A singly ionized helium atom ( He ⁺ ) is initially in its ground state. The minimum energy it needs to absorb in order to subsequently emit radiation in the Paschen series is nearly :

Step-by-step solution

For the Paschen series, the transition must end at n=3 . The minimum excited state from which an electron can fall to n=3 is n=4 . Therefore, the He ⁺ ion must be excited from its ground state ( n=1 ) to at least n=4 . The energy required for this transition is given by: E = 13.6 Z^2 ( 1 n₁^2 - 1 n₂^2 ) For He ⁺ , Z = 2 . Substituting n₁ = 1 and n₂ = 4 : E = 13.6 (2)^2 ( 1 - 1 16 ) E = 54.4 15 16 = 51.0 eV Answer: 51.0 eV