Question

The energy of an electron in the first Bohr orbit of \[H - \]atom is $ - 13.6\,eV$. The energy value of electron in the excited state of $L{i^{2 + }}$ is:
(i) $ - 27.2\,eV$
(ii) $30.6\,eV$
(iii) $ - 30.6\,eV$
(iv) $27.2\,eV$

Answer 1

Hint:After the failure of Rutherford Bohr proposed the planetary model of the atom which pictures electrons orbiting the nucleus in the way that planets orbit the sun. Bohr used this planetary model to develop the first reasonable theory for hydrogen like atoms. In this model energy of the ${n^{th}}$orbit of a hydrogen-like atom is given by the formula: ${E_n}\, = \,{E_ \circ } \times \dfrac{{{Z^2}}}{{{n^2}}}\,\left( {n = 1,\,2,\,3,.....} \right)$.

Complete step-by-step answer:The Bohr model presented by Niels Bohr, is a system consisting of a small, dense nucleus surrounded by orbiting electrons similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity.
Bohr used this planetary model of atom to develop the first reasonable theory for hydrogen-like atoms, i.e. atoms having the same number of electrons like hydrogen.
Bohr correctly proposed that the energy and radii of the orbits of electrons in atoms are quantized, with energy for transitions between orbits given by
$\Delta E = h\nu \, = \,{E_f} - {E_i}$, where $\Delta E$ is the change in energy between the initial and final orbits and $h\nu $ is the energy of an absorbed or emitted photon.
Furthermore, the energies of the ${n^{th}}$orbit of hydrogen-like atoms are given by:
${E_n}\, = \,{E_ \circ } \times \dfrac{{{Z^2}}}{{{n^2}}}\,\left( {n = 1,\,2,\,3,.....} \right)$, where ${E_ \circ }$ correspond to the ground state energy and $Z$ is the atomic number of the atom.
Now, $Li$has an atomic number of $3$, hence when it forms $L{i^{2 + }}$ by losing two electrons it will have only one electron $($as atomic number is equal to the number of electrons present in the atom$)$ left hence it is a hydrogen-like atom.
Also the excited state of $L{i^{2 + }}$ will correspond to $n = 2$.
Hence the energy of excited state of $L{i^{2 + }}$ is, ${E_2}\, = \,{E_ \circ } \times \dfrac{{{3^2}}}{{{2^2}}}$.
Since energy of an electron in first Bohr orbit of \[H - \]atom is $ - 13.6\,eV$, i.e. ${E_ \circ } = - 13.6\,eV$.
Therefore, ${E_2}\, = \, - 13.6\,eV \times \dfrac{{{3^2}}}{{{2^2}}} = \, - 30.6\,eV$.

Hence the correct answer is (iii) $ - 30.6\,eV$.

Note: Many students think $n = 0$ corresponds to the ground state energy of a system and hence $n = 1$ gives the energy of the first excited state. But this is totally wrong as for $n = 0$the energy of the system becomes infinity which is not possible. Hence take a proper note of this fact and do calculations accordingly.