For a photochemical reaction A → B, 1 × 10–5 moles of ‘B’ were formed on absorption of 6.626 × 107 e

Question

For a photochemical reaction A → B, 1 × 10–5 moles of ‘B’ were formed on absorption of 6.626 × 107 erg at 360 nm. The quantum efficiency (molecules of ‘B’ formed per photon) is (NA = 6 × 1023)

Accepted Answer

$E = n.\frac{hc}{\lambda} \Rightarrow 6.626 = n \times \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{360 \times 10^{- 9}}$

∴ Mole of photons absorbed $= \frac{n}{\sim_{A}} = \frac{1.2 \times 10^{19}}{6 \times 10^{23}} = 2 \times 10^{- 5}$

∴ Quantum efficiency = $\frac{1 \times 10^{- 5}}{2 \times 10^{- 5}} = 0.5$

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