Radial distribution function for H atom in ground state

[CAF123]

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.

 

[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
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.

 

[vela]

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.

 

[DrClaude]

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.

 

[CAF123]

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?

 

[vela]

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.

 

[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.)

 

[vela]

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

 

[CAF123]

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.

 

[vela]

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.

 

[CAF123]

I made an arithmetic error and indeed I get the 3a_o².

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

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.

 

[CAF123]

Are these statements along the right lines?

 

[vela]

Yup.