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Hydrogen probably homework

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Griffiths 4.14: What is the most probably value of r, in the ground state of hydrogen? (Hint: First you must figure out the probability that the electron would be found between r and r+dr.

    Here is the posted solution:

    [​IMG]

    I don't understand the approach; why does multiplying the probability density by [tex] \[
    4\pi r^2 dr\] [/tex] give the probability that the electron will be found between r and r+dr? Why do we differentiate p(r)? In general, I thought the way to find the probability of a particle being in a particular energy eigenstate was to take the inner product of the general wavefunction with the energy eigenstate..

    [tex]
    \[
    \left| {c_n } \right|^2 = \int_{ - \infty }^\infty {\Psi ^* (\overrightarrow {\bf{r}},t) \cdot \psi (r){\rm{ }}dr}
    \]
    [/tex]

    But the general wave function is a sum which involves [tex] c_n [/tex].. so I don't understand what governs which approach to take..
     
  2. jcsd
  3. Apr 27, 2008 #2
    I hope this question doesn't fall into the abyss of unanswered questions, forever forgotten by all..
     
  4. Apr 27, 2008 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It takes longer to fall into the abyss of unanswered questions than two hours. Be patient. The solution is doing exactly what the hint proposes. It's finding the probability density of the wave function as a function of r by integrating over the angular coordinates. That's where the 4*pi*r^2 comes from, it's what you get from that integration together with the volume element in spherical coordinates. Then you maximize it, hence the derivative.
     
    Last edited: Apr 27, 2008
  5. Apr 28, 2008 #4
    So I want to figure out at what point the probability density is a maximum; that's why I set the derivative to zero. OK, I get it, thanks!
     
    Last edited: Apr 28, 2008
  6. Apr 28, 2008 #5
    (Also, yes, you are right. I will try to be more patient.)
     
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