# What is the distance of a 2s electron from the nucleus of H atom?

Aug 12, 2017

It is $6$ times the Bohr radius, but you have to suffer through a gauntlet of integration by parts.

The $2 s$ atomic orbital wave function for $\text{H}$ atom is:

${\psi}_{2 s} = \frac{1}{4 \sqrt{2 \pi}} {\left(\frac{1}{a} _ 0\right)}^{3 / 2} \left(2 - \frac{r}{a} _ 0\right) {e}^{- r / 2 {a}_{0}}$

The average value in general is given by

$\left\langlex\right\rangle = {\int}_{\text{allspace}} x p \left(x\right) \mathrm{dx}$,

where $p \left(x\right)$ would be the probability distribution function.

Since ${\psi}^{\text{*}} \psi$ is the probability density, the average distance from the nucleus, or $\left\langler\right\rangle$, the expectation value for the radial position, is found by solving the following integral:

$\left\langler\right\rangle = {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} r \cdot {\psi}_{2 s}^{\text{*}} {\psi}_{2 s} \cdot {r}^{2} \mathrm{dr} \sin \theta d \theta d \phi$

$= \frac{1}{32 \pi} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} {\left[\left(2 - \frac{r}{a} _ 0\right) {e}^{- r / 2 {a}_{0}}\right]}^{2} {r}^{3} \mathrm{dr} \sin \theta d \theta d \phi$

$= \frac{1}{32 \pi} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} \left(4 - \frac{4 r}{a} _ 0 + {\left(\frac{r}{a} _ 0\right)}^{2}\right) {r}^{3} {e}^{- r / {a}_{0}} \mathrm{dr} \sin \theta d \theta d \phi$

$= \frac{1}{32 \pi} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{2 \pi} {\int}_{0}^{\pi} {\int}_{0}^{\infty} \left(4 {e}^{- r / {a}_{0}} - \frac{4 r}{a} _ 0 {e}^{- r / {a}_{0}} + {\left(\frac{r}{a} _ 0\right)}^{2} {e}^{- r / {a}_{0}}\right) {r}^{3} \mathrm{dr} \sin \theta d \theta d \phi$

The integral ${\int}_{0}^{2 \pi} {\int}_{0}^{\pi} \sin \theta d \theta d \phi = 4 \pi$, so...

$= \frac{1}{8} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{\infty} \left(4 {e}^{- r / {a}_{0}} - \frac{4 r}{a} _ 0 {e}^{- r / {a}_{0}} + {\left(\frac{r}{a} _ 0\right)}^{2} {e}^{- r / {a}_{0}}\right) {r}^{3} \mathrm{dr}$

$= \frac{1}{8} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{\infty} 4 {r}^{3} {e}^{- r / {a}_{0}} - \frac{4 {r}^{4}}{a} _ 0 {e}^{- r / {a}_{0}} + {\left(\frac{r}{a} _ 0\right)}^{2} {r}^{3} {e}^{- r / {a}_{0}} \mathrm{dr}$

$= \frac{1}{8} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{\infty} 4 {r}^{3} {e}^{- r / {a}_{0}} \mathrm{dr} - \frac{1}{8} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{\infty} \frac{4 {r}^{4}}{a} _ 0 {e}^{- r / {a}_{0}} \mathrm{dr} + \frac{1}{8} {\left(\frac{1}{a} _ 0\right)}^{3} {\int}_{0}^{\infty} {\left(\frac{r}{a} _ 0\right)}^{2} {r}^{3} {e}^{- r / {a}_{0}} \mathrm{dr}$

I'm not going to go past this though, because it is just ugly integration by parts. Using Wolfram Alpha...

• The first integral gives $3 {a}_{0}$.
• The second gives $12 {a}_{0}$.
• The third gives $15 {a}_{0}$.

So, the result is:

$\textcolor{b l u e}{\left\langler\right\rangle} = 3 {a}_{0} - 12 {a}_{0} + 15 {a}_{0} = \textcolor{b l u e}{6 {a}_{0}}$

And to check, someone on stackexchange got a general formula:

$\left\langler\right\rangle = {a}_{0} / \left(2 Z\right) \left(3 {n}^{2} - l \left(l + 1\right)\right)$

$= {a}_{0} / \left(2 \cdot 1\right) \left(3 {\left(2\right)}^{2} - 0 \left(0 + 1\right)\right) = 6 {a}_{0}$ color(blue)(sqrt"")