Radial distribution function for H atom in ground state

In summary, the normalized energy eigenfunction of the ground state of the hydrogen atom is ##u_{100}(\underline{r}) = C \exp (-r/a_o)##, ##a_o## the Bohr radius. For this state calculate 1)##C##2)The radial distribution function, the probability that the electron is within a sphere of radius ##a_o##3)Expectation values of ##r##, ##V(r)## and the uncertainty ##\Delta r##.
  • #1
CAF123
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Homework Statement


The normalized energy eigenfunction of the ground state of the hydrogen atom is ##u_{100}(\underline{r}) = C \exp (-r/a_o)##, ##a_o## the Bohr radius. For this state calculate
1)##C##
2)The radial distribution function, the probability that the electron is within a sphere of radius ##a_o##
3)Expectation values of ##r##, ##V(r)## and the uncertainty ##\Delta r##

Homework Equations


Normalization of energy eigenfunctions, Expectation values.

The Attempt at a Solution


The radial distribution function gives the probability of finding the electron in a spherical shell of thickness ##dr##. I understand that the RDF is commonly written ##4\pi r^2 R^2 dr##. However, is this the case here? When I do the integral I do not get the ##4\pi## (because of the state in question, the angular part squared gives a ##1/(4\pi)## which cancels)

My integral was $$\int_r^{r+dr} \int_0^{2\pi} \int_0^{\pi} u_{100}^* u_{100} r^2\, \sin \theta \,d\theta\, d\phi\, dr\, = \int_r^{r+dr} R_{10}^* R_{10} r^2\, dr,$$ which gives the probability of finding the electron between ##r## and ##r+dr## and the integrand is the RDF.
Many thanks.
 
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  • #2
I am just trying to make sense of the various descriptions of the radial distribution function I have seen on the net. For example, at the very top of http://www.fordham.edu/images/undergraduate/chemistry/pchem1/radial_distribution_function.pdf
I do not see how they can end up with that integral. If that is a general integral, they have taken the spherical harmonics to equal unity which is obviously not the case.

Then there is http://chemistry.illinoisstate.edu/standard/che362/handouts/362rdf.pdf which gives the same RDF as the one I have.

Finally, there is then http://winter.group.shef.ac.uk/orbitron/AOs/1s/radial-dist.html which simply says the RDF is ##4\pi r^2 \psi^2##.

How does this all come together?
Thanks.
 
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  • #3
It depends on how you decide to normalize the various pieces of the wave function. Regardless of that, however, the answer to (b) is unambiguous because it's asking for a probability of finding the electron in a certain region. That won't depend on your choice of normalization.

The probability that the electron is in an infinitesimal volume bounded by ##r## and ##r+dr##, ##\theta## and ##\theta+d\theta##, and ##\phi## and ##\phi+d\phi## is ##\lvert u_{100}\rvert^2\,r^2\sin\theta\,dr\,d\theta\,d\phi##. Just integrate over the angles.
 
  • #4
My guess is that it depends on how you define the wave function. The only way I can see one getting ##4 \pi r^2 R(r)^2## is if one takes ##\psi (r, \theta \phi) = R(r)##, which is not normalized over ##(r,\theta,\phi)##, and assuming spherical symmetry. Stick to your approach.
 
  • #5
vela said:
It depends on how you decide to normalize the various pieces of the wave function. Regardless of that, however, the answer to (b) is unambiguous because it's asking for a probability of finding the electron in a certain region. That won't depend on your choice of normalization.
So if I suppose the 1/√4π term (from the spherical harmonic) is contained in C then the normalized ground state function can be written as $$u_{100} = \frac{2}{\sqrt{a_o^3}} \exp(-r/a_o)$$
The probability that the electron is in an infinitesimal volume bounded by ##r## and ##r+dr##, ##\theta## and ##\theta+d\theta##, and ##\phi## and ##\phi+d\phi## is ##\lvert u_{100}\rvert^2 \,r^2\sin\theta\,dr\,d\theta\,d\phi##. Just integrate over the angles.
The probability that the electron is located between ##r## and ##r+dr## is given by $$\frac{\int_{r}^{r+dr} |u_{100}|^2 r^2 dr \cdot 4\pi}{\int_{\text{all space}} |u_{100}|^2 dV}$$
Is that correct? I think I can extract the RDF from this to be the integrand ##|u_{100}|^2 4 \pi r^2## which reduces to ##R^2 4\pi r^2## assuming that ##\psi = u = R(r)## as DrClaude mentioned. Is that right?
 
  • #6
CAF123 said:
So if I suppose the 1/√4π term (from the spherical harmonic) is contained in C then the normalized ground state function can be written as $$u_{100} = \frac{2}{\sqrt{a_o^3}} \exp(-r/a_o)$$
I didn't get that constant. The way I interpret the problem statement is that ##u_{100}## is the ground-state eigenfunction. It includes both the radial and angular parts. The problem statement doesn't say it's only the radial part. When you integrate ##u_{100}^2## over all space, not just over ##r##, it should equal 1.

Assuming ##R(r)## and ##Y_{lm}(\theta,\phi)## are normalized separately, you have
$$u_{100}(r,\theta,\phi) = C e^{-r/a_0} = R_{10}(r)Y_{00}(\theta,\phi).$$ If you separate out the factor of ##\frac{1}{\sqrt{4\pi}}##, you have
$$u_{100}(r,\theta,\phi) = \sqrt{4\pi}Ce^{-r/a_0}\frac{1}{\sqrt{4\pi}} = R_{10}(r)Y_{00}(\theta,\phi),$$ from which you can identify ##R_{10}(r)## as being ##\sqrt{4\pi}Ce^{-r/a_0}##, which is what you said equaled ##u_{100}## above.
 
