Question about Radial Distribution Function graphs

[whitesniper001]

Hi, just needed some help in answering a past exam paper question which is as follows:

 

"Draw the Radical Distribution Functions for 1s, 2s and 3s orbitals and point out the important features of these graphs".

 

​After looking through my notes the graphs which I found that correspond to the "Radical Distribution Functions " are :

 

 

 

First question Is are these the right graphs? Even though they are titled "Radial Probability Functions (RPF’s)".

 

​Also there was no information regarding the second part of the question in regards to the "important features of these graphs." I tried looking online and was finding it hard to find anything, can anybody point me in the right direction? Thanks

 

 

 

 

 

[studiot]

I assume the quote from your question was a misprint and you do indeed mean radial, not radical.

That makes sense.

 

Yes they are the correct graphs.

 

 

Important features.

 

Well maxima or minima are usually important featuresof interest in a graph.

 

Alice the camel has 2 humps.

 

How many humps has the 1s, 2s and 3s?

Are they all the same size?

 

What is plotted on the axis

 

X axis?

Hint hence the radial distribution

 

Y axis ?

Note the function plotted is (proportional to) the square of the wave function - What does this signify?

What do you thing the r² is doing in the y axis function?

[whitesniper001]

I assume the quote from your question was a misprint and you do indeed mean radial, not radical.

That makes sense.

 

Yes they are the correct graphs.

 

 

Important features.

 

Well maxima or minima are usually important featuresof interest in a graph.

 

Alice the camel has 2 humps.

 

How many humps has the 1s, 2s and 3s?

Are they all the same size?

 

What is plotted on the axis

 

X axis?

Hint hence the radial distribution

 

Y axis ?

Note the function plotted is (proportional to) the square of the wave function - What does this signify?

What do you thing the r² is doing in the y axis function?

 

Yes sorry that was suppose to say radial and not radical.

 

 

1s has one hump

​2s has two humps

3s has three humps

 

The humps are not all the same size. The nodes present in the graph what do they mean?

 

Is the average radius plotted on the x axis?

 

Does the square of the wave function signify the probability of finding an electron? Also I know that r represent the distance from the centre of the nucleus so would r^​2​ signify the size of the orbital?

 

I have uploaded some more graphs indicating the "average radius", why is it that these lines are not exactly half way between the final hump?

 

[studiot]

You haven't indicated the level of science or maths this question is for.

 

Because the (electrostatic) potential energy of an electron is a function only of its distance from the nucleus; that is the force field is spherically symmetrical, it is possible to express the total wave fucntion as a product of three functions, each involving only one of the three coordinates in a spherical coordinate system.

Two of these coordinates are angles and the third is the radius.

 

Thus

[math]\psi = R\left( r \right)\Theta \left( \theta \right)\Phi \left( \phi \right)[/math]

Where capitals represent the function and lower case represents the variables.

Theta and phi are the angles, r is the radial distance from the nucleus.

 

For all s electrons the angular part is constant for all values of of the angles.

This is because the s orbitals correspond to the zero value for the subsidiary quantum number, l

So all s orbitals have spherical symmetry, so only the radial part R® need be considered.

 

The probability of finding an s electron in a small element of volume dv at a distance between r and (r+dr) from the nucleus is then

 

[math]{\left[ {R\left( r \right)} \right]^2}dv[/math]

 

This is the same regardless of the direction (ie the angles) of the radius vector.

 

The total probability of finding an electron anywhere between r and (r+dr) is found by integrating the expression over a spherical shell between r and (r+dr)

ie replacing dv by the total volume of the shell.

 

This volume is

 

[math]4\pi {r^2}dr[/math]

So the integral now in terms of r becomes

 

[math]4\pi {\left[ {R\left( r \right)} \right]^2}{r^2}dr[/math]

 

This function is known as the radial probability distribution, which is plotted on your y axis.

 

Are you with me so far?

