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. What is the value of n for the intermediate state?

Step-by-step solution

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 = -0.377 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 eV . E_ n' = -0.377 - 1.13 = -1.507 eV . Equating this to the energy formula for the n' -th state: - 13.6 n'^2 = -1.507 n'^2 = 13.6 1.507 9 n' = 3 The intermediate state is n = 3 .