I'm having trouble understanding the following two graphs, which are the radial wavefunction of the hydrogen 1s orbital and the corresponding radial distribution function:

Specifically, why is the radial distribution function zero at $$r = 0$$, but not the radial wavefunction?

Also, why does the radial distribution function have a maximum at the Bohr radius $$a_0$$?

(Mathematica input for the graphs: Plot[{2*Exp[-x], 4*Exp[-2 x]*x^2}, {x, 0, 6}, AxesLabel -> {r/Subscript[a, 0], Subscript[f, "1s"][r]}, PlotRange -> {{0, 6}, {0, 2}}, PlotLegends -> {"Radial wavefunction", "Radial distribution function"}, BaseStyle -> {FontSize -> 14}, ImageSize -> Medium])

• I admit I have a vested interest in this, since I made some nice graphs for it. But... please stop voting to close. Not everything has to be closed as homework. If you foresee a question can have useful answers, consider erring on the side of leniency. And it's already got two, for goodness' sake! – orthocresol Nov 16 '18 at 23:18
• It is actually a quite common question. For beginners it is often not obvious where that $r^2$ term is coming from. – Feodoran Nov 17 '18 at 0:28

The radial wave function $$R(r)$$ is simply the value of the wave function at some radius $$r$$, and its square is the probability of the finding an electron in some infinitesimal volume element around a point at distance $$r$$ from the nucleus.

But, the infinitesimal volume of space at radius $$r$$ is $$4\pi r^{2} dr$$ (it's a spherical shell with thickness $$dr$$ at radius $$r$$). That means that probability of finding an electron at radius $$r$$ is proportional to $$R^{2}(r)4\pi r^{2}$$. But the behavior of this function is such that the probability of finding the electron at radius 0 is also 0.

The radial distribution has a different form due to integration over the angles:

If we take the absolute square of the wave function $$\Psi$$ and integrate over the whole volume, we get the Norm of the wave function

$$$$N = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ \int \limits _{r=0}^{\infty } |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi$$$$

where $$r^{2}\sin \theta \,\mathrm {d} r\,\mathrm {d} \theta \,\mathrm {d} \varphi$$ comes from transforming the infinitesimal volume element from cartesian coordinates ($$\mathrm{d} V = \mathrm {d} x\,\mathrm {d} y\,\mathrm {d} z$$) to spherical coordinates.

For the radial probability distribution $$P(r)$$ we do the same, but omit the integral over $$r$$

$$$$P(r) = \int \limits _{\varphi =0}^{2\pi }\ \int \limits _{\theta =0}^{\pi }\ |\Psi(r,\theta ,\varphi )|^2r^{2}\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi$$$$

which we can rearrange to

$$$$P(r) = |R(r)|^2r^{2}\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi$$$$

Where $$R(r)$$ is the radial part of the wave function $$\Psi(r,\theta ,\varphi)=R(r)Y(\theta ,\varphi)$$. For the $$1s$$ orbital we $$Y(\theta ,\varphi)=1$$ and straight forward integration over $$\theta$$ and $$\varphi$$ yields $$$$P(r) = |R(r)|^2r^{2} 4\pi$$$$

So no matter what orbital $$\Psi$$ represents, $$P(r)$$ always has a factor $$r^2$$, which means we always have $$P(r=0)=0$$.

Your given example is the $$1s$$ orbital $$\Psi_{1s}\propto \exp(-r)$$. In contrast to all other orbitals, it does not have a polynomial in $$r$$. Therefore it has no node at $$r=0$$.

• Why did we omit the integral over $r$ to find $P(r) ?$ – Nikola Alfredi Apr 19 '20 at 7:52
• @NikolaAlfredi Because we are interested in how the probability changes over $r$. If we integrate over $r$, the resulting function is no longer a "radial distribution". – Feodoran Apr 19 '20 at 9:18
• It is a bit difficult to understand as I am a highschool student.. Please elaborate a little more. – Nikola Alfredi Apr 19 '20 at 16:13
• @NikolaAlfredi I am not sure what exactly it is you don't understand. The question is about radial distribution, by definition this excludes integration over $r$. If we integrate over a variable, the result no longer explicitly depends on that variable. For example if we integrate a simple function $f(x)$ over $x$, the result is a scalar number, not a function of $x$, e.g. $\int_0^2 x \mathrm{d}x = 2$. – Feodoran Apr 19 '20 at 18:12
• Ok, I got it.. thank you. – Nikola Alfredi Apr 20 '20 at 5:19