I Demerits radial distribution functions

[kenyanchemist]

i have a question, why is the plot of r²(Ψ2p)² not a good representation of the probability of finding an electron at a distance r from the nucleus in a 2p orbital

 

[vanhees71]

Well, in spherical coordinates you have
$$\mathrm{d}^3 \vec{x} = \mathrm{d} r \mathrm{d} \vartheta \mathrm{d} \varphi r^2 \sin \vartheta.$$
The wave function squared is the probability distribution. So the probability for finding a particle described by the wave function at a distance $r$ is
$$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, r^2 \sin \vartheta |\psi(r,\vartheta,\varphi)|^2.$$
Note that sometimes one writes
$$\psi(r, \vartheta,\varphi)=\frac{1}{r} u(r,\vartheta,\varphi),$$
because this ansatz simplifies the solution of the Schrödinger equation in spherical coordinates. In terms of $u$ the factor $r^2$ cancels in the above integral.
$$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, \sin \vartheta |u(r,\vartheta,\varphi)|^2.$$

 

[kenyanchemist]

Well, in spherical coordinates you have
$$\mathrm{d}^3 \vec{x} = \mathrm{d} r \mathrm{d} \vartheta \mathrm{d} \varphi r^2 \sin \vartheta.$$
The wave function squared is the probability distribution. So the probability for finding a particle described by the wave function at a distance $r$ is
$$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, r^2 \sin \vartheta |\psi(r,\vartheta,\varphi)|^2.$$
Note that sometimes one writes
$$\psi(r, \vartheta,\varphi)=\frac{1}{r} u(r,\vartheta,\varphi),$$
because this ansatz simplifies the solution of the Schrödinger equation in spherical coordinates. In terms of $u$ the factor $r^2$ cancels in the above integral.
$$P(r)=\int_0^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \, \sin \vartheta |u(r,\vartheta,\varphi)|^2.$$

What do you mean by cancels?
I apologize, I am relatively new at quantum mechanics

 

[kenyanchemist]

and thanks so much... you have really helped me alot