A hydrogen sample is prepared in a particular state A photon of energy `2.55eV` get observed into the sample to take some of the electron in a farther excited state B find the quantum numbers of the state A and B

Explanation

The allowed energies of hydrogen atoms are
`E_(1) = - 13.6eV`
`E_(2) = - 3.4eV`
`E_(3) = - 1.5eV`
`E_(4) = - 0.85eV`
`E_(5) = - 0.54eV`
We see that a difference of `2.55eV` can only be observed in transition`n = 2 to n = 4` so the state A has quantum number`2` and the state B has quantum number`4`