- #1

Gregg

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## Homework Statement

The function

[itex] \Psi(r) = A(2-{Zr\over a})e^-{Zr\over 2a} [/itex]

gives the form of the quantum mechanical wavefunction representing the electron

in a hydrogen-like 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.

## The Attempt at a Solution

[itex] {\int \int \int}_R |\Psi|^2 dV = 1[/itex]

[itex] \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[/itex]

Which implies

[itex] \int _0^{\infty }A^2e^{-\frac{\text{Zr}}{a}} \left(2-\frac{\text{Zr}}{a}\right)^2dr = {1\over 2\pi^2}[/itex]

This turns out to be

[itex] \frac{2aA^2}{Z} = \frac{1}{2\pi^2} [/itex]

[itex] A = \pm \frac{\sqrt{\frac{z}{a}}}{2\pi} [/itex]

This is wrong though?

Is the problem the fact that Psi(r) isnt a function of theta or phi?