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Radial distribution function for H atom in ground state

  1. Feb 17, 2014 #1

    CAF123

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    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. jcsd
  3. Feb 18, 2014 #2

    CAF123

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    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
  4. Feb 18, 2014 #3

    vela

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    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.
     
  5. Feb 18, 2014 #4

    DrClaude

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    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.
     
  6. Feb 18, 2014 #5

    CAF123

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    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?
     
  7. Feb 18, 2014 #6

    vela

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    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.
     
  8. Feb 18, 2014 #7

    CAF123

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    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.)
     
  9. Feb 18, 2014 #8

    vela

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    Yeah, and the integral on the bottom should equal 1 since ##u_{100}## is normalized.
     
  10. Feb 18, 2014 #9

    CAF123

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    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.
     
  11. Feb 18, 2014 #10

    vela

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    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.
     
  12. Feb 19, 2014 #11

    CAF123

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    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.
     
  13. Feb 20, 2014 #12

    CAF123

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    Are these statements along the right lines?
     
  14. Feb 21, 2014 #13

    vela

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    Yup.
     
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