Answer:
20.1 eV.
Explanation:
Energy of n the orbital of a hydrogen like atom in eV having atomic no Z is given by the relation
Eₙ = -Z² x 13.6 / n² eV
For B⁴⁺ , Z = 5.
For B⁴⁺ having only one electron ( like hydrogen ) energy of n=5 orbital
E₅ = - 5² X 13.6 / 5² = - 13.6 eV.
And for n = 3
E₃ = - 5² X 13.6 / 3²
= - 37.77 eV.
Difference = 33.77-13.6 = 20.17 eV.
Light energy will be emitted when electron transits from n = 5 to n = 3. The light will have energy equal to the difference of energy of 20.1 eV.
