Question

The ratio between the ionization energy of Li^2- ion and that of H-atom is-
1. 2 ∶ 1
2. 3 ∶ 1
3. 9 ∶ 1
4. 4 ∶ 1

Answer 1

Correct Answer - Option 3 : 9 ∶ 1

Concept:

Ionization energy:

• It is the energy that an electron in a gaseous atom or ion has to absorb to come out of the influence of the nucleus.
• It is typically measured in the unit electron volt (eV). 
• Ionization is the process by which electrically neutral atoms or molecules are converted to electrically charged atoms or molecules.

According to Bohr’s model of an atom:

• An electron can absorb energy and jump to the next higher orbits of higher energy. 
• If more energy is available and the electron absorbed, it will come out of force of attraction of the nucleus, which means out of the atom.

The energy of an electron in the n^th orbit is calculated by the Bohr model of an atom as

\(E_n = - 13.6 × \frac{z^2}{n^2} \) eV

Where,

Z = Atomic number, n = positive integer (n = 1 at ground state) 

Calculation:

We know that, ionation energy is given by,

\(E_n = - 13.6 × \frac{z^2}{n^2} \) eV

Let n = 1 

For Li^2-, Z = 3 therefore,

(E_n)₁ = -13.6 × 9    .....(1)

For H, Z = 1, therefore

(En)2 = -13.6 × 1   .....(2)

(1) divided by (2)

\(\frac{(E_n )_1}{(E_n )_2} = \frac{9}{1}\)

Hence, the ratio between the ionization energy of Li2- ion and that of H-atom is 9:1.