- #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) isn't a function of theta or phi?