Q1(p). Energy of one einstein at 600 nm and 300 nm

Answer in brief

E(600 nm) = 199.4 kJ mol⁻¹ and E(300 nm) = 398.8 kJ mol⁻¹; E(600) : E(300) = 1 : 2 (ratio 0.5).

Key points

• One einstein is one mole (NA) of photons: E = NAhc/λ.
• E(600) = (6.022 × 10²³ × 6.626 × 10⁻³⁴ × 2.998 × 10⁸)/(600 × 10⁻⁹ m) = 1.994 × 10⁵ J mol⁻¹ = 199.4 kJ mol⁻¹.
• E(300) = 0.11963 J m mol⁻¹/(3.00 × 10⁻⁷ m) = 3.988 × 10⁵ J mol⁻¹ = 398.8 kJ mol⁻¹.
• E(600)/E(300) = 300/600 = 0.5, i.e. 1 : 2; energy ∝ 1/λ (per photon 2.07 eV and 4.13 eV).
• 300 nm UV (399 kJ mol⁻¹) exceeds many bond energies (C–C ≈ 348 kJ mol⁻¹), so UV causes photolysis; 600 nm light cannot break such bonds.