Question

$Li^{+2} \longrightarrow Li^{+3} + e^-$ :
Energy required for this process.
Given ionisation energy for ground state of hydrogen atom is $2.17 \times 10^{-18} \text{ J}$.

• A. $19.53 \times 10^{-18} \text{ J}$
• B. $19.54 \times 10^{-18} \text{ J}$
• C. $19.56 \times 10^{-18} \text{ J}$
• D. $19.55 \times 10^{-18} \text{ J}$

Answer 1

The Correct Option is A

The given reaction is the ionization of \(Li^{+2}\) to \(Li^{+3}\) by losing an electron:

\(Li^{+2} \longrightarrow Li^{+3} + e^-\)

We are given the ionization energy for the ground state of a hydrogen atom as \(2.17 \times 10^{-18} \text{ J}\).

To find the energy required for this process, we need to consider the energy levels of the lithium ion. The formula for the ionization energy of a hydrogen-like ion is:

\(E = Z^2 \times E_H\)

Where:

• \(E_H = 2.17 \times 10^{-18} \text{ J}\) (the ionization energy of a hydrogen atom in the ground state)
• \(Z\) is the atomic number of the ion

Lithium has an atomic number \(Z = 3\).

For \(Li^{+2}\), the effective nuclear charge is \(3\) because it is similar to a hydrogen-like species but with three protons. Therefore, the ionization energy for removing one electron from \(Li^{+2}\) is calculated as:

\(E = 3^2 \times E_H = 9 \times 2.17 \times 10^{-18} \text{ J}\)

Calculating this gives us:

\(E = 9 \times 2.17 \times 10^{-18} \text{ J} = 19.53 \times 10^{-18} \text{ J}\)

Therefore, the energy required for the ionization process \(Li^{+2} \longrightarrow Li^{+3} + e^-\\) is \(19.53 \times 10^{-18} \text{ J}\).

Hence, the correct answer is:

$19.53 \times 10^{-18} \text{ J}$