a) The physical interpretation of the wavefunction and the fact that it is a solution of the Schroedinger equaton, which is a 2nd order differential equation, causes many restrictions on an acceptable wave function solution: (i) it must be single valued; (ii) it must be continuous; (iii) its slope must be continuous; and (iv) it must be normalizable or normalized. Sketch the following functions and check whether they can be wave function. Explain your answers. 1) Y(x) =, for 0 < x < ∞ 2) 4(x) = | sin(nx)|, for 0 < x < 2 3) Y(x) = In (x), for 1 < x < ∞ (0 for – 1 < x < 0 1+x - 4) ¥(x) = {1 for 0

a) The physical interpretation of the wavefunction and the fact that it is a solution of the Schroedinger equaton, which is a 2nd order differential equation, causes many restrictions on an acceptable wave function solution: (i) it must be single valued; (ii) it must be continuous; (iii) its slope must be continuous; and (iv) it must be normalizable or normalized. Sketch the following functions and check whether they can be wave function. Explain your answers. 1) Y(x) =, for 0 < x < ∞ 2) 4(x) = | sin(nx)|, for 0 < x < 2 3) Y(x) = In (x), for 1 < x < ∞ (0 for – 1 < x < 0 1+x - 4) ¥(x) = {1 for 0

Physical Chemistry

2nd Edition

ISBN:9781133958437

Author:Ball, David W. (david Warren), BAER, Tomas

Publisher:Ball, David W. (david Warren), BAER, Tomas

Chapter12: Atoms And Molecules

Section: Chapter Questions

Problem 12.52E

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