Q 7: An electronic system with 12 л-electrons (all of them in conjugation) can be treated as a particle in a 2-D box of size of 3 Å x 6 Å. Calculate the wavelength corresponding to the lowest energy electronic transition for this system. The expression for the energy of a 2-D box with sides of length a and b is h² (n² Enxny=8me + where nx and ny are quantum numbers (1, 2, 3, 4, etc.). These two quantum numbers are independent of each other and do not have to be equal, though they will be equal for some energy levels, for example the lowest energy state (the ground state) is E11. Some of the energy levels will degenerate.

A quantum system has a ground state with energy Eo = 0 meV and a 2-fold degenerate excited state with energy E₁ = 50 meV. E1 Calculate the probability of finding the system in its ground state when it is at T = 300 K. Select one: O a. 0.78 O b. 0.22 O c. 1 O d. 0.87

2-d particle in a box problem. Benzene can be treated as a 2-d box with sides of length 3.5 Å and containing 6 7-electrons. What wavelength of light would be required to promote an electron from the ground state to the first excited electronic state? Answer 135 nm Hint: Use the result for the energy of a particle in a 2-d box: 7²h? Y + (5.1) Nx,Ny 2me L? L3 1,2, 3... and те 9.8 × 10-31kg You need with ng = 1,2, 3... and ny the first 3 energy levels for the 6e¯, and the transition energy will be the energy between the HOMO and the LUMO (the 4th energy level).

Physics 1. Derive the expression ε(ω)=1- ωp 2 / ω2 , ωp2 =ne2 /ε0m for the dielectric constant as a function of ω for a free electron gas of number density n.   2. Show clearly that metals are opaque to light for which ω is less than ωp.   3. Calculate the wavelength cutoff for Na metal if the volume of a primitive unit cell in Na is 35×10-30 m3   how to solve this problem?