In atension of state `n` from a state of excitation energy given is `10.19 eV`, hydrogen atom emits a photon with wavelength `4890 Å` . Determine the binding energy of the initial state.

Answer

`(13.6)/((4)^(2))=0.85eV(n=4 "to" n=2)`

Explanation

Let initially atom is in `n^(th)` energy excited state
Energy of `n^(th)` level
`:. E_(n)=(13.6)/(n^(2))`
Excitation energy is the magnitude of the difference of the energy of the energy level so Excitation energy `=E_(n)-E_(1)`
`rArr 13.6-(13.6)/(n^(2))=10.19`
`rArr (13.6)/(n^(2))=13.6-10.19=3.41`
So, finally atom is in `1^(st)` excited state during the transtion from higher one.Now energy corresponding to `4890Å` photon is
Now, `DeltaE=(hc)/(lambda)=(6.6xx10^(-34)xx3xx10^(8))/(4890xx10^(-10)xx1.6xx10^(-19))eV`
Now, let atom makes transition from `n(th)` state to `n=2` and emitting `2.53 eV` photon
`:. E_(n)-E_(2)=2.53 eV " "("initially at"n_(i)=2)`
or `-(13.6)/(n^(2))+(13.6)/(2^(2))=2.53=0.87`
`rArr n^(2)=(13.6)/(0.87)~~16 rArr n=4 " " rArr "initial state"`
`:.` Transition is from `n=4` "to" `n=2`.
Binding energy of the initial state `(n=4)=(13.6)/(4^(2))=0.85 eV`.