# What is the ionization energy of a hydrogen atom that is in the n = 6 excited state?

Jun 4, 2016

$36.46 \text{ ""kJ/mol}$

#### Explanation:

The Rydberg Expression is given by:

$\frac{1}{\lambda} = R \left[\frac{1}{n} _ {1}^{2} - \frac{1}{n} _ {2}^{2}\right]$

$\lambda$ is the wavelength of the emission line

${n}_{1}$ is the principle quantum number of the lower energy level

${n}_{2}$ is the principle quantum number of the higher energy level

$R$ is the Rydberg Constant $1.097 \times {10}^{7} \text{ } {m}^{- 1}$

The energy levels converge and coalesce:

At higher and higher values of ${n}_{2}$ the term $\frac{1}{n} _ {2}^{2}$ tends to zero. Effectively ${n}_{2} = \infty$ and the electron has left the atom, forming a hydrogen ion.

The Rydberg Expression refers to emission where an electron falls from a higher energy level to a lower one, emitting a photon.

In this case we can use it to find the energy required to move an electron from $n = 6$ to $n = \infty$.

The expression now becomes:

$\frac{1}{\lambda} = R \left[\frac{1}{n} _ {1}^{2} - 0\right]$

$\therefore \frac{1}{\lambda} = \frac{R}{n} _ {1}^{2}$

Since ${n}_{1} = 6$ this becomes:

$\frac{1}{\lambda} = \frac{R}{36}$

$\therefore \lambda = \frac{36}{R} = \frac{36}{1.097 \times {10}^{7}} = 32.816 \times {10}^{-} 7 \text{ ""m}$

To convert this into energy we use the Planck expression:

$E = h f = \frac{h c}{\lambda}$

$\therefore E = \frac{6.626 \times {10}^{- 34} \times 3 \times {10}^{8}}{32.816 \times {10}^{- 7}} = 6.0575 \times {10}^{- 20} \text{J}$

You'll notice from the graphic that the energy of the $n = 6$ electron is $- 0.38 \text{eV}$.

This means the energy to remove it will be $+ 0.38 \text{eV}$.

To convert this to Joules you multiply by the electronic charge $= 0.38 \times 1.6 \times {10}^{-} 19 = 6.08 \times {10}^{- 20} \text{ ""J}$

As you can see my calculated value is very close to this.

This is the energy required to ionise a single atom. To get the energy required to ionise a mole of atoms we multiply by the Avogadro Constant:

$E = 6.0575 \times {10}^{-} 20 \times 6.02 \times {10}^{23} \text{ ""J/mol}$

$E = 36.461 \times {10}^{3} \text{ ""J/mol}$

$E = 36.46 \text{ ""kJ/mol}$