Question

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

Answer 1

`36.46" ""kJ/mol"`

Explanation:

The Rydberg Expression is given by:

`1/lambda=R[1/n_1^2-1/n_2^2]`

`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 `1.097xx10^7" ""m^(-1)`

The energy levels converge and coalesce:

www.stmary.ws

At higher and higher values of `n_2` the term `1/n_2^2` tends to zero. Effectively `n_2=oo` 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=oo`.

The expression now becomes:

`1/lambda=R[1/n_1^2-0]`

`:.1/lambda=R/n_1^2`

Since `n_1=6` this becomes:

`1/lambda=R/36`

`:.lambda=36/R=36/(1.097xx10^7)=32.816xx10^-7" ""m"`

To convert this into energy we use the Planck expression:

`E=hf=(hc)/lambda`

`:.E=(6.626xx10^(-34)xx3xx10^(8))/(32.816xx10^(-7))=6.0575xx10^(-20)"J"`

You'll notice from the graphic that the energy of the `n=6` electron is `-0.38"eV"`.

This means the energy to remove it will be `+0.38"eV"`.

To convert this to Joules you multiply by the electronic charge `=0.38xx1.6xx10^-19=6.08xx10^(-20)" ""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=6.0575xx10^-20xx6.02xx10^23" ""J/mol"`

`E=36.461xx10^3" ""J/mol"`

`E=36.46" ""kJ/mol"`