Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Probability of finding an electron in a cylinder. . .

  1. Feb 24, 2010 #1
    For,[tex] \Psi_{321} [/tex], calculate the probabily of finding the electron inside the region defined by a thin cylinder of radius .10a and length 10a, where a is the Bohr radius.

    Little lost on this one.

    What I do know/think, is that because it is a thin cylinder, dr is negligible, so r can just be substituited for [tex] \Psi_{321} [/tex]. But what about [tex] \phi [/tex] ? Can [tex]2\pi[/tex] just be substituited also? And I'm assuming dz gets integrated from 0 to 10a.

    Please shed some light. :)
  2. jcsd
  3. Feb 24, 2010 #2
    Ohh, and to add, this is in a hydrogen atom. Stupid not to mention that.
  4. Feb 24, 2010 #3
    Also, I derived [tex] \Psi_{321} [/tex] in a previous section to be:

    [tex] \Psi_{321}= \sqrt{\frac{1}{9(120)^3}} \ e^{\frac{-r}{3a}}\left(\frac{320r^3}{a^3}\right) \sqrt{\frac{15}{8\pi}}\ sin{\theta}\ cos{\theta}\ e^{i\phi} [/tex]

    Can I apply this function to solve this problem? My function is in spherical coords, and question seems to be in cylindrical. I'm not sure how to derive in cylindrical I don't think. Do I have to?
  5. Feb 24, 2010 #4


    User Avatar
    Homework Helper
    Gold Member

    First, let me ask you a couple of basic questions:

    (1) What is the mathematical relationship between probability and probability density?
    (2) How is probability density related to the wavefunction?

    Second, upon answering these questions you should realize that you will need to perform a volume integral over the volume of the cylinder. To make that integration easier, you will want to transform your expression for [itex]\Psi_{321}[/itex] to cylindrical coordinates. So, what is the relationship between spherical coordinates [itex]\{r,\theta,\phi\}[/itex] and cylindrical coordinates [itex]\{\rho,\phi,z\}[/itex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook