Average electron distance (GRE practice problem)

The probability of finding the electron in a region of thickness dr at distance r is proportional to ψ(r)^2 4πr^2dr, which gives the probability density at r. The wavefunction is only showing you the amplitude of the electron in that region, but you have to sum over all angles to get the probability of finding the electron at r.In summary, the question is asking for the most likely distance that the electron is from the nucleus in the ground state of hydrogen. This is found by finding the maximum of the radial probability density P(r) = 4πψ^2(r)r^2, which is given by the wave function ψ(r) = 2/(4π)^(1/
  • #1
VantagePoint72
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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 [itex]\psi_{100}=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}[/itex], where [itex]a_0[/itex] is the Bohr radius. What is the most likely distance that the electron is from the nucleus?
(A) 0
(B) [itex]a_0/2[/itex]
(C) [itex]a_0/\sqrt{2}[/itex]
(D) [itex]a_0[/itex]
(E) [itex]2a_0[/itex]

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.
 
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  • #2
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[itex]\pi[/itex]r2dr.

Can you see that this leads to P(r) = 4[itex]\pi[/itex]ψ2(r)r2 ?

They are asking for the value of r that maximizes P(r). Can you find the value of r?
 
  • #3
In that case, would it not have been more appropriate to write the wave function as [itex]\psi_{100}(r,\theta,\phi)=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}[/itex]? 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.
 
  • #4
LastOneStanding said:
In that case, would it not have been more appropriate to write the wave function as [itex]\psi_{100}(r,\theta,\phi)=\frac{2}{(4\pi)^{1/2}a_0^{3/2}}e^{-r/a_0}[/itex]? 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 [itex]\theta[/itex] and [itex]\phi[/itex] appear as arguments on the left but don't appear anywhere on the right!] But I see your point.
 
  • #5
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.
 

FAQ: Average electron distance (GRE practice problem)

1.

What is the concept of average electron distance?

The average electron distance is a measure of the average distance of an electron from the nucleus in an atom. It is typically used to describe the size of an atom, as electrons are considered to be located in orbitals around the nucleus.

2.

How is average electron distance calculated?

The average electron distance is calculated by dividing the total distance of all electrons from the nucleus by the number of electrons in the atom. This value is then compared to the size of the entire atom to determine its relative size.

3.

What factors affect the average electron distance in an atom?

The average electron distance can be affected by the number of electrons in the atom, as well as the energy level of the electrons. The more energy an electron has, the further it can be from the nucleus, resulting in a larger average electron distance.

4.

How does average electron distance relate to atomic size?

Generally, atoms with larger average electron distances are considered to be larger in size. This is because the electrons in these atoms are located further from the nucleus, resulting in a larger overall size.

5.

What is the purpose of using average electron distance in GRE practice problems?

In GRE practice problems, average electron distance may be used to test a student's understanding of atomic structure and how it relates to the size of an atom. It can also be used to compare and contrast different atoms and their relative sizes based on their average electron distance.

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