Calculate the energy required for the following process

{\mathrm{Li}}^{2+}\to {\mathrm{Li}}^{3+}+e^{-}

{}^{}

Given: Ground state energy of Hydrogen is –13.6 eV/atom

$$$$$$$$

The correct answer is 122.4 eV/atom.

Step-by-Step Explanation:

The given process is:
${\mathrm{Li}}^{2+}\to {\mathrm{Li}}^{3+}+e^{-}$
This process represents the ionization of a lithium ion (Li²⁺) to form Li³⁺. This is the removal of the third electron (which is the last remaining electron in Li²⁺, making it a hydrogen-like species).

For a hydrogen-like species with atomic number Z, the energy of an electron in the n-th orbit is given by Bohr's formula:
$E_n=E_1\left(H\right)\times \frac{Z^2}{n^2}$
where:
• $E_1\left(H\right)$ is the ground state energy of hydrogen = -13.6 eV/atom.
• Z is the atomic number. For Lithium (Li), Z = 3.
• n is the principal quantum number. Since the electron is in the ground state of Li²⁺, n = 1.

Let's calculate the energy of the electron in the ground state (n = 1) of Li²⁺:
$E_1=-13.6\times \frac{3^2}{1^2}$
$E_1=-13.6\times 9$
$E_1=-122.4\text{eV/atom}$

The energy required for the ionization process (removal of the electron to infinity, where $E_{\infty }=0$) is the ionization energy (IE):
$\Delta E=E_{\infty }-E_1$
$\Delta E=0-(-122.4\text{eV/atom})$
$\Delta E=122.4\text{eV/atom}$

Therefore, the energy required for the given process is 122.4 eV/atom.