In the CO2 laser, the pumping action involves exciting N2 molecules to the v=1 state. If ~ve=2358.57cm^-1 and ~vexe=14.324cm^-1, what is the energy difference between this transition and the transition in the CO2 from the ground to 1 quanta of energy in the asymmetric stretch.© BrainMass Inc. brainmass.com November 29, 2021, 11:54 pm ad1c9bdddf
dE = (2358.57 - 14.324)*100 * h*c
h = Plank's constant = 6.63*10^(-34) J-s
c = speed of light = 3*10^8 m/s
h*c = 6.63*10^(-34) * 3*10^8 ...
The energy difference is found. The asymmetric stretch of energy is examined. Solution provides equations, calculations and answer.