Calculating Wavelength of Photons Emitted in H2 Molecule Transitions

[alfredbester]

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 [tex]I = $$0.5mr^2$$, where m = 938Mev/c^2 and r = 0.074nm.<br /> <br /> 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.<br /> Then E<span style="font-size: 10px">2 - E<span style="font-size: 10px">1 = $$\triangle E.$$<br /> <br /> $$\triangle E = hf, v = \lambda f.$$<br /> <br /> => $$\lambda = hv / \triangle E = hc / \triangle E$$<br /> <br /> <br /> I've no idea if I'm on the right track, couldn't find anything similar in the textbook.</span></span>[/tex]

 

[member 553083]

Yes, this is the correct approach. The equation you derived is:λ = hc / ΔEwhere h is Planck's constant, c is the speed of light, and ΔE is the energy difference between the two states. Plugging in the values for I and m, you can calculate the rotational energy for each state and find ΔE. Then, you can use the equation above to calculate the wavelength of the photons emitted in the transition.