# For the Hydrogen atomic 2s wave function given by psi = 1/(2sqrt(2pi)) sqrt(1/a_0) (2 - r/a_0)e^(-r//2a_0), at what radial distance away from the nucleus can no electrons be found?

Feb 11, 2016

${r}_{0} = 2 \cdot {a}_{0}$

#### Explanation:

The key to this problem lies with what characterizes a radial node.

Basically, the wave function, $\Psi \left(x\right)$, is simply a mathematical function used to describe a quantum object.

The wave function that describes an electron in an atom is actually a product between the radial wave function, which is of interest in your case, and the angular wave function.

The radial wave function depends only on the distance from the nucleus, $r$.

Now, a node occurs when a wave function changes signs, i.e. when its passes through zero. A radial node occurs when a radial wave function passes through zero.

The important thing to remember about nodes is that an electron has zero probability of being located at a node. The probability of an electron being located at a particular point is given by the square of the absolute value of the wave function, $| \Psi \left(x\right) {|}^{2}$.

Since you have zero probability of locating an electron at a node, you can say that you have

$\textcolor{b l u e}{| \Psi \left(x\right) {|}^{2} = 0} \to$ this is true at nodes

So, you are given the wave function of a 2s-orbital

${\Psi}_{2 s} = \frac{1}{2 \sqrt{2 \pi}} \cdot \sqrt{\frac{1}{a} _ 0} \cdot \left(2 - \frac{r}{a} _ 0\right) \cdot {e}^{- \frac{r}{2 {a}_{0}}}$

and told that at $r = {r}_{0}$, a radial node is formed. Right from the start, this tells you that you have

$| {\Psi}_{2 s} {|}^{2} = 0$

Now, take a look at the wave function again. The only way to get the square of its absolute value equal to zero is if you have

$\left(2 - \frac{r}{a} _ 0\right) = 0$

since

${\Psi}_{2 s} = {\overbrace{\frac{1}{2 \sqrt{2 \pi}} \cdot \sqrt{\frac{1}{a} _ 0}}}^{\textcolor{p u r p \le}{> 0}} \cdot \left(2 - \frac{r}{a} _ 0\right) \cdot {\overbrace{{e}^{- \frac{r}{2 {a}_{0}}}}}^{\textcolor{p u r p \le}{> 0}}$

This means that you have

$2 - \frac{r}{a} _ 0 = 0 \implies r = 2 \cdot {a}_{0}$

At $r = {r}_{0}$, you will have

$\textcolor{g r e e n}{{r}_{0} = 2 \cdot {a}_{0}}$

Here's how the wave function for the 2s-orbital looks like A 2s-orbital is characterized by the fact that it has no directional properties - you get the exact same value for its wave function regardless of the value of $r$.

This is why the 2s-orbital is spherical in shape. Moreover, this tells you that the wave function changes signs at the same distance from the nucleus in all directions, which is why a nodal surface is formed.

The wave function of a 2s-orbital changes signs once, so you only have one nodal surface here.

Feb 11, 2016

Here's an alternate approach.

Since the wave function shown has no time variable, let us define $\Psi = \psi$ where $\psi$ is the time-independent wave function. (Normally, $\Psi$ is the time-dependent wave function.)

SEPARATION OF VARIABLES GIVES RADIAL AND ANGULAR COMPONENTS

From the wave function you gave, you are showing ${\psi}_{2 s}$, which for hydrogen is defined in spherical coordinates via separation of variables to give a radial and angular component:

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

$\setminus m a t h b f \left({\psi}_{2 s} \left(r , \theta , \phi\right) = {R}_{20} \left(r\right) {Y}_{0}^{0} \left(\theta , \phi\right)\right)$

Now, the radial component in general is (Inorganic Chemistry, Miessler):

color(green)(R_(20)(r) = 2(Z/(2a_0))^"3/2"(2-(Zr)/a_0)e^(-Zr"/"2a_0))

And the angular component in general is $\textcolor{g r e e n}{\frac{1}{2 \sqrt{\pi}}}$. Fortunately there is no $\theta$ or $\phi$ term to complicate things here.

You can also tell that if you substitute in $Z = 1$, you get your posted function.

NODES ARE FOUND WHEN THE PROBABILITY DENSITY IS 0

Now, you have a node wherever ${\psi}^{\text{*}} \psi$ (for real numbers, ${\psi}^{2}$), the probability density, as a whole is $0$.

For real numbers, $\setminus m a t h b f {\left({\psi}^{2} \left(r , \theta , \phi\right) = {R}_{n l}^{2} \left(r\right) \left({Y}_{l}^{{m}_{l}} \left(\theta , \phi\right)\right)\right)}^{2}$.

On the radial distribution graph for ${a}_{0} {r}^{2} {R}_{n l}^{2} \left(r\right)$ vs. $r \text{/} {a}_{0}$ (always positive), which plots probability density vs. viewing-window radius $r$, it touches $R = 0$ when $r = 0$ or ${R}_{n l}^{2} \left(r\right) = 0$.

However, $r = 0$ doesn't count as a node because we would be looking at nothing with a viewing window of $r = 0$. Once you expand your viewing window from the center of the orbital, then you start seeing the electron density come into play. Since ${Y}^{2} = \frac{1}{2 \sqrt{\pi}} \ne 0$, we need to find ${R}^{2} = 0$. Since if $R = 0$, ${R}^{2} = 0$, let us just find $R = 0$.

Just from looking at the graph, we should see it close to $2 {a}_{0}$ or $3 {a}_{0}$.

0 = cancel(2)^(ne 0)cancel((Z/(2a_0))^"3/2")^(ne 0)(2-(Zr)/a_0)cancel(e^(-Zr"/"2a_0))^(>0)

$\textcolor{g r e e n}{Z r = 2 {a}_{0}}$

Now, since we are talking about the hydrogen atom, $\textcolor{b l u e}{r = 2 {a}_{0}}$ for the radial node in the $2 s$ orbital of hydrogen.

We would know that that is the only one because the total number of radial nodes is $\textcolor{b l u e}{n - l - 1} = 2 - 0 - 1 = \textcolor{b l u e}{1}$