# What is an excited state of scandium?

Aug 2, 2017

Well, the ground-state electron configuration of $\text{Sc}$ is

$\left[A r\right] 3 {d}^{1} 4 {s}^{2}$

$\text{Sc}$ has many excited states, but let's choose an intuitive one... one where a valence electron is promoted to a clearly higher energy level.

We'll assume the $4 s \to 4 p$ transition. That would give one of the excited states as:

$\left[A r\right] 3 {d}^{1} 4 {s}^{\textcolor{red}{2}} \to \overline{\underline{| \stackrel{\text{ ")(" "[Ar] 3d^1 4s^1 4p^color(red)(1)" }}{|}}}$

(There are other lower-lying states, but we are ignoring them for simplicity.)

Due to spin-orbit coupling...

(Notice how the ""^3 states have three energy levels.)

...the destination state (a ""^4 F), which was quadruply-degenerate in the absence of a magnetic field, now splits into four energy levels of this same configuration, with term symbols ""^4 F_(3//2), ""^4 F_(5//2), ""^4 F_(7//2), and ""^4 F_(9//2), in order of increasing energy.

Assuming we start at the ground-state energy level of ${\text{0 cm}}^{- 1}$, this transition thus requires an input of any of the following frequencies:

${\text{15 672.58 cm}}^{- 1}$

${\text{15 756.57 cm}}^{- 1}$

${\text{15 881.75 cm}}^{- 1}$

${\text{16 026.62 cm}}^{- 1}$