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

1 Answer
Jun 4, 2016

Answer:

#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:

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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"#