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QM- Relative probability of finding an electron

  1. Dec 4, 2005 #1
    Hey!

    I'm having a bit of trouble with solving the following problems, so help would be appreciated a lot!

    Consider a hydrogen atom in the ground state. Calculate the relative probability of finding the electron in the area described:

    a) In a sphere of volume 1.0 E-3 pm3 centered at the nucleus

    Well this wasn't really hard: the volume is 1.0 E-3 pm3 = the volume of a sphere = 4/3 pi r3.
    Dividing by 4/3 pi and taking the 3-rd square root gave me r= 0.0620350491 E -12 m.

    The psi2 function of the 1s orbital is:

    (1/pi) (1/a0)^3 e-2r/a0

    The radial probability distribution is obtained when the function above is multiplied with the formula for the surface of a sphere given by 4 pi r2. Integrating between 0 and the radius 0.0620350491 E -12 m gives a probability of 2.2E -9 .


    Now the hardest part. I don't know how to get to a right answer with these questions:

    Calculate the relative probability of finding the electron in the area described:

    b) In a sphere of volume 1.0 E-3 pm3 centered on a point 1.0 E-11 m from the nucleus.

    c) In a sphere of volume 1.0 E-3 pm3 centered on a point 53 pm from the nucleus.
     
    Last edited: Dec 4, 2005
  2. jcsd
  3. Dec 4, 2005 #2
    It's really urgent! Please help.. :(
     
  4. Dec 4, 2005 #3

    Physics Monkey

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    Hi Lisa, in principle the probablity that the electron is to be found in any volume V is simple the integral [tex] \int_V d^3 x | \psi|^2 [/tex] over the relevant volume. Now, in the case of parts a), d), and e) the integral is easy to do because the region is spherically symmetric. In the case of b) and c) you would have a little more trouble evaluating the integral by hand, though you could always do it numerically. This is however, not the way to approach this problem. The key lies in the realization that the volumes of interest are very small compared to the typical length scale of the wavefunction, in other words, the volume V is small compared to the Bohr radius cubed a^3. What does this mean? It means that the wavefunction is effectively constant over the region of integration. Each of the problems is designed to be attacked approximately. How might you evaluate the integral for the probability approximately given this information?
     
  5. Dec 4, 2005 #4

    Physics Monkey

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    Be careful Leonhard, the volume is given as 10^(-3) pm^3 so Lisa is correct to find the radius of the sphere by using the formula for the volume of a sphere. Given this radius you can then integrate the wavefunction between 0 and r (and over angles), but r doesn't equal 10^(-3) pm.
     
  6. Dec 4, 2005 #5
    Gee thanks Physics Monkey! I found the correct answers to a, d and e now (checked them in my book)..... but I really don't see how to get to the correct answers in b & c because I don't really know how to handle the asymmetrical cases....
     
  7. Dec 4, 2005 #6
    EDIT: I solved it! Woohoo! Thanks again!!!
     
  8. Nov 24, 2011 #7
    How you can solve it? Please tell me, I am very curious :smile:
     
  9. Nov 25, 2011 #8
    Please help me!!! Many thanks
     
  10. Nov 25, 2011 #9
    I can not find out the correct answers. How can you integrate to find the relative possibilities?
     
  11. Nov 26, 2011 #10
    Woo, I've solved it, too :smile:
     
  12. Nov 26, 2011 #11
    Two questions:

    (i) when you did the integral you just did

    [tex] \frac{1}{\pi a_0^3} \int r^2 e^{-\frac{2r}{a_0}} r^2 \sin{\theta} dr d\theta d\phi [/tex]

    which comes out if you integrate by parts twice and then you substitute the value for [tex]a_0[/tex] right?

    (ii) i still don't get how to do b and c when they aren't centred on the origin.

    Surely we can do the integral, we just have to alter the ranges of integration. i.e. integral over r will now run from -r to +r, integral over theta is still from 0 to pi but integral over phi (azimuthal) will now be 0 to pi rather than 0 to 2 pi since we have doubled the radial integral we must half the azimuthal integral, right?
     
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