Average electron distance (GRE practice problem)

[VantagePoint72]

Homework Statement

Question 26 of the physics GRE from the practice test available here: https://www.ets.org/gre/subject/about/content/physics/

The normalized ground state wave function of hydrogen is \psi_{100}=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}, where a_0 is the Bohr radius. What is the most likely distance that the electron is from the nucleus?
(A) 0
(B) a_0/2
(C) a_0/\sqrt{2}
(D) a_0
(E) 2a_0

Homework Equations

Given in question.

The Attempt at a Solution

This seems like a really poorly worded question to me (at least, based on what answer they're looking for). The answer, according to the booklet, is (D). This (obviously) is the expectation value for the radial distance. However, the most likely distance that the electron is from the nucleus is just the distance corresponding to maximum value for the wavefunction's squared norm. This would be (A). So, is my reading of the question right? And if it's right, is this horrible abuse of language standard? If it's not standard, then the question is just plain wrong, and it certainly doesn't inspire much confidence in the physics GRE if that's the case.

 

[TSny]

Hi. You might want to review the idea of "radial probability density", P(r). P(r) is defined such that if you take a thin spherical shell of radius r then the probability of finding the electron in this shell is P(r)dr. So, P(r)dr is the probability of "finding the electron at distance r" within the range r and r + dr.

For a spherically symmetric wave function (like the ground state of hydrogen) the probability of finding the electron anywhere in this shell would be the square of the wavefunction multiplied by the volume of the shell: 4\pir²dr.

Can you see that this leads to P(r) = 4\piψ²(r)r² ?

They are asking for the value of r that maximizes P(r). Can you find the value of r?

 

[VantagePoint72]

In that case, would it not have been more appropriate to write the wave function as \psi_{100}(r,\theta,\phi)=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}? This would explicitly demonstrate that the implicit angular dependence still needs to be integrated out. I understand your explanation (thank you!), but it (the question) still seems poorly worded.

 

[TSny]

In that case, would it not have been more appropriate to write the wave function as \psi_{100}(r,\theta,\phi)=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}? This would explicitly demonstrate that the implicit angular dependence still needs to be integrated out. I understand your explanation (thank you!), but it (the question) still seems poorly worded.

Yes, probably it would be clearer to include the angles in the arguments of the wavefunction. [I'm sure that there would be other students who would then wonder why \theta and \phi appear as arguments on the left but don't appear anywhere on the right!] But I see your point.

 

[vela]

I don't think it's poorly worded. If the question asked for where the electron would most likely be, then your answer, that you'd most likely find it at the origin, would be correct. The question, however, is asking for the most likely distance, which means you need to sum the probabilities of all points at the given distance.