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GR0177 #81
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Quantum Mechanics$\Rightarrow$}Angular Momentum

Recall the angular momentum eigen-equations, $L^2 \psi = \hbar^2 l(l+1) \psi$ and $L_z \psi = m\hbar \psi$.

The problem wants $L^2 \psi = 6\hbar^2 \psi$ and $L_z \psi = -\hbar \psi$. Matching coefficients with the above equations, one finds that

$l(l+1) = 6$ and $m=-1$. Solving, one finds that $l=2,-3$. Any one of the spherical harmonics with $Y^{-1}_{2}$ or $Y^{-1}_{-3}$ would work. So, since the second spherical harmonic isn't listed, take choice (B).

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jmason86
2009-07-03 15:08:49
How did you know to use 6? The problem said to use 3S and there are 2 electrons.

3 = 2s+1, which can be solved for spin. But where does the 6 come from?
 tankonst2011-09-03 06:26:43 The eigenvalues of $\L^2$ are $\l(l+1)$. So one can solve $\l(l+1)$ to get l=2 or l=-3. That why the 6 is given.
 tankonst2011-09-03 06:30:19 The eigenvalues of $\ L^2$ are . So one can solve l(l+1)=6 to get l=2 or l=-3. That why the 6 is given.
Mexicana
2007-10-02 11:34:01
A BIG correction to Yosun in here. That is regarding the possibility of having the orbital quantum number $l=-3$. This is never possible since $l$ is defined to be ALWAYS POSITIVE and in the range of $0,1,2,...,n-1$; where $n$ is the principal quantum number. So it is not as if the choice of $Y^{-1}_{-3}$ is not part of the multiple choice, but even if it was, you should always know $l$ can't have negative values.
 sidharthsp2008-10-20 03:23:08 BINGO!!
 gravity2010-11-10 02:47:36 I had an oh shit moment. Whew!
 sina22013-09-15 05:49:30 Hey Yosun, please revise this.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$