The energy of the first Lyman line of the hydrogen atom is E . The energy of the second Balmer line of the Li ²⁺ ion is :

Step-by-step solution

For a hydrogen-like species, the energy of a transition is given by E = 13.6 Z^2 ( 1 n₁^2 - 1 n₂^2 ) eV. For the first Lyman line of the H atom ( Z=1 ), the transition is from n=2 to n=1 : E = 13.6 (1)^2 ( 1 1^2 - 1 2^2 ) = 13.6 3 4 For the second Balmer line of the Li ²⁺ ion ( Z=3 ), the transition is from n=4 to n=2 : E' = 13.6 (3)^2 ( 1 2^2 - 1 4^2 ) E' = 13.6 9 ( 1 4 - 1 16 ) = 13.6 9 3 16 Taking the ratio of the two energies: E' E = 13.6 9 3 16 13.6 3 4 = 27 16 4 3 = 9 4 = 2.25 Thus, the energy of the second B