UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q10. Fluorescence and phosphorescence from a Jablonski diagram
Question
Using a Jablonski diagram, explain the following: (i) Why fluorescence lifetimes are typically much shorter than phosphorescence lifetimes. (ii) Why phosphorescence is usually observed at longer wavelengths. (iii) Why fluorescence intensity decreases in the presence of heavy atoms.
Answer in brief
(i) S₁→ S₀ is spin-allowed (10⁻⁹–10⁻⁸ s); T₁→ S₀ is spin-forbidden (10⁻³–10 s). (ii) T₁ lies below S₁, so the emitted photon has less energy. (iii) Heavy atoms raise the spin–orbit coupling, so S₁→ T₁ intersystem crossing speeds up and fluorescence is quenched.
Key points
- (i) Lifetime (3 marks): fluorescence is S₁→ S₀ with Δ S=0: allowed, k_f 10⁷–10⁹ s⁻¹, τ_f 1–10 ns. Phosphorescence is T₁→ S₀ with Δ S≠0: forbidden, relaxed only weakly by spin–orbit coupling, kₚ 10⁻¹–10³ s⁻¹, τₚ from ms to s.
- (ii) Wavelength (3 marks): the triplet T₁ is lower than S₁ by twice the exchange integral K (Hund's rule: parallel spins keep the electrons apart and lower the repulsion). The phosphorescence photon hνₚ=E(T₁)-E(S₀) is smaller…
- (iii) Heavy atoms (4 marks): the spin–orbit coupling constant ζ∝ Z⁴. A heavy atom (Br, I, Pb), in the molecule or in the solvent, mixes singlet and triplet character, so k_(ISC)∝ζ² rises. Then Φ_f=k_fk_f+k_(IC)+k_(ISC) falls…
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
- (i) Lifetime (3 marks): fluorescence is S₁→ S₀ with Δ S=0: allowed, k_f 10⁷–10⁹ s⁻¹, τ_f 1–10 ns. Phosphorescence is T₁→ S₀ with Δ S≠0: forbidden, relaxed only weakly by spin–orbit coupling, kₚ 10⁻¹–10³ s⁻¹, τₚ from ms to s.
- (ii) Wavelength (3 marks): the triplet T₁ is lower than S₁ by twice the exchange integral K (Hund's rule: parallel spins keep the electrons apart and lower the repulsion). The phosphorescence photon hνₚ=E(T₁)-E(S₀) is smaller than hν_f=E(S₁)-E(S₀), so λₚ>λ_f (typically 50–100 nm).
- (iii) Heavy atoms (4 marks): the spin–orbit coupling constant ζ∝ Z⁴. A heavy atom (Br, I, Pb), in the molecule or in the solvent, mixes singlet and triplet character, so k_(ISC)∝ζ² rises. Then Φ_f=k_fk_f+k_(IC)+k_(ISC) falls as k_(ISC) grows: the heavy-atom effect. The triplet population increases, so phosphorescence increases (and τₚ falls).
CoversJablonski diagramFluorescence and phosphorescence from a Jablonski diagram
All parts of Question 10
- 10 Fluorescence and phosphorescence from a Jablonski diagram10 marks