Question

A hydrogen atom is in a state that has $L_{z}=2 h .$ In the semiclassical vector model, the angular momentum vector $\overrightarrow{\boldsymbol{L}}$ for this state makes an angle $\theta_{L}=63.4^{\circ}$ with the $+z$ -axis. (a) What is the $l$ quantum number for this state? (b) What is the smallest possible $n$ quantum number for this state? 

          A hydrogen atom is in a state that has $L_{z}=2 h .$ In the semiclassical vector model, the angular momentum vector $\overrightarrow{\boldsymbol{L}}$ for this state makes an angle $\theta_{L}=63.4^{\circ}$ with the $+z$ -axis. (a) What is the $l$ quantum number for this state? (b) What is the smallest possible $n$ quantum number for this state?
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A hydrogen atom is in a state that has $L_{z}=2 h .$ In the semiclassical vector model, the angular momentum vector $\overrightarrow{\boldsymbol{L}}$ for this state makes an angle $\theta_{L}=63.4^{\circ}$ with the $+z$ -axis. (a) What is the $l$ quantum number for this state? (b) What is the smallest possible $n$ quantum number for this state?
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Transcript

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00:01 All right, so let's say the projection of the quantum angular momentum operator on the z -axis gives us a value of 2h -bar for a particular result.
00:13 So that means like our m -quantum number is 2.
00:18 And we're told that this makes an angle of 63 .4 degrees with the angular momentum vector.
00:24 So basically what we have is like l dot, you know, z -hat is equal to, the, you know, h -bar times the square root of l times l plus 1, times the cosine of theta, and this should be equal to 2h -bar.
00:45 So we have 2 -h -bar equals h -bar times the square root of l times l -plus -1 times the cosine of this angle.
00:53 So the h -bars go away...
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