Radial distribution function for H atom in ground state

1. Feb 17, 2014

CAF123

1. The problem statement, all variables and given/known data
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 $\Delta r$

2. Relevant equations
Normalization of energy eigenfunctions, Expectation values.

3. 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.

2. Feb 18, 2014

CAF123

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 [Broken]
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.

Last edited by a moderator: May 6, 2017
3. Feb 18, 2014

vela

Staff Emeritus
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. Feb 18, 2014

Staff: Mentor

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. Feb 18, 2014

CAF123

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 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. Feb 18, 2014

vela

Staff Emeritus
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. Feb 18, 2014

CAF123

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. Feb 18, 2014

vela

Staff Emeritus
Yeah, and the integral on the bottom should equal 1 since $u_{100}$ is normalized.

9. Feb 18, 2014

CAF123

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. Feb 18, 2014

vela

Staff Emeritus
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).

You made a mistake somewhere. What did you get for $\langle r^2 \rangle$? It should be 3a02.

11. Feb 19, 2014

CAF123

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. Feb 20, 2014

CAF123

Are these statements along the right lines?

13. Feb 21, 2014

vela

Staff Emeritus
Yup.