The ionization energy of a hydrogen-like ion is I . If an electron in this ion transitions from the third excited state to the first excited state, the energy of the emitted photon is :

Step-by-step solution

The ionization energy I of a hydrogen-like ion is the energy required to remove an electron from the ground state ( n=1 ) to infinity ( n= ). Therefore, I = 13.6 Z^2 ( 1 1^2 - 1 ^2 ) = 13.6 Z^2 . The principal quantum number for an excited state is given by n = state + 1 . First excited state corresponds to n₁ = 2 . Third excited state corresponds to n₂ = 4 . The energy of the photon emitted during the transition from n=4 to n=2 is: E = 13.6 Z^2 ( 1 n₁^2 - 1 n₂^2 ) E = I ( 1 2^2 - 1 4^2 ) E = I ( 1 4 - 1 16 ) = I (