Question

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?

Answer 1

`r_0 = 2 * a_0`

Explanation:

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

Basically, the wave function, `Psi(x)`, 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(x)|^2`.

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

`color(blue)(|Psi(x)|^2 = 0) ->` this is true at nodes

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

`Psi_(2s) = 1/(2sqrt(2pi)) * sqrt(1/a_0) * (2 - r/a_0) * e^(-r/(2a_0))`

and told that at `r = r_0`, a radial node is formed. Right from the start, this tells you that you have

`|Psi_(2s)|^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

`(2 - r/a_0) = 0`

since

`Psi_(2s) = overbrace(1/(2sqrt(2pi)) * sqrt(1/a_0))^(color(purple)(>0)) * (2 - r/a_0) * overbrace(e^(-r/(2a_0)))^(color(purple)(>0))`

This means that you have

`2 - r/a_0 = 0 implies r = 2 * a_0`

At `r = r_0`, you will have

`color(green)(r_0 = 2 * 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.

Answer 2

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_(2s)`, which for hydrogen is defined in spherical coordinates via separation of variables to give a radial and angular component:

`psi(r,theta,phi) = R_(nl)(r)Y_l^(m_l)(theta,phi)`

`\mathbf(psi_(2s)(r,theta,phi) = R_(20)(r)Y_0^(0)(theta,phi))`

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 `color(green)(1/(2sqrtpi))`. 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^"*"psi` (for real numbers, `psi^2`), the probability density, as a whole is `0`.

For real numbers, `\mathbf(psi^2(r,theta,phi) = R_(nl)^2(r)(Y_l^(m_l)(theta,phi)))^2`.

On the radial distribution graph for `a_0r^2R_(nl)^2(r)` vs. `r"/"a_0` (always positive), which plots probability density vs. viewing-window radius `r`, it touches `R = 0` when `r = 0` or `R_(nl)^2(r) = 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 = 1/(2sqrtpi) 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 `2a_0` or `3a_0`.

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

`color(green)(Zr = 2a_0)`

Now, since we are talking about the hydrogen atom, `color(blue)(r = 2a_0)` for the radial node in the `2s` orbital of hydrogen.

We would know that that is the only one because the total number of radial nodes is `color(blue)(n - l - 1) = 2 - 0 - 1 = color(blue)(1)`