  • #7
Yes, so then ##C = 1/\sqrt{\pi a_o^3}##. Was the rest of my post ok? (I think I did not need to divide by the integral over all space though - if the RDF is interpreted as a probability density (prob per unit vol) then by integrating over some infinitesimal region of thickness ##dr##, the units cancel and I obtain a probability.)
 
  • #8
Yeah, and the integral on the bottom should equal 1 since ##u_{100}## is normalized.
 
  • #9
vela said:
Yeah, and the integral on the bottom should equal 1 since ##u_{100}## is normalized.

Okay, I worked the rest of the problem. For the probability of the electron being in a sphere of radius ##a_0##, I computed ## \frac{4}{a_0^3}\int_0^{a_0} e^{-2r/a_o}r^2dr## which by integration by parts gives a probability of around 0.3.

For 3), the expected value of ##r## was ##3/2 a_o##. I thought to obtain the expected value of ##V(r)##, I could do ##\langle V(r) \rangle = \frac{-e^2}{4\pi \epsilon_o \cdot \langle r \rangle} = \frac{-e^2}{6 \pi \epsilon_o a_o}## but if I do it by computing ##\langle u| V(r)| u \rangle## I get the same result but with a factor 1/4 rather than a 1/6. However, if I do ##\langle 1/r \rangle##, then put this into expression for V(r), I also get the result with the factor 1/4. So I suppose my question boils down to why I cannot use ##\langle r \rangle##.

Finally ##(\Delta r)^2 = \langle r^2 \rangle - \langle r \rangle^2 = 0##, although I am not sure how to interpret this. Thanks.
 
  • #10
CAF123 said:
For 3), the expected value of ##r## was ##3/2 a_o##. I thought to obtain the expected value of ##V(r)##, I could do ##\langle V(r) \rangle = \frac{-e^2}{4\pi \epsilon_o \cdot \langle r \rangle} = \frac{-e^2}{6 \pi \epsilon_o a_o}## but if I do it by computing ##\langle u| V(r)| u \rangle## I get the same result but with a factor 1/4 rather than a 1/6. However, if I do ##\langle 1/r \rangle##, then put this into expression for V(r), I also get the result with the factor 1/4. So I suppose my question boils down to why I cannot use ##\langle r \rangle##.
The simple answer is because it doesn't work that way.

Suppose you have a random variable X which is equal to a half the time and to b half the time. The expected value of X would be <X> = (a+b)/2. The expected value of <1/X> would be
$$\big\langle\frac{1}{X}\big\rangle = 0.5\times\frac{1}{a} + 0.5\times\frac{1}{b} = \frac{a+b}{2ab}.$$ Clearly, that's not equal to 1/<X> = 2/(a+b).

Finally ##(\Delta r)^2 = \langle r^2 \rangle - \langle r \rangle^2 = 0##, although I am not sure how to interpret this. Thanks.
You made a mistake somewhere. What did you get for ##\langle r^2 \rangle##? It should be 3a02.
 
  • #11
I made an arithmetic error and indeed I get the 3ao2.

It seems non-intuitive that the expectation value of r is 3/2 ao, yet the expected value of V(r) is -e2/4πεoao.

I have seen in many cases in probability, for example, where to compute an expectation of a random variable there has also been a division, e.g say $$\langle E \rangle = \frac{\int_0^{\infty} EN(E) dE}{\int_0^{\infty} N(E)dE}$$ I was just wondering why this would be the case. My thoughts are that in QM, the way the expectation is defined means we have a probability density integrated over all space, so the resulting answer is already dimensionless.

If the wavefunction was not normalized, then I think it would be appropriate to divide by the integral over all space. The result is then unchanged since factors in the numerator/denominator, which would be the reciprocal of the norm. constant, would cancel.

Thanks.
 
  • #12
Are these statements along the right lines?
 
  • #13
Yup.
 
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Related to Radial distribution function for H atom in ground state

1. What is the radial distribution function for the hydrogen atom in its ground state?

The radial distribution function for the hydrogen atom in its ground state is a mathematical function that describes the probability of finding the electron at a certain distance from the nucleus. It is often represented by the symbol g(r) and is used to understand the spatial distribution of electrons in an atom.

2. How is the radial distribution function calculated?

The radial distribution function is calculated by taking the square of the wavefunction for the electron in the hydrogen atom and normalizing it. This involves solving the Schrödinger equation for the hydrogen atom and using the resulting wavefunction to calculate the probability density at different distances from the nucleus.

3. What does the radial distribution function tell us about the hydrogen atom?

The radial distribution function provides information about the spatial distribution of electrons in the hydrogen atom. It shows that the electron is most likely to be found at a distance of a0 (the Bohr radius) from the nucleus, but there is also a small probability of finding the electron at larger distances.

4. How does the radial distribution function change for different energy levels in the hydrogen atom?

The radial distribution function changes for different energy levels in the hydrogen atom because the wavefunction for the electron is different for each energy level. This means that the probability of finding the electron at different distances from the nucleus also changes.

5. Why is the radial distribution function important in studying atoms?

The radial distribution function is important in studying atoms because it provides a way to visualize and understand the distribution of electrons in an atom. It also allows us to compare the distribution of electrons in different atoms and molecules, which is crucial in understanding their chemical and physical properties.

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