Answer to Question #247682 in Inorganic Chemistry for an2

Question

The first ionization energies of cesium and gold are 3.89 and 9.23 eV, respectively. Calculate the ionization energies of an H atom and account for the differences in these three values. 

Answer 1

Given first IE of Cs=3.89eV (IE=Ionization energy)

First IE of Au=9.23eV

Electronic configuration of Cs(55) : [Xe] 6s¹

Electronic configuration of Au(79) : [Xe] 4f¹⁴ 5d¹⁰ 6s¹

in both elements outermost electron in 6^th orbit , so principal quantum number n=6.

According to Bhor theory , Ionization energy of H if electron in n^th orbit

$E_n = \frac{13.6}{n^2} \\ n=6 \\ E_n = \frac{13.6}{36} = 0.38 \;eV$

Difference in ionization energy of Cesium(Cs) to Hydrogen(H) =3.89-0.38 =3.51eV

Difference in ionization energy of Gold(Au) to Hydrogen(H)= 9.23-0.38 = 8.85eV

Difference in ionization energy of Gold(Au) to Cesium(Cs)= 9.23-3.89 = 5.34eV