A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV . Find the energy and wavelength of photon emitted.

Explanation

Let `n_(2)` be the final excitation state of the electron, Since excitation energy is always measured with respect to the ground state, therefore
`DeltaE=13.6[1-(1)/(n_(2)^(2))]` here `DeltaE=10.2eV`, therefore,
`10.2=13.6[1-(1)/(n_(2)^(2))]` or `n_(2)=2` Thus, the electron jumps from `n_(1)=4` to `n_(2)=2`.