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GR0177 #30 |
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Problem
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This problem is still being typed. |
Atomic }Hydrogen Atom
The hydrogen atom wave-functions all have to do with an exponentially decreasing radial function. Thus, choose choice A. (FYI: Only the spherical harmonic angular parts have trig involved, and thus choice.)
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Alternate Solutions |
vroomfondel
2006-12-01 01:16:43 | A/r blows up at 0
ASin(br) does not go to 0 at infinity
only 1 satisfies both. |  | scottopoly
2006-10-29 22:14:45 | I think the only way to very correctly say that "I" is the only correct one is to realize two facts: the solution from the QM problem is a decaying exponential (you would need this memorized I think), exponentials do not form a complete set of functions, so you cannot build other functions from them (choice II and III are wrong then)
Let me know if I am correct here. | |
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Comments |
eliasds
2008-09-20 14:49:28 | Choice II DOES work iif you choose A and b to be imaginary numbers. What do you all say to that!?!?
a19grey2
2008-11-05 21:51:43 |
ETS doesn't use imaginary numbers on this test unless they are very explicit about it.
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|  | DanS510
2008-09-20 14:47:05 | Choice II DOES work iif you choose A and b to be imaginary numbers. What do you all say to that!?!? |  | Furious
2007-10-19 18:34:03 | I'm not sure if people are even still checking. But wouldn't the function blowing up at zero be a good reason for thinking its a wave function?
I mean I thought as r goes to zero we get hardshell repulsion, and that it's physically impossible for the electron to be in the very center of the nucleus, hence it blowing up as r -> 0.
But I mean, I have the solutions right here, and it says that its only A. I'd like to know why that rationale is wrong.
Gaffer
2007-10-20 12:00:11 |
Since the wave function squared is associated with the probability of finding a particle at some location x, blowing up at zero is bad. An infinite value for the wave function implies that there is an "infinite" probability of the particle being at that location, it throws off all the other finite location values- ie, for 1/r, the electron couldn't be ANYWHERE except at the middle of the nucleus, which is obviously untrue.
The decaying exponential fits the bill since it has a finite value at r = 0 (so the probability distribution is normalizable) and probability goes to 0 for large r. Non-normalization is also why the sin(r) form does not follow. r is unrestricted and so the sin just keeps oscillating. With the angular functions, you limit your variable to pi etc. not all angles and you can normalize you probability.
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greed
2008-08-24 15:57:35 |
As far as I know, wave functions that blow up a the origin are fine provided that they are square-integrable. You could have /r^2) which is divergent at  but  is finite
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| | vroomfondel
2006-12-01 01:16:43 | A/r blows up at 0
ASin(br) does not go to 0 at infinity
only 1 satisfies both. | | scottopoly
2006-10-29 22:14:45 | I think the only way to very correctly say that "I" is the only correct one is to realize two facts: the solution from the QM problem is a decaying exponential (you would need this memorized I think), exponentials do not form a complete set of functions, so you cannot build other functions from them (choice II and III are wrong then)
Let me know if I am correct here.
Walter
2008-08-12 16:47:49 |
The question does not say that the wavefunction is for a hydrogen atom, just "an electron in an atom", so the inference that just because the hydrogen radial wavefunctions have a decaying exponential then the decaying exponential answer is the only correct answer is a bad inference. It could be a multielectron atom or whatever.
This idea about 'complete set of functions' is way off as well. In the case of finding an analytical solution to the differential equation then any damn function will do, it doesn't have to be something like a trig function or complex exponential that you can use to generate infinite series to represent arbitrary other functions.
The vroomfondel solution is correct for the reasons given.
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| | tera
2006-09-13 13:46:14 | But A/r disappears in infinity there so why it is not the correct answer?
askewchan
2008-11-07 15:17:25 |
Because as  we have  . The electron can't have an infinite wave function at the center of the nucleus.
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| | keflavich
2005-11-10 11:47:48 | It looks like you stopped before the end of the sentence here. Choice II doesn't work because it's a spherical harmonic, not a radial WF. You could also eliminate II and III by taking the limit as r-> and realizing the WF must disappear out there. | |
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