# How do you calculate the ionization energy (in kJ/mol) of the He+ ion?

## A H- like ion is an ion containing only one electron. The energies of the electron in a hydrogen-like ion are given by $E \left(n\right) = - \left(2.18 \cdot {10}^{-} 18 J\right) {Z}^{2} \left(\frac{1}{n} ^ 2\right)$ where $n$ is the principal quantum number and $Z$ is the atomic number of the element.

Aug 18, 2016

You can do it like this:

#### Explanation:

You are given the expression for the energy of the electron:

$\textsf{{E}_{n} = - \left(2.18 \times {10}^{- 18}\right) {Z}^{2} \left[\frac{1}{n} ^ 2\right]}$

When the electron is in its ground state, the value of the principle quantum number $\textsf{n}$ is 1.

$\textsf{n}$ can take values 1, 2, 3, 4..... etc.

For these higher energy levels you can see that the value of $\textsf{\frac{1}{n} ^ 2}$ will get smaller and smaller.

The means that the gap in energy between successive energy levels gets less and less such that they converge and eventually coalesce.

At this point as $\textsf{n}$ tends to infinity so $\textsf{\frac{1}{n} ^ 2}$ tends to zero.

An electron excited to this point can be regarded as ionised i.e it has left the atom.

The difference in energy between these levels will represent the ionisation energy.

sf(I.E.=-(2.18xx10^(-18))Z^2xx0)-[-2.18xx10^(-18)xxZ^2[1/1^2]

For $\textsf{H {e}^{+}}$ there are 2 protons in the nucleus $\therefore$$\textsf{Z = 2}$

$\therefore$$\textsf{I . E . = \frac{2.18 \times {10}^{- 18} \times {2}^{2}}{1} = 8.72 \times {10}^{- 18} \textcolor{w h i t e}{x} J}$

To convert to kJ divide by 1000$\textsf{\Rightarrow}$

$\textsf{I . E . = 8.72 \times {10}^{- 21} \textcolor{w h i t e}{x} k J}$

This is the energy required to ionise a single $\textsf{H {e}^{+}}$ ion.

To find the energy required to ionise a mole of ions you need to multiply by The Avogadro Constant which is $\textsf{6.02 \times {10}^{23} \textcolor{w h i t e}{x} {\text{mol}}^{- 1}}$.

$\textsf{I . E = 8.72 \times {10}^{- 21} \times 6.02 \times {10}^{23} = 5250 \textcolor{w h i t e}{x} \text{kJ/mol}}$