A molecule with angular momentum L and moment of inertia I has a rotational energy [tex]E = L^2 / 2I[/tex]. Since angular momentum is quantized, find the wavelength of the photons emitted in n=2 to n=1 transition of the H2 molecule. This molecule has a moment of inertia [tex]I = [tex]0.5mr^2[/tex], where m = 938Mev/c^2 and r = 0.074nm. My attempt is to say [tex]L = [[l(l+1)]^.5}\hbar[/tex] and use l =2 for n=2 state and l = 1 for n=1. Put these values for L into the equation for E. Then E2 - E1 = [tex] \triangle E. [/tex] [tex] \triangle E = hf, v = \lambda f. [/tex] => [tex] \lambda = hv / \triangle E = hc / \triangle E [/tex] I've no idea if I'm on the right track, couldn't find anything similar in the textbook.