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Topic: electron probability  (Read 4962 times)

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Offline Sophia7X

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electron probability
« on: July 01, 2012, 07:48:38 PM »
Evaluate the probability of finding an electron in a small region of a hydrogen 2s orbital at distances from 0 to 3ao in increments of 0.10ao.

I multipled R20(r) by Y00(θ,Φ), simplified, squared to get Ψ2 = (1/16)(1/2παo3)(2 - r/ao)2e-r/ao

I'm kinda confused here, do I use Ψ2 to find the probability because that doesn't really work ???



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also, could someone give me a pointer for this problem? Thanks.

The data Millikan collected included the following series of charges found on oil drops: 9.60e-10 esu, 1.92e-9 esu, 2.4e-9, 2.88e-9, 4.80e-9. What is the likely charge on the electron in esu?
« Last Edit: July 01, 2012, 08:00:58 PM by Sophia7X »
Entropy happens.

Offline XGen

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Re: electron probability
« Reply #1 on: July 01, 2012, 10:03:17 PM »
Not really sure on the first part, but the oil drops must each have a whole number of electrons. What does this tell you about the charges?

Offline Jorriss

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Re: electron probability
« Reply #2 on: July 01, 2012, 10:40:39 PM »
Why do you say that does not work?

In one dimension the probability of measuring a particle between 0 and a is the integral from 0 to a of the wave function squared. In three dimensions, it's the same principle. You integrate the radial component of the wave function squared from 0 to 3a while integrating out the angles.

Offline Sophia7X

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Re: electron probability
« Reply #3 on: July 02, 2012, 10:16:51 AM »
Why do you say that does not work?

In one dimension the probability of measuring a particle between 0 and a is the integral from 0 to a of the wave function squared. In three dimensions, it's the same principle. You integrate the radial component of the wave function squared from 0 to 3a while integrating out the angles.

When I try to plug in for r, for example 0.10 ao, the units won't all cancel out. Probability density shouldn't have any units. Sorry, I have no knowledge of calculus except derivatives, the maximum math I've taken so far is precalc. But I believe this problem can be done without calc because all the calc-based problems are marked and I skipped all of those.




Not really sure on the first part, but the oil drops must each have a whole number of electrons. What does this tell you about the charges?

If the oil drops must have whole numbers of electrons, then it looks like they are all multiples of 4.80e-10 esu, so that must be the charge?
Entropy happens.

Offline XGen

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Re: electron probability
« Reply #4 on: July 04, 2012, 03:55:27 PM »
That is what I assume, but don't quote me on it :P that's just what seems the most intuitive.

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