Edited May 1, 2016 by studiot

[whitesniper001]

You haven't indicated the level of science or maths this question is for.

 

Because the (electrostatic) potential energy of an electron is a function only of its distance from the nucleus; that is the force field is spherically symmetrical, it is possible to express the total wave fucntion as a product of three functions, each involving only one of the three coordinates in a spherical coordinate system.

Two of these coordinates are angles and the third is the radius.

 

Thus

[math]\psi = R\left( r \right)\Theta \left( \theta \right)\Phi \left( \phi \right)[/math]

 

Where capitals represent the function and lower case represents the variables.

Theta and phi are the angles, r is the radial distance from the nucleus.

 

For all s electrons the angular part is constant for all values of of the angles.

This is because the s orbitals correspond to the zero value for the subsidiary quantum number, l

So all s orbitals have spherical symmetry, so only the radial part R® need be considered.

 

The probability of finding an s electron in a small element of volume dv at a distance between r and (r+dr) from the nucleus is then

 

[math]{\left[ {R\left( r \right)} \right]^2}dv[/math]

 

 

This is the same regardless of the direction (ie the angles) of the radius vector.

 

The total probability of finding an electron anywhere between r and (r+dr) is found by integrating the expression over a spherical shell between r and (r+dr)

ie replacing dv by the total volume of the shell.

 

This volume is

 

[math]4\pi {r^2}dr[/math]

 

So the integral now in terms of r becomes

 

[math]4\pi {\left[ {R\left( r \right)} \right]^2}{r^2}dr[/math]

 

 

This function is known as the radial probability distribution, which is plotted on your y axis.

 

Are you with me so far?

 

 

Should have indicated this before asking the question, but I am a first year undergraduate chemistry student. I do follow you so far as it similar to what my notes contain. So the y axis is used to calculate the total probability of finding an electron anywhere between r and (r+dr) from what I gather.

[studiot]

Here is some bedtime reading then, (from Physical Chemisry by Moore)

 

 

 

 

 

[whitesniper001]

Thanks for the posted pages, have read them and referred to another book also which has helped make the issue alot clear, thank you.

[studiot]

One thing I didn't make clear is that the probability is not a point function, it is cumulative.

 

It is not the probability at a point, but the probabilty that the electron will be found within a certain region of space.

The region is defined by the limits of the integration over the volume.

For spherical symmetry this is particularly easy since the limits are from r = 0 to r = r and define a ball or spherical shell anywhere within which the electron may be found.

(Note the volume integral is the area of a spherical surface (4pir²) times a small radial distance dr)

 

So as we move out from r = 0 the cumulative probability is the integral described for a sphere of radius r.

 

The significance of the shape is that the probability starts off low and rises until you get to the hump.

A bit before the hump it is significantly less so the value of r at the hump contributes the most to the cumulative probability.

 

Your library should have a copy of Moore, it is one of the best textbooks at this level.

 

Sleep well.

[whitesniper001]

Thanks for the clarification I understand now what you mean by it being cumulative and not a point function. I've being reading "Chemical, The Central Science by Brown" which seems like a more introductory book. I will try getting hold of a copy by Moore also. Thanks

[studiot]

I'm sorry I put that rather badly, I was watching a 4 beer DVD at the time.

 

The cumulative probability for the integral over all space is obviously 1 (The electron must be somewhere).

r then runs from zero to infinity and these are the limits of a cumulative integration.

 

 

The formula

 

[latex] 4\pi {\left[ {R\left( r \right)} \right]^2}{r^2}dr [/latex]

can be broken down into a point probability P at r = r multiplied by an expanding volume dr.

That is Pdr, whre P is

[latex] 4\pi {\left[ {R\left( r \right)} \right]^2}{r^2}[/latex]

This is what is plotted on your graphs.

The y value at some r give the point value of P at that r.

So the hump corresponds to the max P and the Bohr radius for the 1s orbital.

 

[whitesniper001]

Thanks for the reply, just read up on the topic some more and you input helped.