
#1
Mar110, 12:57 PM

P: 463

1. The problem statement, all variables and given/known data
The function [latex] \Psi(r) = A(2{Zr\over a})e^{Zr\over 2a} [/latex] gives the form of the quantum mechanical wavefunction representing the electron in a hydrogenlike atom of atomic number Z when the electron is in its first allowed spherically symmetric excited state. Here r is the usual spherical polar coordinate, but, because of the spherical symmetry, the coordinates θ and φ do not appear explicitly in Ψ. Determine the value that A (assumed real) must have if the wavefunction is to be correctly normalised, i.e. the volume integral of Ψ^2 over all space is equal to unity. 3. The attempt at a solution [latex] {\int \int \int}_R \Psi^2 dV = 1[/latex] [latex] \int _0^{\infty }\int _0^{2\pi }\int _0^{\pi }A^2e^{\frac{Zr}{a}} \left(2\frac{Zr}{a}\right)^2d\phi d\theta dr = 1[/latex] Which implies [latex] \int _0^{\infty }A^2e^{\frac{\text{Zr}}{a}} \left(2\frac{\text{Zr}}{a}\right)^2dr = {1\over 2\pi^2}[/latex] This turns out to be [latex] \frac{2aA^2}{Z} = \frac{1}{2\pi^2} [/latex] [latex] A = \pm \frac{\sqrt{\frac{z}{a}}}{2\pi} [/latex] This is wrong though? Is the problem the fact that Psi(r) isnt a function of theta or phi? 



#2
Mar110, 01:16 PM

HW Helper
P: 482

Looks to me you forgot the r^{2} sin term from the differential volume element (note that the way you did it, the volume of a unit ball would come out as 2pi^{2}). See http://en.wikipedia.org/wiki/Spheric...rdinate_system



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