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Probabilty of finding the electron of the hydrogen atom in

  1. Jan 7, 2012 #1
    1. The problem statement, all variables and given/known data
    The average function of the H-atom in its ground state is ψ([itex]\vec{r}[/itex])=(1/(πa03)1/2exp(-r/a0)

    a0: Bohr radius

    a.What is the probability
    i. P([itex]\vec{r}[/itex])d3[itex]\vec{r}[/itex] that the electron will be found in the volume
    d3[itex]\vec{r}[/itex] around [itex]\vec{r}[/itex]?
    ii. Pdr that the electron will be found within the infinitesimal spherical shell of radius r and thickness dr ?
    b. Calculate the rms uncertainty [itex] \langle[/itex]r-[itex] \langle[/itex]r[itex] \rangle[/itex]2[itex] \rangle[/itex] 1/2



    Let me update this post a little later. It took me some time to write the whole thing with the notations.
     
  2. jcsd
  3. Jan 7, 2012 #2

    Simon Bridge

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    Welcome to PF.
    Look forward to seeing your reasoning and attempt.
     
  4. Jan 7, 2012 #3
    So going from the last to the first..

    I guess there was a mistake in the b. part which asks for rms uncertainty, a typing error in the question?
    It should be asking for √[itex]\langle[/itex](r-[itex]\langle[/itex]r[itex]\rangle[/itex])2[itex]\rangle[/itex] which actually clears that part. If that is the case, then the variance [itex]\langle[/itex](r-[itex]\langle[/itex]r[itex]\rangle[/itex])2[itex]\rangle[/itex] can be found through computing [itex]\langle[/itex]r2[itex]\rangle[/itex] -[itex]\langle[/itex]r[itex]\rangle[/itex]2 and this through substituting ∫rψψ*dV from ∫r2ψψ*dV where ψ* is the complex conjugate of ψ and the limits of the integral are from 0 to infinity. Do correct me if i am mistaken.

    About a. ii)
    It should be simply computing the integral ∫ψψ*dV from 0 to infinity, right?

    What I'm confused about is what a. i) is asking for. What is the physical meaning of a volume P([itex]\vec{r}[/itex])d3[itex]\vec{r}[/itex] around [itex]\vec{r}[/itex]? This sound completely meaningless to me. An infinitesimal volume around [itex]\vec{r}[/itex]?
    I could try something like ∫∫∫ψψ*drdrdr with all the limits from r to r+dr but this seems senseless because what is natural for a system like this is to use spherical coordinates or taking dV= 4∏r2dr to calculate an integral such as the ones above. Or since dr is infinitesimal I could just multiply P(r) with d3r, I feel there is something wrong with this but what is it?
     
    Last edited: Jan 7, 2012
  5. Jan 8, 2012 #4

    Simon Bridge

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    May be a bit of a nitpick but...[tex]
    \langle r \rangle = \int \psi^* r \psi dr
    [/tex]... the order can be important so it is a useful discipline.
    (above would be for the expectation of the radial coordinate - for volume integrals you want spherical polar coordinates... you can exploit the fact that the wavefunction varies only radially.)

    It would be the volume probability density at [itex]\vec r[/itex] multiplied by the volume element there. It's the first step in formally constructing the integral. The sum of all of these for all space would be one.

    It is useful to think of the integration elements as having a physical extent - although a very small one. Helps you construct the integral properly. I think the question wants to know if you can construct dV in useful coordinates... I'd pick dV=4πr2dr myself (note - lower-case pi) since ψ has no angular dependence... which is the answer to ii. Now do you get it?

    The whole thing is just getting you to do the integrations so you will have the core understanding when you deal with the simpler, even more abstract, notation later. YOu are doing fine, just translate the instructions into an integral and crunch the numbers. After a while you'll be doing these integrals in your head! :) Amaze your friends!

    (niggle: what's wrong with bold-face for vectors I don't know...)
     
    Last edited: Jan 8, 2012
  6. Jan 8, 2012 #5
    Thank you for the effort.

    I guess the ordering is important when you use operators?

    Concerning the first part.. Can't really think of any other way than this equality left as a differential. P([itex]\vec{r}[/itex])d3[itex]\vec{r}[/itex]= ψψ*d3[itex]\vec{r}[/itex]

    Is there a point in integrating from r to r+dr?

    I also don't understand why the question asks for P([itex]\vec{r}[/itex])d3[itex]\vec{r}[/itex] in vector form, how could the volume be in vector form and if a volume dV is defined as d3r the dr's that make it up wouldn't be in the same direction anyway
     
  7. Jan 8, 2012 #6

    Simon Bridge

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    r is the radial position operator. Some operators it is important.


    Consider if you were to find the mass of a sphere, with a radial mass density [itex]\rho(r)[/itex]

    step 1: the mass of the volume element at r is [itex]\rho(r)r^2\sin(\theta)drd\theta d\phi[/itex]
    step 2: the mass of a spherical shell thickness dr, radius r is just the area of the surface times dr (the volume) times the density at r (since the density only varies radially).
    step 3: mass of the whole sphere is the sum of masses of each concentric shell from r=0 to the radius of the sphere.

    Now you are doing the same sort of thing, only with a probability density rather than a mass density. Got it now?

    Don't let the notation throw you - d3r will come out as a scalar volume.
    see these notes, p12.

    You are spending a long time worrying about this instead of just doing it.
    You'll soon find out if you're right :)
     
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