Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Photochemistry · Topic: Photochemistry – General · Marks: 2 · Difficulty: Medium
Substance 'A' is exposed to 600 nm, 100 W light source for 6626 s, with $50 \%$ of the incident light being absorbed. 'A' decomposes according to the reaction A → 2B . As a result of irradiation, 0.2 mol B is produced. The quantum yield of the reaction is closest to
(a)$6 \times 10^{6}$
(b)$6 \times 10^{-4}$
(c)$6 \times 10^{-2}$
(d)6
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
Absorbed energy = 0.5 × 100 W × 6626 s = 3.313 × 10⁵ J. Photon energy is 3.313 × 10⁻¹⁹ J, so 1.66 mol of photons were absorbed. A consumed = 0.1 mol, giving Φ ≈ 0.06 = 6 × 10⁻² (c).