# Ionization Energy Formula

## Ionization Energy Formula

The energy essential to take away an electron from a gaseous atom A or a gaseous molecule AB is titled as ionization energy. Referring to the following equation. A mass spectrometer can determine the ionization energy. The ionization energy or potential is therefore sometimes also called the “threshold” or “appearance” energy or potential.

Energy + A →→ A + e-

Energy + AB →→ AB + e-

is called as the ionization energy of the molecule or atom.

How tightly an atom holds onto its electrons is measured by the ionization energy. The tighter an electron is held, the higher is its ionization energy. The developments in ionization energy are just the reverse of those for atomic radii. Usually, as the atomic radii get bigger, ionization energies get lesser and vice versa.

Solved Problems – Ionization Energy

Problem 1: in both copper and potassium the outer electron is to be removed from the 4s-orbital. Why is the first ionization energy for copper (745 kJ mol-1) higher than that of potassium (418 kJ mol-1).

In both these elements the external electron is in the 4s level, but however in potassium the nucleus’s positive charge (Z=19) is curtained by the argon core: 1s22s22p63s3p6, in copper the nuclear charge (z=29) is curtained by 3d electrons and the argon core.

The ten extra electrons in copper are in the 3d level. The diffused, and strongly directional d-electrons protect the 4s electron somewhat poorly from the nuclear charge, it thus experiences a highly effective nuclear charge, and the ionization energy of copper is far greater than that of potassium.

Problem 2: Compute the energy essential to convert all the Al atoms to Al3+ ions existent in 0.720g of Al vapors.The ionization energies first, second and third respectively of Al are 578, 1817 and 2745kJ mol-1.

Total energy required for the change Al → Al3+

= IE1 + IE2 + IE3

= 578 + 1817 + 2745 kJ mol-1

= 5140 kJ mol-1

Number of mole in 0.720 gm of Al =

$\frac{0.720}{27}=0.03&space;mole$

= 3 × 10-2 mole

3 mole of Al needs for ionization = 5140 kJ

Therefore, 3 × 10-2 mole of Al will need = 5140 × 10-2 kJ

= 51.40 kJ

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