If the ionization enthalpy of He ⁺ ion is 54.4 eV , what is the energy of an electron in the second Bohr orbit of Li ²⁺ ion?

Step-by-step solution

The ionization enthalpy is the energy required to remove an electron from the ground state ( n=1 ). Ionization energy of He ⁺ = 54.4 eV . Therefore, the energy of the electron in the first orbit of He ⁺ is E₁( He ⁺) = -54.4 eV . The energy of an electron in the n^ th Bohr orbit of a hydrogen-like species is given by: E_n = E₁( H ) Z^2 n^2 For He ⁺ ( Z=2, n=1 ): E₁( He ⁺) = E₁( H ) 2^2 1^2 = 4 E₁( H ) = -54.4 eV E₁( H ) = -13.6 eV For the second orbit of Li ²⁺ ( Z=3, n=2 ): E₂( Li ²⁺) = E₁( H ) Z^2 n^2 E₂( Li ²⁺) =