In atension from state `n` to a state of excitation energy `10.19 eV`, hydrogen atom emits a `4890 Å` photon. Etermine the binding energy of the initial state.

Explanation

The energy of the emitted photon is
`hupsilon=(hc)/(lambda)=(12.40xx10^(3)eVÅ)/(4.89xx10^(3)Å)=2.54eV`
The excitation energy `(E_(x))` is the energy to excite the atom to a level above the ground state. Therefore, the energy of the level is
`E_(l)=E_(1)+E_(x)=-13.6eV+10.19eV=-3.41eV`
The photon aries from the transition between energy states such that `E_(u)-E_(1)=hupsilon`, hence
`E_(u)-(-3.41eV)=2.54eV or E_(u)=-0.87eV`
Therefore, the binding energy of an electron in the state is 0.87 eV.
Note that the transition corresponds to
`n_(u)=sqrt((E_(1))/(E_(u)))=sqrt((13.6eV)/(0.87eV))=4 and n_(l)=sqrt((E_(1))/(E_(l)))=sqrt((13.6eV)/(3.141eV))=2`