Question

(a) What is the ionization energy of a hydrogen atom that is in the n = 4 excited state? (b) For a hydrogen atom, determine the ratio of the ionization energy for the n = 4 excited state to the ionization energy for the ground state.

Answer 1

As we know that;

= Energy of the hydrogen atom = -13.6 eV/n^2

= Ionization energy En = 13.6 eV/n^2

So,

(a)

The ionisation energy of a hydrogen atom that is in the n = 4 excited state is,

= 13.6 eV / 4^2

= 13.6 eV / 16

= 0.85eV

Then,

b)

The ionization energy of a hydrogen atom that is in the ground state,

= 13.6 eV/1^2

= 13.6 eV

So, the ratio of the ionization energy for the n = 4 excited state to the ionization energy for the ground state,

= 0.85 eV / 13.6 eV

= 0.0625