Probabilty of finding the electron of the hydrogen atom in

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Homework Help Overview

The discussion revolves around the probability of finding the electron in a hydrogen atom, specifically focusing on the mathematical expressions related to the wave function in its ground state. Participants are examining the implications of the probability density function and its integration over specified volumes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the interpretation of the probability density function P(\vec{r})d3\vec{r} and its physical meaning. There are questions about the integration limits and the appropriateness of using spherical coordinates for the calculations. Some participants are also clarifying the mathematical expressions related to the rms uncertainty and variance.

Discussion Status

The discussion is active, with participants providing insights and raising questions about the mathematical setup and physical interpretations. Some guidance has been offered regarding the integration process and the use of spherical coordinates, although there is no explicit consensus on all points raised.

Contextual Notes

Participants are navigating potential ambiguities in the problem statement, particularly regarding the notation and the definitions of the variables involved. There is an acknowledgment of possible typing errors in the original question, which may affect the interpretation of the tasks.

Yhaz
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Homework Statement


The average function of the H-atom in its ground state is ψ(\vec{r})=(1/(πa03)1/2exp(-r/a0)

a0: Bohr radius

a.What is the probability
i. P(\vec{r})d3\vec{r} that the electron will be found in the volume
d3\vec{r} around \vec{r}?
ii. Pdr that the electron will be found within the infinitesimal spherical shell of radius r and thickness dr ?
b. Calculate the rms uncertainty \langler-\langler\rangle2\rangle 1/2
Let me update this post a little later. It took me some time to write the whole thing with the notations.
 
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Welcome to PF.
Look forward to seeing your reasoning and attempt.
 
So going from the last to the first..

I guess there was a mistake in the b. part which asks for rms uncertainty, a typing error in the question?
It should be asking for √\langle(r-\langler\rangle)2\rangle which actually clears that part. If that is the case, then the variance \langle(r-\langler\rangle)2\rangle can be found through computing \langler2\rangle -\langler\rangle2 and this through substituting ∫rψψ*dV from ∫r2ψψ*dV where ψ* is the complex conjugate of ψ and the limits of the integral are from 0 to infinity. Do correct me if i am mistaken.

About a. ii)
It should be simply computing the integral ∫ψψ*dV from 0 to infinity, right?

What I'm confused about is what a. i) is asking for. What is the physical meaning of a volume P(\vec{r})d3\vec{r} around \vec{r}? This sound completely meaningless to me. An infinitesimal volume around \vec{r}?
I could try something like ∫∫∫ψψ*drdrdr with all the limits from r to r+dr but this seems senseless because what is natural for a system like this is to use spherical coordinates or taking dV= 4∏r2dr to calculate an integral such as the ones above. Or since dr is infinitesimal I could just multiply P(r) with d3r, I feel there is something wrong with this but what is it?
 
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May be a bit of a nitpick but...<br /> \langle r \rangle = \int \psi^* r \psi dr<br />... the order can be important so it is a useful discipline.
(above would be for the expectation of the radial coordinate - for volume integrals you want spherical polar coordinates... you can exploit the fact that the wavefunction varies only radially.)

What is the physical meaning of a volume P(r⃗ )d3r⃗ around r⃗ ?
It would be the volume probability density at \vec r multiplied by the volume element there. It's the first step in formally constructing the integral. The sum of all of these for all space would be one.

It is useful to think of the integration elements as having a physical extent - although a very small one. Helps you construct the integral properly. I think the question wants to know if you can construct dV in useful coordinates... I'd pick dV=4πr2dr myself (note - lower-case pi) since ψ has no angular dependence... which is the answer to ii. Now do you get it?

The whole thing is just getting you to do the integrations so you will have the core understanding when you deal with the simpler, even more abstract, notation later. YOu are doing fine, just translate the instructions into an integral and crunch the numbers. After a while you'll be doing these integrals in your head! :) Amaze your friends!

(niggle: what's wrong with bold-face for vectors I don't know...)
 
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Thank you for the effort.

I guess the ordering is important when you use operators?

Concerning the first part.. Can't really think of any other way than this equality left as a differential. P(\vec{r})d3\vec{r}= ψψ*d3\vec{r}

Is there a point in integrating from r to r+dr?

I also don't understand why the question asks for P(\vec{r})d3\vec{r} in vector form, how could the volume be in vector form and if a volume dV is defined as d3r the dr's that make it up wouldn't be in the same direction anyway
 
I guess the ordering is important when you use operators?
r is the radial position operator. Some operators it is important.Consider if you were to find the mass of a sphere, with a radial mass density \rho(r)

step 1: the mass of the volume element at r is \rho(r)r^2\sin(\theta)drd\theta d\phi
step 2: the mass of a spherical shell thickness dr, radius r is just the area of the surface times dr (the volume) times the density at r (since the density only varies radially).
step 3: mass of the whole sphere is the sum of masses of each concentric shell from r=0 to the radius of the sphere.

Now you are doing the same sort of thing, only with a probability density rather than a mass density. Got it now?

Don't let the notation throw you - d3r will come out as a scalar volume.
see these notes, p12.

You are spending a long time worrying about this instead of just doing it.
You'll soon find out if you're right :)
 

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