Question

Assuming the Voo transition occurs at 280.38 nm and an anharmonicity parameter (xe = 0.0086 cm') calculate the number of vibrational energy levels below the dissociation limit and the equilibrium oscillation frequency (@c) in the excited state of CIO. 279.7 nm = 33590.86 cm-1 Voo 280.38 nm = 35665.88 cm-1 Intersity At Cris) 280.2 280.-1 279.8 280.0 Wavelength(nm)

I'm okay for the first part of the question, I only need to obtain the oscillation frequency for the excited state

Answer 1

120€ Vou 2 280.38 nm - Wosc = o voo = 35665.88 cm emharmonicity constant = 0.00 Bocu'? The = equilibrium oscillation frequency I wase - we (utl) - We xe(0 +1² ground state vzo 85 665.88 = we (o.5-0.00215) - 0.4g785 we 11639.811 an Cena equilibrium frequency of enited state = = wese - we . 3 wexexa 4.

DATE: 1 Cosas we [3 - 0.0048375 e x 1.49 S1628 WoSL = 107113.158 cm required answer