# Wave function problem

Gregg

## Homework Statement

The function

$\Psi(r) = A(2-{Zr\over a})e^-{Zr\over 2a}$

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

${\int \int \int}_R |\Psi|^2 dV = 1$

$\int _0^{\infty }\int _0^{2\pi }\int _0^{\pi }A^2e^{-\frac{Zr}{a}} \left(2-\frac{Zr}{a}\right)^2d\phi d\theta dr = 1$

Which implies

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

This turns out to be

$\frac{2aA^2}{Z} = \frac{1}{2\pi^2}$

$A = \pm \frac{\sqrt{\frac{z}{a}}}{2\pi}$

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