alfredbester
Feb28-06, 05:43 PM
A molecule with angular momentum L and moment of inertia I has a rotational energy E = L^2 / 2I. 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 I = [tex]0.5mr^2, where m = 938Mev/c^2 and r = 0.074nm.
My attempt is to say L = [[l(l+1)]^.5}\hbar 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 = \triangle E.
\triangle E = hf, v = \lambda f.
=> \lambda = hv / \triangle E = hc / \triangle E
I've no idea if I'm on the right track, couldn't find anything similar in the textbook.
My attempt is to say L = [[l(l+1)]^.5}\hbar 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 = \triangle E.
\triangle E = hf, v = \lambda f.
=> \lambda = hv / \triangle E = hc / \triangle E
I've no idea if I'm on the right track, couldn't find anything similar in the textbook.