UPSC Combined Geo-Scientist, Mains · 2018 · Mains Paper II
Q1(m). Anthracene photodimerization
Question
Anthracene in benzene undergoes 2C₁₄H₁₀+hν→C₂₈H₂₀. What quantum yield is expected ideally? Explain an observed value of 0.5.
Answer in brief
Ideally one absorbed photon produces one dimer and consumes two anthracene molecules: Φ_(dimer)=1 or Φ_(disappearance)=2, depending on the stated convention.
Key points
- Absorption gives A*; a productive encounter with ground-state anthracene forms the dimer: A+hν→ A*, then A*+A→ A₂.
- Not every excited molecule reaches a productive encounter. Fluorescence, internal conversion, intersystem crossing and quenching compete with dimer formation, reducing the chemical yield.
- If 0.5 denotes anthracene disappearance per photon, the productive fraction is 0.5/2=0.25.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

CoversAnthracene photodimerizationPhotochemistry
All parts of Question 1
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- 1(b) Fugacity under repulsion5 marks
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- 1(e) Joule–Thomson cooling5 marks
- 1(f) Gibbs and Helmholtz changes5 marks
- 1(g) Equilibrium-constant conventions5 marks
- 1(h) Very dilute hydrochloric acid5 marks
- 1(i) Pressure controlling boiling5 marks
- 1(j) Primary kinetic salt effect5 marks
- 1(k) Heat of spontaneous adsorption5 marks
- 1(l) Conductivity as an intensive property5 marks
- 1(m) Anthracene photodimerization5 marks
- 1(n) Particle-in-a-box eigenvalue5 marks
- 1(o) Second Balmer line5 marks
- 1(p) Normal-mode counts5 marks