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?

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

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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.

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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.

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.

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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)#