Question

How many cl atom can you ionise in this process cl--> cl++e- by the energy liberated for the process cl+e--> cl- for one avogadro number of atoms. given I.P=130eV and EA=3.60eV
Please answer me urgently... And answer appropriately...

Answer 1

Answer:

Explanation:

I.P = 130eV (Given)

EA = 3.60eV (Given)

Ionization means the energy required to remove one or more electrons from any neutral atom to form a positively charged ion. This is a physical property that in return influences the chemical behaviour of the atom.

Thus,

Energy liberated = 3.60 × 6.02 × 10`23

By 130 eV the energy number of atoms that ionize will be one.

Thus,  

=  3.60 × 6.02 ×10`23 eV,

Number of atoms that ionize = 1/13 × 3.60 × 6.02 × 10`23

≈ 2×10`23

Answer 2

Explanation:

Let $\rm n$ atoms be ionised.

$\rm 6.022×10^{23}×EA=n×IP$

$\rm n = \dfrac{6.022×10^{23}×3.60}{13}$

$\rm n = \bf 1.667×10^{23} \: atoms$