Problem 9.5 3. In the textbook the Ziman band gap calculation uses a potential with positive and negative parts, where the positive part has a width of w, and the total function is periodic with period a. However, it is easier to use a periodic potential function that is zero except for a region of width w. a. Using this simplified potential function calculate the first 3 band gaps (i.e. n = 1, 2, 3) when w = a/4. b. For which value of n is the gap reduced to 10 % of its value for n = 1. w / 2 () = Sy* Vidx = SHIV. do - w / 2 nd = 2 a si o X 0- 900 six-1 Note different functions for symmetric and antisymmetric An = 29 wavefunctions due to choice of origin 4 = Acco nix , Ys = Asin a SW cooke ) k : 27 - nn Normalize Juz A' c hix dy = J. - a/z Vo = 0.5 a ) b n = 1 , gap = 0, 225 A' Z = 1 7 A = 12 n gap W/ z 0. =35 1090 = 0.025 ard x 2 0. 159 n =0 3 0. 075 4 O VA VS gap = Vs Va { {0. 23754 , 0 . 204577, 0.162513, 0.125) , {0. 0124605, 0. 0454225, 0. 0874868, 0.125) , (0. 225079, 0.159155, 0. 0750264,0.}}Using the Ziman model, the band gap was approximated by comparing the expectations of the potential energies for symmetric and antisymmetric wavefunctions. In class, the case of a relatively wide barrier was worked out, and the gap was found to rapidly decrease with increasing band number. Now consider a very narrow barrier. a. Approximately how narrow should the barrier be (compared to the well width) so that the band gap is nearly constant up to the 10th band? b. What should the barrier height be in this case so that the tunneling between wells is the same as for the wider band considered in class? c. Forthe barrier width and height computed above, approximately what is the band gap for the lowest band