# Where does the maximum electron density occur for 2s and 2p orbitals in hydrogen atom?

Dec 28, 2015

For hydrogen, we have to use spherical harmonics, so our dimensions are written as $\left(r , \theta , \phi\right)$. The wave function is defined as follows, via separation of variables:

$\textcolor{g r e e n}{{\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right) = {R}_{n l} \left(r\right) {Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)}$

${R}_{n l} \left(r\right)$ is the radial component of the wave function ${\psi}_{n l {m}_{l}} \left(r , \theta , \phi\right)$, ${Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)$ is the angular component, $n$ is the principal quantum number, $l$ is the angular momentum quantum number, and ${m}_{l}$ is the projection of the angular momentum quantum number (i.e. $0 , \pm l$). The wave function represents an orbital.

If you don't understand all of that, that's fine; it was just for context.

To get the maximum electron density, you have to look at probability density curves.

If we plot $4 \pi {r}^{2} {R}_{n l} {\left(r\right)}^{2}$ against $r$, we get the probability density curves for an atomic orbital.

The $2 s$ orbital's plot looks like this:

From this, you can tell that the maximum electron density occurs near $5 {a}_{0}$ (with ${a}_{0} \approx 5.29177 \times {10}^{- 11} \text{m}$, the Bohr radius) from the center of the atom, and $4 \pi {r}^{2} {R}_{20} {\left(r\right)}^{2}$ is about $2.45$ or so.

From this similar diagram, we can compare the $2 s$ with the $2 p$ orbital:

Here, you should see that the $2 p$ orbital has a maximum electron density near about $4 {a}_{0}$ from the center of the atom, and the value of $4 \pi {r}^{2} {R}_{21} {\left(r\right)}^{2}$ is perhaps around $2.5$.

This should make more sense once you realize what the probability density plots of the $2 s$ and $2 p$ orbitals look like:

2s

2p

"The density of the [dark spots] is proportional to the probability of finding the electron in that region" (McQuarrie, Ch. 6-6).