An electron approaches a fixed proton from a large distance with a kinetic energy of 12.2 eV. The electron gets captured by the proton to form a hydrogen atom in an excited state and a photon of wavelength `lambda_(1) = 796 Å` is emitted. Later, the hydrogen atom de-excites by emitting a single photon of wavelength `lambda_(2)`. Find `lambda_(2)`. [Take `h = 6.62 xx 10^(– 34) Js, c = 3 xx 108 m//s`]

Answer

`1217 Å`