Question

Potential energy of electrons present in $L{i^{2 + }}$ is?
(A) $- \dfrac{{{e^2}}}{{2\pi {\varepsilon _ \circ }r}}$
(B) $- \dfrac{3}{2}\dfrac{{{e^2}}}{{\pi {\varepsilon _ \circ }r}}$
(C) $- \dfrac{3}{4}\dfrac{{{e^2}}}{{\pi {\varepsilon _ \circ }r}}$
(D) $- \dfrac{1}{2}\dfrac{{{e^2}}}{{\pi {\varepsilon _ \circ }r}}$

Answer 1

To solve this question; we have to know about the potential energy for the orbital, i.e. the energy that is stored by the body due to the position possessed by the body. After finding it, we will substitute the atomic number of the given element.
Potential energy, $E = - \dfrac{{{Z^2}m{e^4}}}{{4{n^2}{h^2}{\varepsilon _ \circ }^2}}\,$
Where $Z$ is the atomic number, $e$ is the charge of the particle i.e. $1.602 \times {10^{ - 19}}C$ , $m$ is the mass of the electron.
$r = \dfrac{{{n^2}{h^2}{\varepsilon _ \circ }}}{{Z\pi m{e^2}}}$
Here, $r$ is the radial distance of the electron from the nucleus.
By multiplying $E$ and $r$ we will get:
$Er = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _ \circ }}}$ .

Complete Step By Step Answer:
Potential energy is the energy that is stored in a body due virtue of its position related to another body. In the atoms, electrons are held together by the electric force of attraction to nuclei, the zero reference for potential energy is the distance from the nucleus. It is so great that the electric force is not detectable. Here the bound electrons have negative potential energy.
To calculate the potential energy we will use the Bohr potential energy formula. This formula derived under the assumption that the orbital of electrons is circular but afterward it was found that orbitals are elliptical. The Bohr potential energy formula is:
$Er = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _ \circ }}}$
The atomic number of $Li$ is $3$ , thus:
$Er = - \dfrac{{3{e^2}}}{{4\pi {\varepsilon _ \circ }}}$
$\Rightarrow E = - \dfrac{{3{e^2}}}{{4\pi {\varepsilon _ \circ }r}}$
Thus, option (C) is correct.

Note:
The potential energy is greater than the Kinetic energy of the electron but it is negative. With the increase in radii of the orbit, the potential energy of the electron decreases. The Bohr energy formula is only valid for hydrogen atoms. It has poor predictions for larger atoms.