Indian Forest Service (Mains), Chemistry optional · 2024 · Paper I
Q1(c). Primary quantum yield
Question
An absorbing substance is exposed to light of wavelength 400 nm using a 100 mW laser for 50 minutes. The transmitted intensity is 40% of incident intensity. As a result, 0.4 mmol of the substance decomposes. Determine the primary quantum yield.
Answer in brief
Assuming the missing 60% of light is absorbed and the decomposition measures the primary process, Φ ≃ 0.665.
Key points
- Absorbed power is Pabs = 0.100(1 − 0.40) = 0.060 W. Over 50(60) = 3000 s, Eabs = 180 J.
- One photon has E = hc/λ = (6.626 × 10⁻³⁴)(3 × 10⁸)/(400 × 10⁻⁹) = 4.9695 × 10⁻¹⁹ J.
- One mole of photons carries 2.9931 × 10⁵ J/mol, so moles of photons absorbed = 180/(2.9931 × 10⁵) = 6.014 × 10⁻⁴ mol.
- The quantum yield is Φ = ndecomposed/nphotons absorbed = (0.4 × 10⁻³)/(6.014 × 10⁻⁴) = 0.665.
- Use absorbed, not incident, photons in the denominator. Reflection and scattering are assumed negligible.
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.
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All parts of Question 1
- 1(a) Free-particle waves and de Broglie relation8 marks
- 1(b) Entropy of water-vapour formation at 373 K8 marks
- 1(c) Primary quantum yield8 marks
- 1(d) Components in ammonium chloride dissociation8 marks
- 1(e) Aluminium–magnesium galvanic cell8 marks