SME
Physical Chemistry II: Quantum & Spectroscopy
Multi-electron atoms and term symbols
Use this for “The ³D term has L = 2 and S = 1. Which J levels does it split into, and how many states does it hold?”
Table
One level (J)
States in that level (gJ)
This is an example. Type your own values to start a new problem.
L
Total orbital angular momentum
example
S
Total spin
example
2S + 1
Multiplicity
calculated
Jₘᵢₙ
Lowest J
worked out (not typed)
Jₘₐₓ
Highest J
worked out (not typed)
nJ
Number of levels
worked out (not typed)
N
States in the term
worked out (not typed)
J
One level
example
gJ
States in that level
calculated
Formulas
2
S
+
1
=
2
×
S
+
1
3 = 2 × 1 + 1
Jₘᵢₙ
=
|
L
−
S
|
0 = |1 − 1|
Jₘₐₓ
=
L
+
S
2 = 1 + 1
nJ = Jₘₐₓ − Jₘᵢₙ + 1
3 = 2 − 0 + 1
N
=
(2
×
L
+
1)
×
(2
×
S
+
1)
9 = (2 × 1 + 1) × (2 × 1 + 1)
gJ
=
2
×
J
+
1
1 = 2 × 0 + 1
Assumptions
•
Russell–Saunders (LS) coupling, as in lighter atoms: the orbital momenta of the electrons add to L and their spins add to S.
•
The term is written ²ˢ⁺¹L, with L = 0, 1, 2, 3, 4, 5, 6 named S, P, D, F, G, H, I.
•
Spin-orbit coupling splits the term into levels J = |L − S|, …, L + S, and level J holds 2J + 1 states.
•
Hund’s rules: the ground term has the largest S, then the largest L; its lowest J lies lowest when the subshell is less than half full, its highest J when more.
Step-by-step
WE KNOW
L = 1, S = 1, J = 0
FIND
multiplicity (2S + 1), lowest J (Jₘᵢₙ), highest J (Jₘₐₓ), number of levels (nJ), states in the term (N), states in that level (gJ)
Step 1 · Find multiplicity (2S + 1)
Use 2S + 1 = 2 × S + 1
The
multiplicity
counts
the
2
S
+
1
ways
the
total
spin
can
point.
2
S
+
1
=
2
×
1
+
1
2
S
+
1
=
3
Step 2 · Find lowest J (Jₘᵢₙ)
Use Jₘᵢₙ = |L − S|
Orbital and spin momenta as nearly opposite as they can be: the smaller is taken from the larger.
Jₘᵢₙ = |1 − 1|
Jₘᵢₙ = |0|
Jₘᵢₙ = 0
Step 3 · Find highest J (Jₘₐₓ)
Use Jₘₐₓ = L + S
The two lined up as nearly as they can be: their sizes add.
Jₘₐₓ = 1 + 1
Jₘₐₓ = 2
Step 4 · Find number of levels (nJ)
Use nJ = Jₘₐₓ − Jₘᵢₙ + 1
Count
the
J
values
from
Jₘᵢₙ
to
Jₘₐₓ
in
steps
of
1,
both
ends
included.
nJ = 2 − 0 + 1
nJ = 3 → ³P splits into ³P₀, ³P₁ and ³P₂. Less than half full, ³P₀ lies lowest; more than half full, ³P₂.
Step 5 · Find states in the term (N)
Use N = (2 × L + 1) × (2 × S + 1)
Every
orbital
orientation
with
every
spin
orientation;
the
2
J
+
1
of
the
levels
add
up
to
the
same.
N
=
(2
×
1
+
1)
×
(2
×
1
+
1)
N
=
3
×
3
N
=
9
Step 6 · Find states in that level (gJ)
Use gJ = 2 × J + 1
Its
MJ
run
from
−
J
to
J
in
steps
of
1.
gJ = 2 × 0 + 1
gJ = 1
Check
Put the numbers back in. Both sides should match.
3 = 2 × 1 + 1 ✓
0 = |1 − 1| ✓
2 = 1 + 1 ✓
3 = 2 − 0 + 1 ✓
9 = (2 × 1 + 1) × (2 × 1 + 1) ✓
1 = 2 × 0 + 1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.