Question #118727
Determine the total angular momentum quantum number if the minimum angle between the total angular momentum vector (in a single electron atom) and the z axis is 28.13°.
1
Expert's answer
2020-05-28T12:01:29-0400

The minimum angle between the  total angular momentum vector (in a single electron atom) and the z axis is 

cosθ=l(h2π)l(l+1)(h2π)cos\theta=\frac{l (\frac{h}{2\pi})}{\sqrt{l(l+1)}(\frac{h}{2\pi})}

cos28.13°=l(h2π)l(l+1)(h2π)cos28.13°=\frac{l (\frac{h}{2\pi})}{\sqrt{l(l+1)}(\frac{h}{2\pi})}

cos28.13°=ll(l+1)cos28.13°=\frac{l }{\sqrt{l(l+1)}}

0.88=ll(l+1)0.88=\frac{l }{\sqrt{l(l+1)}}

on solving the above equation, we get l=3.43=3(Approximately)l=3.43=3 (Approximately)

So the total angular momentum is

L=l(l+1)(h2π)=3(3+1)(h2π)=12(h2π)L=\sqrt{l(l+1)}(\frac{h}{2\pi})=\sqrt{3(3+1)}(\frac{h}{2\pi})=\sqrt{12}(\frac{h}{2\pi})

L=12(h2π)=12(6.62610342π)=3.651034J.sL=\sqrt{12}(\frac{h}{2\pi})=\sqrt{12}(\frac{6.626*10^{-34}}{2\pi})=3.65*10^{-34} J.s


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