A hydrogen atom initially in the n = 6 state undergoes two successive transitions to reach the ground state. During the first transition, a photon of 1.13 eV is emitted. Determine the energy of the photon emitted in the second transition.

Step-by-step solution

The energy of an electron in the n^ th orbit of a hydrogen atom is given by E_n = - 13.6 n^2 eV . For the initial state n = 6 , the energy is: E₆ = - 13.6 6^2 = -0.378 eV During the first transition, a photon of energy 1.13 eV is emitted. The energy of the intermediate state n' is: E_ n' = E₆ - 1.13 = -0.378 - 1.13 = -1.508 eV Comparing this with the standard energy levels of hydrogen, we have E₃ = - 13.6 3^2 = -1.51 eV . Thus, the intermediate state is n' = 3 . The second transition occurs from n' = 3 to the groun