Q1(o). Why radiative transitions are drawn vertical

Answer in brief

Franck–Condon principle: a photon is absorbed or emitted in about 10⁻¹⁵ s, far faster than nuclei move (10⁻¹³ s); the nuclear geometry does not change, so the transition is vertical.

Key points

• Electronic transition time ≈ 10⁻¹⁵ s; one nuclear vibration ≈ 10⁻¹³ s; the heavy nuclei are effectively stationary during the jump (Born–Oppenheimer separation).
• The internuclear distance is the same before and after, so the transition is vertical on an energy–nuclear-coordinate diagram; the Jablonski diagram keeps this convention.
• Only the energy changes, by exactly hν; the molecule lands in the vibrational level of best overlap (Franck–Condon factor).
• Radiationless steps involve no photon: internal conversion and intersystem crossing are horizontal wavy arrows; vibrational relaxation (10⁻¹² s) is a wavy arrow down the ladder.
• Emission starts from v = 0 of S₁ (fluorescence) or T₁ (phosphorescence) (Kasha), again vertically.