In: Physics
1. Write the Schrodinger equation for particle on a ring, and rearrange it until you have the following: ? 2? ??2 = − 2?? ℏ 2 ? …
a) Assuming that ?? 2 = 2?? ℏ 2 , (where ml is a quantum number and has nothing to do with mass), show that the following is a solution for the Schrodinger equation you obtained: ?(?) = ? ? ????…
b)Now think about bounds of variable ?. Using that argue that ?(?) = ? ( ?+2pi), and prove that ml can be 0, ±1, ±2, ±3, ±4, Normalize that ?(?)
c)Prove that particle on the ring will have discrete energy levels described by the following equation, ??? = ?? 2ℏ 2 2? …
d) Now we will apply these solutions to Benzene molecule. Each double bond in benzene is 1.4Å, so you know the circumference of benzene. Now calculate the radius of benzene using the circumference. There are six -electrons on benzene, which are free to move around the ring due to conjugated double bonds. Make an energy level diagram using equation 6, and calculate the wavelength for the lowest energy electronic transition. Experimentally observed transition for benzene is at 200 nm. (Hint: The ground state will be ml = 0, and then rest of the energy levels are going to be doubly degenerate.)