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 `hf = (12400)/(4890) = 2.54 e V`
The excitation energy is the energy to excite the atom to a leval above the ground state. Therefore the energy of the level to which the transition occurs is :
`E_(n) = - 13.6 + 10.19 = - 3.41 e V`
The photon arises from the transition between energy of an electron in the state (say `n to x`), then
`E_(n) - E_(s) = hf, hence E_(n) - (- 3.41) = hf`
`E_(n) - (- 3.41) = 2.54 or E_(n) = - 0.87 e V`
Therefore, the binding energy of an electron in the state is `0.87 e V`
`n = sqrt(E_(1)/(E_(n))) = sqrt ((13.6)/(0.87)) = 4 to x = sqrt(E_(1)/(E_(s))) = sqrt ((13.6)/(3.41)) = 2`