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:

Graphs of radial wavefunction and radial distribution functions of 1s orbital

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])

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    $\begingroup$ 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! $\endgroup$ Commented Nov 16, 2018 at 23:18
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    $\begingroup$ It is actually a quite common question. For beginners it is often not obvious where that $r^2$ term is coming from. $\endgroup$
    – Feodoran
    Commented Nov 17, 2018 at 0:28

2 Answers 2


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

\begin{equation} 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 \end{equation}

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 $r$

\begin{equation} 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 \end{equation}

which we can rearrange to

\begin{equation} 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 \end{equation}

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 \begin{equation} P(r) = |R(r)|^2r^{2} 4\pi \end{equation}

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 $r=0$.

  • $\begingroup$ Why did we omit the integral over $r$ to find $P(r) ? $ $\endgroup$ Commented Apr 19, 2020 at 7:52
  • $\begingroup$ @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". $\endgroup$
    – Feodoran
    Commented Apr 19, 2020 at 9:18
  • $\begingroup$ It is a bit difficult to understand as I am a highschool student.. Please elaborate a little more. $\endgroup$ Commented Apr 19, 2020 at 16:13
  • $\begingroup$ @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. $\int_0^2 x \mathrm{d}x = 2$. $\endgroup$
    – Feodoran
    Commented Apr 19, 2020 at 18:12

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