Question

The ionization energy for potassium is 419 kJ/mol. The wavelength of light emitted when an excited K-atom undergoes the 4s ← 4p transition is approximately 769 nm. Using this information, calculate the energies of the 4s and 4p orbitals in potassium.​

Answer 1

wave length = = 769 nm

energy = h C /

            = 6.625 x 10^-34 x 3 x 10^8 / 769 x 10^-9

            = 2.58 x 10^-19 J

the ionization of K for 1 molecule is ionization energy divided by Avogadro's number

energy = 419 x 10^3/ 6.023 x 10^23

           =6.96 x 10^-19 J   (1)
the energy of the 4s orbital is the opposite of (1)

E(4s) = - 6.96 x10^-19J

the energy of 4p is

E(4p) = E(4s) + 2.58 x10^-19

          = - 6.96 x10^-19 + 2.58 x 10^-19

           =- 4.38 x 10^-19J