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.