Question

ao pts.) The energy eigenvalues for some quantum particle are descr expression e, - ap where p is the quantum num values including zero (i.e, .p o, 1, 2,) and a-5 10 for some quantum particle are described by the p where p is the quantum number that can take on positive integer the analytical expression for the molecular partition function qr To receive eredit, you must clearly state any approximations or assumptions that you make. (b) At T 300 K, how many of these particles would be found in the first excited state if there are 1o* total particles in the system?

Answer 1

Energy e_p = ap with p = 0,1,2,.... the position function z = Sigma_p = 0^infinity e^-(elementof p)beta z = Sigma_p = 0^infinity exp[-ap beta] z = 1 + e^-a beta + e^-2 a beta + e^- 3a beta+..... it is Gematric progression series whose sum unto infinite term us given by S = a/1 - r a = flirt term r = common ratio So, z = 1/1 - r^-a beta This is the required position function part rightarrow (b) for first excited state p = 1 elementof_1 ap = a So, probability in first existed state = (e^-app/z) = e^-a beta/1/1 - e^-a beta p(elementof_1) = (e^-a beta)(1 - e^- a beta) P(elementof_1) = (e^-a beta - e^-2 a beta) a = 5 times 10^-22 beta = 1/KT = 1/1.38 times 10^-23 times 300 a beta = 5 times 10^-22/1.38 times 10^-21 times 3 = 120 So P(elementof_1) = e^-(.12) - e^-(.24) p(elementof_1) = (.886 - .786) = 0.1 p(elementof) = 0.1 hence number of particle in first excited state will be = P(elementof_1) times total particle = 0.1 times 10^4 = 10^3 atom 10^3 particle will be