What is the correct A value for normalizing the given wave function?

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SUMMARY

The correct normalization constant A for the wave function Ψ(r) = A(2 - (Zr/a))e^(-Zr/2a) representing an electron in a hydrogen-like atom is determined through a volume integral of |Ψ|^2. The integral must equal unity, leading to the equation ∫_0^∞ A^2 e^(-Zr/a) (2 - (Zr/a))^2 dr = 1. The final result for A is A = ±√(Z/a)/(2π), confirming that the initial calculation was incorrect due to the omission of the r² sin(θ) term in the volume element.

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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) isn't a function of theta or phi?
 
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