Radiation emitted from the n = 2 to n = 1 transition of hydrogen atoms is incident on helium ions that are in the n = 1 and n = 2 states. Determine the possible transitions of the helium ions as they absorb energy from this radiation.

Step-by-step solution

The energy of the photon emitted during the transition of a hydrogen atom from n = 2 to n = 1 is given by: E = 13.6 1^2 ( 1 1^2 - 1 2^2 ) = 13.6 3 4 eV This energy is absorbed by a helium ion ( He^+ ) with atomic number Z = 2 . Let the transition be from state n₁ to n₂ . The energy required for this transition is: E' = 13.6 2^2 ( 1 n₁^2 - 1 n₂^2 ) = 13.6 4 ( 1 n₁^2 - 1 n₂^2 ) Equating the two energies, E = E' : 13.6 3 4 = 13.6 4 ( 1 n₁^2 - 1 n₂^2 ) 3 16 = 1 n₁^2 - 1 n₂^2 The helium ions are initially in n₁ = 1 or n