\,eV}=\frac{(6.63\times {{10}^{-34}})\times (3\times {{10}^{8}})}{108.8\times 1.6\times {{10}^{-19}}}m\]                         \[=11425\overset{\text{o}}{\mathop{\text{A}}}\,\]