Vanadium Ionization Energy

The Vanadium Ionization Energy is the energy required to remove from atom one mole of electrons with subsequent production of positively charged ion of Vanadium.
V -> V⁺ + e^-
This process can be repeated many times, but the energy cost is increased dramatically. The general equation for the Vanadium is:
V^N+ -> V^(N+1)+ + e^-

V -> V¹⁺ = 650 kJ/mol = 155.50 Kcal/mol
V¹⁺ -> V²⁺ = 1414 kJ/mol = 338.28 Kcal/mol
V²⁺ -> V³⁺ = 2828 kJ/mol = 676.56 Kcal/mol
V³⁺ -> V⁴⁺ = 4507 kJ/mol = 1078.23 Kcal/mol
V⁴⁺ -> V⁵⁺ = 6294 kJ/mol = 1505.74 Kcal/mol
V⁵⁺ -> V⁶⁺ = 12362 kJ/mol = 2957.42 Kcal/mol
V⁶⁺ -> V⁷⁺ = 14489 kJ/mol = 3466.27 Kcal/mol
V⁷⁺ -> V⁸⁺ = 16760 kJ/mol = 4009.57 Kcal/mol
V⁸⁺ -> V⁹⁺ = 19860 kJ/mol = 4751.20 Kcal/mol
V⁹⁺ -> V¹⁰⁺ = 22240 kJ/mol = 5320.57 Kcal/mol