(a) In observing the Raman spectrum of a sample using 3637Å as the exciting line, one gets stoke line at 3980Å. Deduce the Raman…

(a) The Raman shift is the difference between the wavenumbers of the incident and Stokes-scattered radiation. For λ₀ = 3637 Å and λₛ = 3980 Å,

ν̄ = 1/λ, so

Δν̄ = 1/λ₀ − 1/λₛ = 1/3637 − 1/3980 Å⁻¹ = (3980 − 3637)/(3637 × 3980) Å⁻¹ = 343/14,475,260 Å⁻¹.

Since 1 Å⁻¹ = 10¹⁰ m⁻¹,

Δν̄ = (343/14,475,260) × 10¹⁰ m⁻¹ = 2.36956 × 10⁵ m⁻¹ = 2369.6 cm⁻¹.

Thus the Raman shift is Δν̄ = 2.37 × 10⁵ m⁻¹ = 2369.6 cm⁻¹.

Now take the exciting line λ₀ = 6465 Å. The same shift in Å⁻¹ is

a = 2.36956 × 10⁻⁵ Å⁻¹.

For the Stokes line,

1/λₛ = 1/6465 − a = 1.54679 × 10⁻⁴ − 2.36956 × 10⁻⁵ = 1.30983 × 10⁻⁴ Å⁻¹.

Hence

λₛ = 1/(1.30983 × 10⁻⁴) Å = 7634.55 Å.

For the anti-Stokes line,

1/λₐ = 1/6465 + a = 1.54679 × 10⁻⁴ + 2.36956 × 10⁻⁵ = 1.78375 × 10⁻⁴ Å⁻¹.

Hence

λₐ = 1/(1.78375 × 10⁻⁴) Å = 5606.18 Å.

Therefore, for exciting line 6465 Å, Stokes line λₛ ≈ 7635 Å and anti-Stokes line λₐ ≈ 5606 Å.

(b) Spin-orbit coupling arises from the interaction of the intrinsic spin magnetic moment of an electron with the magnetic field produced by its orbital motion in the electric field of the nucleus. In the electron’s rest frame, the nuclear Coulomb field appears partly as a magnetic field, and this field couples to the electron spin. The interaction is relativistic in origin and produces fine-structure splitting.

The spin-orbit Hamiltonian is written as

H_SO = ξ(r) L·S,

where

ξ(r) = (1/(2mₑ²c²))(1/r)(dV/dr).

For the hydrogen atom,

V(r) = −e²/(4π ε₀ r),

so

dV/dr = e²/(4π ε₀ r²)

and hence

ξ(r) = e²/(8π ε₀ mₑ² c² r³).

In the coupled representation, L and S combine to give J = L + S. Using

L·S = ½(J² − L² − S²),

the first-order energy shift is

E_SO = (ħ²/2) ξ̄ [j(j+1) − l(l+1) − s(s+1)],

where ξ̄ is the radial average of ξ(r). For one electron, s = 1/2, so j = l ± 1/2 for l ≠ 0; for l = 0, L·S has zero expectation value and there is no spin-orbit splitting. Thus each level with l ≠ 0 splits into two fine-structure levels, j = l + 1/2 and j = l − 1/2. For hydrogen, the j = l + 1/2 level lies higher in energy than the j = l − 1/2 level.

Because the upper and lower atomic levels involved in a spectral transition are split differently, a single spectral line splits into several closely spaced components. The electric dipole selection rules are

Δl = ±1, Δj = 0, ±1,

with j = 0 ↔ j = 0 forbidden. For example, the Balmer Hα line splits into a multiplet because the n = 3 and n = 2 levels have different j values. The spin-orbit splitting is very small, of order α² compared with the gross energy. This discussion is valid in first-order perturbation theory; for hydrogen, the complete fine structure also includes relativistic kinetic-energy and Darwin corrections, but spin-orbit coupling alone gives the j-dependent splitting.

(c)(i) In L-S coupling, the individual orbital angular momenta couple to give total L, and the spins couple to give total S.

Here l₁ = 4 and l₂ = 2. Therefore

L = |l₁ − l₂|, …, l₁ + l₂ = 2, 3, 4, 5, 6.

Also s₁ = s₂ = 1/2, so

S = 0 or 1.

For S = 0, J = L, so

J = 2, 3, 4, 5, 6.

For S = 1, J = |L − 1|, …, L + 1. Thus:

• L = 2 gives J = 1, 2, 3
• L = 3 gives J = 2, 3, 4
• L = 4 gives J = 3, 4, 5
• L = 5 gives J = 4, 5, 6
• L = 6 gives J = 5, 6, 7

Hence the possible L values are L = 2, 3, 4, 5, 6, and the overall possible J values are J = 1, 2, 3, 4, 5, 6, 7.

(c)(ii) In j-j coupling, each electron first couples its own l and s to give j.

For electron 1:

l₁ = 4, s₁ = 1/2,

so

j₁ = 4 ± 1/2 = 7/2 or 9/2.

For electron 2:

l₂ = 2, s₂ = 1/2,

so

j₂ = 2 ± 1/2 = 3/2 or 5/2.

Now couple j₁ and j₂ to give total J:

J = |j₁ − j₂|, …, j₁ + j₂.

Possible cases:

• j₁ = 9/2, j₂ = 5/2: J = 2, 3, 4, 5, 6, 7
• j₁ = 9/2, j₂ = 3/2: J = 3, 4, 5, 6
• j₁ = 7/2, j₂ = 5/2: J = 1, 2, 3, 4, 5, 6
• j₁ = 7/2, j₂ = 3/2: J = 2, 3, 4, 5

Therefore the possible J values in j-j coupling are J = 1, 2, 3, 4, 5, 6, 7.