Question

Determine the energy of light of the photon of an B4+ cation from the n = 5 energy level to n = 3? Is the energy being absorbed or emitted?

Answer 1

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.