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Homework Help: Schaum's Outline of Quantum Mechanics 8.12

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the most probable value of r for an electron in the state n, l = n - 1 in a hydrogen-like atom?

    2. Relevant equations
    Eqn (8.31)
    [tex]R_{n, n-1}(\rho) = Ae^{\rho/2}\rho^{n-1}L_{2n-1}^{2n-1}(\rho)[/tex]

    3. The attempt at a solution
    Although I have not proved it, I suspect [itex]L_m^m = (-1)^m m![/itex] as in the three cases provided on page 306. That is, L is a constant that can be absorbed into A.

    [tex]0 = A \frac{d}{d\rho}Z\rho e^{-Z\rho/2}(Z\rho)^{n-1}[/tex]
    [tex]= A \frac{d}{d\rho}e^{-Z\rho/2}(Z\rho)^n[/tex]
    [tex]= A (-\frac{Z}{2}(Z\rho)^n + nZ^n\rho^{n-1})e^{-Z\rho/2}[/tex]
    [tex]= Ae^{-Z\rho/2} Z^n\rho^{n-1} (-\frac{Z}{2}\rho + n)[/tex]
    [tex]0 = -\frac{Z}{2}\rho + n[/tex]
    [tex]\rho = 2n/Z[/tex]

    The book has [itex]n^2/Z[/itex]

    I tend to believe that the book is correct and I am wrong because according to eqn (8.42) with l = n - 1, the average value of r is [itex](n^2 + n/2)/Z[/itex] and I would expect the average value to be near the most probable value.
  2. jcsd
  3. Oct 22, 2012 #2


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    You are right about one thing: [itex] L_{2n-1}^{2n-1} (\rho) [/itex] is a numerical constant, i.e. independent of [itex] \rho [/itex].

    Shouldn't you be squaring R before calculating its extremal point ? (Check out page 149, point c) from problem 8.5).
  4. Oct 22, 2012 #3
    Actually, I followed the example of problem 8.5 part c. Note that since [itex]\psi[/itex] is real, the extrema of [itex](\rho\psi)^2[/itex] and [itex]\rho\psi[/itex] are the same. Also note that he uses [itex]\rho\psi[/itex] in his calculations in part c. However, in retrospect, I believe that the author is not correct. The most probable value for [itex]\rho[/itex] is the extremum of [itex]|\psi|^2[/itex] which is the same as the extremum of [itex]\psi[/itex]. In this case I get

    [tex]0 = \frac{d}{d\rho}e^{-Z\rho/2}Z^{n - 1}\rho^{n-1}[/tex]
    [tex]= (-\frac{Z}{2}\rho + (n - 1))Z^{n - 1}\rho^{n-2}[/tex]
    [tex]\rho = \frac{2(n - 1)}{Z}[/tex]

    As before, I do not think this result is correct. Do you agree that the most probable value of [itex]\rho[/itex] is the extremum of [itex]P(\rho) = |\psi(\rho)|^2[/itex]?
  5. Oct 22, 2012 #4


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    Your first result is correct; you just need to write it in terms of r now.
  6. Oct 22, 2012 #5
    [itex]\rho = r/ a_0[/itex], so in units of [itex]a_0[/itex], my result is [itex]r = 2n/Z[/itex]. The answer in the book is given as [itex]r = n^2/Z[/itex] also in units of [itex]a_0[/itex]. However, what do you think about my question in post #3? Did the author make a mistake by finding the extremum of [itex](\rho\psi)^2 = \rho^2P(\rho)[/itex] rather than finding the extremum of [itex]P(\rho)[/itex]?

    Also, if my result is correct, why is there such a discrepency between the most probable value of r and the average value of r?
  7. Oct 22, 2012 #6


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    I think you might have looked up the n=2 case for ##\rho##. It's generally defined as ##\rho = 2r/na_0##.

    The author is correct. The probability is the product of ##P(\rho)## and the size of the sample space, which is proportional to ##r^2##. For example, for the 1s orbital, the P attains a maximum at the origin, but you're more likely to find the electron away from the origin. The probability of finding the electron at any single point where ##r=a_0## is less than finding it at the origin, but when you sum the probabilities over all points for which ##r=a_0##, the total probability of ##r=a_0## is greater than the probability of finding the electron at ##r=0##.
  8. Oct 22, 2012 #7
    Thanks vela, I see it now.
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