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 hv = `hc//lambda`
Now, hc = `(6.63 xx 10^(-34) Js)xx(3 xx 10^(8) m//s) = 19.89 xx 10^(-26) J-m` =`19.89 Xx 10^(-16) JA`
`( 19.89xx10^(16))/(1.6xx10^(-19))eVÀ = 12.4xx 10^(3) eVẢ` 
  `therefore`    Energy of photon =`( hc)/(lambda)=( 12,4 xx 10^(3)eVA)/(4.89xx10^(3)A) =2.54eV`
The excitation energy E, is the energy to excite the atom to a level above the ground state. Now `E_(n)=E_(1)+E_(x) = -13.6 eV + 10.19 eV = -3.41 eV`.
Let the emission of photon occur due to the transition from a higher energy state `E_(n)` to the lower energy state `E_(L) ( =E_(n))`.
Now `E_(L) = -3.41 eV` and the difference`therefore`  ` E_(h)=2.54+E_(L) eV.
`therefore`  `E_(n) = 2.54 + E_(L) = (2.54 -3.41) eV = -0.874 eV`.
  Hence binding energy of electron in the initial state is 0.874 eV
  The nature of the transition is n ton state where `n_(h)=sqrt((E_(1))/(E_(h)))=sqrt((13.6eV)/(0.87eV))cong4`and `n_(L)=sqrt((E_(1))/(E_(L)))=sqrt((E_(1))/(E_(L)))=sqrt((13.6ev)/(3.41ev))cong2`