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angular momentum |
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| Aug16-12, 07:52 AM | #1 |
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angular momentum
1. The problem statement, all variables and given/known data
a classical electron moves in a circle of radius 0.5mm with velocity 20ms-1 what is the value of the quantum number L which gives a quantised angular momentum close to the angular momentum of this classical electron? 2. Relevant equations L=r * p 3. The attempt at a solution L=r*p 500e-6 * 20=0.010 |
| Aug16-12, 08:29 AM | #2 |
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You need to throw in the equation for quantum angular momentum.
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| Aug16-12, 08:36 AM | #3 |
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is it L=[itex]\sqrt{l(l+1)hbar}[/itex]
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| Aug16-12, 09:03 AM | #4 |
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angular momentum
is it possible to find the value of the quantum number "l" (azimuthal quantum number)?
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| Aug16-12, 09:09 AM | #5 |
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Check the dimensions in that formula.
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| Aug16-12, 09:16 AM | #6 |
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do i have to presume that n=1 before i continue the calculation, because there is no mention of principal quantum number in the question?
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| Aug16-12, 09:55 AM | #7 |
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The description says "classical electron". So I guess you should use Bohr's model here. What is the angular momentum in Bohr's model?
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| Aug16-12, 10:22 AM | #8 |
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the lowest value for n is 1, this gives the smallest orbital radius 0.0529nm(bohr radius)
L=r*p=mvr m=9.1e-31, v=20m/s r=0.5nm mvr=nhbar L=n h/2pi=nhbar |
| Aug16-12, 10:36 AM | #9 |
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I think you should use the latter formula to determine n that gives the closest match of L to that of the classical electron.
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| Aug16-12, 10:39 AM | #10 |
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can you explain that again please
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| Aug16-12, 10:52 AM | #11 |
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You can compute the angular momentum from the radius and velocity given.
You have the formula for the angular momentum in Bohr's model. What n gives the closest fit between the two? You could also consider the other formula, involving the square root of l(l + 1). For large n, and correspondingly large l, what can be said about the results given by these two equations? |
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