Question

The energy of an electron in the first Bohr order of H atom is ${\text{ - 13}}{\text{.6 eV}}$. Find the energy value of the electron in the state of ${\text{L}}{{\text{i}}^{{\text{2 + }}}}$.
A.${\text{ - 30}}{\text{.6 eV}}$
B.${\text{ - 27}}{\text{.2 eV}}$
C.${\text{27}}{\text{.2 eV}}$
D.${\text{30}}{\text{.6 eV}}$

Answer 1

This question is based on Bohr’s model of hydrogen atom. In the above question, we have to find out the energy value of electron in the state of ${\text{L}}{{\text{i}}^{{\text{2 + }}}}$. We can use mathematical expression of Bohr’s theory for finding out the energy in the ${{\text{n}}^{{\text{th}}}}$orbit.
Formula Used-
${{\text{E}}_{\text{n}}}{\text{ = }}\dfrac{{{\text{ - 13}}{{.6 \times }}{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$
where ${{\text{E}}_{\text{n}}}$= energy in the ${{\text{n}}^{{\text{th}}}}$order
Z= nuclear charge
n= no. of the electron order

Complete step by step answer:
From the Bohr’s postulate, it is possible to calculate the energy of hydrogen atoms and other one electron species.
We know that ${{\text{E}}_{\text{n}}}{\text{ = }}\dfrac{{{\text{ - 13}}{{.6 \times }}{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$
The number of electrons present in lithium is 3 and hence, the number of protons is 3. Nucleus consists of protons and neutrons, so the nuclear charge is 3.
The presence of ${\text{2 + }}$ on the Li atom indicates that two electrons have been lost. So, now, the number of electrons present is one and hence, we can apply the Bohr’s postulate.
Bohr states that only a fixed number of electrons can stay in an orbit. Mathematically, it is represented as:
Number of electron in nth orbit =${\text{2}}{{\text{n}}^{\text{2}}}$
It indicates that only 2 electrons can stay in the first orbit. Hence, the 3rd electron will go to the next orbit. Therefore, 2 orbits are present.
Hence, the nuclear charge of ${\text{L}}{{\text{i}}^{{\text{2 + }}}}$is 3 and the number of electron orders is 2. Now, substituting these values in the Bohr’s equation, we get:
${{\text{E}}_{\text{n}}}{\text{ = }}\dfrac{{{\text{ - 13}}{{.6 \times }}{{\text{Z}}^{\text{2}}}}}{{{{\text{n}}^{\text{2}}}}}$
which is now equal to
${{\text{E}}_{\text{n}}}{\text{ = }}\dfrac{{{\text{ - 13}}{{.6 \times }}{{\text{3}}^{\text{2}}}}}{{{{\text{2}}^{\text{2}}}}}{\text{ = - 30}}{\text{.6eV}}$
$\therefore$The energy of electron in state of ${\text{L}}{{\text{i}}^{{\text{2 + }}}}$is ${\text{ - 30}}{\text{.6eV}}$.

So, option A is correct.

Note:
Bohr’s atomic model is valid for single electron system like hydrogen, ${\text{H}}{{\text{e}}^{\text{ + }}}$,${\text{L}}{{\text{i}}^{{\text{2 + }}}}$ etc. But it does not work well for spectra of more complex atoms. The limitation of Bohr’s atomic model is that it does not give any valid explanation for intense spectral lines at one place and less intense ones at other places.