Schaum's Outline of Quantum Mechanics 8.12

[Jimmy Snyder]

Homework Statement

What is the most probable value of r for an electron in the state n, l = n - 1 in a hydrogen-like atom?

Homework Equations

Eqn (8.31)
R_{n, n-1}(\rho) = Ae^{\rho/2}\rho^{n-1}L_{2n-1}^{2n-1}(\rho)

The Attempt at a Solution

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

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

The book has n^2/Z

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 (n^2 + n/2)/Z and I would expect the average value to be near the most probable value.

 

[dextercioby]

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

Shouldn't you be squaring R before calculating its extremal point ? (Check out page 149, point c) from problem 8.5).

 

[Jimmy Snyder]

Actually, I followed the example of problem 8.5 part c. Note that since \psi is real, the extrema of (\rho\psi)^2 and \rho\psi are the same. Also note that he uses \rho\psi in his calculations in part c. However, in retrospect, I believe that the author is not correct. The most probable value for \rho is the extremum of |\psi|^2 which is the same as the extremum of \psi. In this case I get

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

As before, I do not think this result is correct. Do you agree that the most probable value of \rho is the extremum of P(\rho) = |\psi(\rho)|^2?

 

[vela]

Your first result is correct; you just need to write it in terms of r now.

 

[Jimmy Snyder]

Your first result is correct; you just need to write it in terms of r now.

\rho = r/ a_0, so in units of a_0, my result is r = 2n/Z. The answer in the book is given as r = n^2/Z also in units of a_0. However, what do you think about my question in post #3? Did the author make a mistake by finding the extremum of (\rho\psi)^2 = \rho^2P(\rho) rather than finding the extremum of P(\rho)?

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?

 

[vela]

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 $r=0$.

 

[Jimmy Snyder]

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 $r=0$.

Thanks vela, I see it now.