Quantum Mechanics - Spectroscopic Notation

A 3p electron is found in the ³P_3/2 energy level of a hydrogen atom. Which of the following is true about the electron in this state?

A. It is allowed to make an electric dipole transition to the ²S_1/2 level
B. It is allowed to make an electric dipole transition to the ²P_1/2 level
C. It has quantum numbers l = 3, j = 3/2, s = 1/2
D. It has quantum numbers n = 3, j = l, s = 3/2
E. It has exactly the same energy as it would in the ³D_3/2 level

(GR9677 #41)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

For ³P_3/2  : 

P → l = 1
j = 3/2

→ (C) and (D) are FALSE

Selection rules (for electron transition):

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆ml = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

(A) TRUE
Transition for ∆l = 0 is allowed.

(B) FALSE
Transition for ∆l = 0 is not allowed.

(E) FALSE
Electron can not move to different energy level (P to D) while maintaining the same energy.

Answer: A

Nuclear & Particle Physics - Spectroscopic Notation

In a ³S state of the helium atom, the possible values of the total electronic angular momentum quantum number are

A. 0 only
B. 1 only
C. 0 and 1 only
D. 0, 1/2, and 1
E. 0, 1, and 2

(GR9277 #31)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

• N = the principal quantum number and will often be omitted
• s = the total spin quantum number
• L = the orbital angular momentum quantum number, l but is written as S, P, D, F, … for l = 0, 1, 2, 3,⋯
• j = the total angular momentum quantum number

For ³S 
S → l = 0
3 → 3 = 2s + 1 → s = 1
j = l + s = 0 + 1 = 1

Answer: B 

Nuclear & Particle Physics - Spin

The ground state of the helium atom is a spin

A. singlet
B. doublet
C. triplet
D. quartet
E. quintuplet

(GR9277 #59)

Solution:

Spectroscopic notation: N^2s+1 L_j

2s + 1 = multiplicity

Singlet: 2s + 1 = 1, s = 0
Doublet: 2s + 1 = 2, s = 1/2
Triplet: 2s + 1 = 3, s = 1

Electron configuration of Helium: 1s²

Pauli Exclusion Principle: no two electrons can have exactly the same quantum number.

m_s1 =  ½  and m_s2 =  −½
Total spin quantum number, s = ½ + (−½) = 0
Multiplicity, 2 · 0 + 1 = 1 → Singlet

Answer: A 

Nuclear & Particle Physics - Spectroscopic Notation

The configuration of three electrons is 1s2p3p has which of the following as the value of its maximum possible total angular momentum quantum number?

A. 7/2
B. 3
C. 5/2
D. 2
E. 3/2

(GR9277 #76)

Solution:

In Spectroscopic Notation, the total angular momentum quantum number is j = l + s, where
s = the total spin quantum number → 2n+1= multiplicity, the number of spin states
l = the orbital angular momentum quantum number

Since none of the electron sub-shells are filled, we add the individual angular momentum quantum number:
1s → l₁ = 0, s₁ = 1/2 → j₁ = 1/2
2p → l₂ = 1, s₂ = 1/2 → j₂ = 1+1/2 = 3/2
3p → l₃ = 1, s₃ = 1/2 → j₃ = 1 + 1/2 = 3/2

The maximum possible total angular momentum quantum number j = j₁ + j₂ + j₃ = 7/2

Answer: A

Nuclear & Particle Physics - Spectroscopic Notation

Sodium has eleven electrons and the sequence in which energy levels fill in atom is 1s, 2s, 2p, 3s, 3p, 4s, 3d, etc. What is the ground state of sodium in the usual notation ^2s+1L_j?

A. ¹S₀
B. ²S_1/2
C. ¹P₀
D. ²P_1/2
E. ³P_1/2

(GR8677 #84)

Solution:

11 electrons → 1s², 2s², 2p⁶, 3s¹

3s¹ → 1 electron with spin ½ → s = 1/2 → multiplicity = 2s + 1 = 2(1/2) + 1 = 2

3s¹ → Subshell s → l = 0 → j = l + s = 0 + 1/2 = 1/2

3s¹ → Subshell s → L = S

Spectroscopic Notation: ^2s+1L_j → ²S_1/2

Answer: B