ATM-31

Term degeneracy

Russell–Saunders term.

Reading speed

Atomicterm

Governing equation

g=(2S+1)(2L+1)

$L \in \left[0, 6\right]$

where

S: S (—)
L: L (—)
g: g (—)

The figure, part by part

Each numbered marker wears the color of its card. Hover a card to light the marker. Tap a marker to jump to that card. The slider walks a red point along the curve; Animate drawing traces it like a pen.

00:0 / 00:08

①

Governing law — red

g=(2S+1)(2L+1)

$L \in \left[0, 6\right]$

$L = 0$ $g = 3$

Along the curve

②

Domain of the variable — blue

L_{\min} = 0,\quad L_{\max} = 6

$L \in \left[0, 6\right]$

$L = 0$ $L = 0$

Along the curve

③

Response curve — green

g=(2S+1)(2L+1)

$L \in \left[0, 6\right]$

$L = 0$ $g = 3$

Along the curve

④

Rate of change — amber

\frac{\mathrm{d}g}{\mathrm{d}L}

$L \in \left[0, 6\right]$

$L = 0.038$ $g^{\prime} = 6$

Along the curve

Lecture brief

Historical brief

Balmer, Rydberg, Bohr (1913), fine structure, Zeeman and Stern–Gerlach built the spectrum of one atom. The lab computes levels, selection and magnetic splitting. This sheet (ATM-31 — Term degeneracy) is the form associated with term. Working symbols: $S$ , $L$ $\rightarrow$ $g$ . Russell–Saunders term.

Purpose

Purpose: compute

g

from

S

L

in Atomic physics via

g=(2S+1)(2L+1)

Russell–Saunders term. Use it when a real atomic physics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

S = 1.000\,\mathrm{—}

L = 1.000\,\mathrm{—}

, the governing relation

g=(2S+1)(2L+1)

yields

g = 9\,\mathrm{—}

. Russell–Saunders term. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

S S1 —L L1 —

Outputs

g g9 —

Reading speed

Watch on YouTube

Free library

Full library

Free PDF / open book

YouTube channels

00:0 / 00:08

Narration of this film

Russell–Saunders term.

Reading speed

Watch on YouTube