# How do you calculate the ionization energy of an element?

Use the Rydberg Equation $\Delta {E}_{i}$ = $A \left(\frac{1}{n} _ {f}^{2} - \frac{1}{n} _ {i}^{2}\right)$ where $A = 2.18 \times {10}^{-} 18 \text{Joules}$; ${n}_{f} = \infty$; ${n}_{i} = \text{starting energy level} .$
For the Hydrogen Atom, ionization from the ground state where ${n}_{i} = 1$ => $\Delta {E}_{i z n} = 2.18 \times {10}^{-} 18 J \left(\frac{1}{\infty} ^ 2 - \frac{1}{1} ^ 2\right)$ = $- 2.18 \times {10}^{-} 18 \text{Joules}$ to remove the electron from n = 1 energy level to $n = \infty$. The negative indicates that on 'gain' of electron, $2.18 \times {10}^{-} 18$Joules energy is released, but on removal, $\Delta E$ should be greater than zero indicating the gain of energy to remove the electron. $\Delta {E}_{r e m o v a l}$=$\Delta {E}_{\text{gained}}$.
The $2.18 \times {10}^{-} 18$ Joules is per atom#
=> $\text{Joules"/"mole}$ = $\left(2.18 \times {10}^{-} 18 \frac{j}{\text{atom}}\right)$$\left(6.02 \times {10}^{23} \text{atoms"/"mole}\right)$ = $1 , 351 , 600 \text{joules"/"mole} =$1352$\text{Kj"/"mole}$. This corresponds favorably with the published 1st ionization energy of the Hydrogen atom.