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:
(A)- $-27.2\,eV$
(B)- $30.6\,eV$
(C)- $-30.6\,eV$
(D)- $27.2\,eV$

Answer 1

Hint: Atoms that are hydrogen-like, that is, with only one valence electron, the energy of the electron in the ${{n}^{th}}$ orbit is given by Bohr’s hydrogen atom model. According to which the electron revolves around the nucleus in fixed orbits and on excitation they jump to higher energy level or orbit.

Complete step by step solution:
It is given that the energy of the electron in the first Bohr H-atom is $-13.6\,eV$. We have the $L{{i}^{2+}}$ atom, which has atomic number, Z = 3 and one valence electron in its ground state, where n = 1. That is, similar to the hydrogen atom which has Z = 1.
So, for the energy of the electron in the ${{n}^{th}}$ orbit from Bohr’s model is:
${{E}_{n}}=\dfrac{-13.6{{Z}^{2}}}{{{n}^{2}}}$
Then, for the $L{{i}^{2+}}$ atom in its first excited energy state, with the electron in n = 2 orbit or shell, the energy of the excited state will be:
${{E}_{2}}=\dfrac{-13.6\times {{3}^{2}}}{{{2}^{2}}}=-30.6\,eV$

Therefore, the energy value of the electron in the excited state of $L{{i}^{2+}}$ is option (C)- $-30.6\,eV$.

Additional information:
There are few other atoms as well which are hydrogen-like such as $H{{e}^{+}}$, $B{{e}^{3+}}$, ${{B}^{4+}}$ and so on. All having only one electron, thus are isoelectronic to the hydrogen atom.
This energy required ($\Delta E$) to cause the excitation of the electron, can be calculated by the difference in the energy of the electron in the ground state and the excited state.

Note: It must be noted that the ground state of the atom is the lowest energy state in which the electron is present in its most stable state, whereas the excited state is when the electron jumps to the higher energy level on providing energy.