Question

Write out the complete electron configuration of scandium (element 21) and compute the K-alpha wavelength for its characteristic x-ray spectrum.

Answer 1

Electron Configuration of scandium

1s2 2s2 2p6 3s2 3p6 3d1 4s2

K-alpha wavelength

K-alpha rdiation comes from the emission of a photon that occurs when an excited atom with a "hole" in its core 1s shell experiences a 2p - 1s transition. In a hydrogen atom, such a transition occurs to give off a photon that is (3/4)of the ionization energy of the H-atom, which is (3/4)(13.6 eV) = 10.2 eV.

The factor of (3/4) comes from the Rydberg formula:

E = {(1/n1^2)-(1/n2^2)}*13.6eV

E = {(1/1^2)-(1/2^2)}*13.6 = (3/4)*13.6

There is a formula called "Moseley's law" that adapts this to many-electron atoms

"the frequency of the line is adequately calculated to 2-digit accuracy by the use of Moseley's law. This formula is 10.2 eV multiplied by the square of a quantity one less than the atomic number of the element in question (atomic number minus 1)."

K-alpha for scandium (Z = 21) is thus estimated as (10.2 eV)(20)² = 4080 eV.

Use E = hc/lamda

lamda = 0.304nm = 3.04 angstrom