Angles made by angular momentum vector with magnetic field

1. Mar 21, 2007

Amith2006

1. The problem statement, all variables and given/known data

1) For l=3, find the possible angles that the orbital angular momentum vector(L) makes with the z axis? Here the magnetic field acts along the z axis. l is the orbital quantum number.

2. Relevant equations

3. The attempt at a solution

I solved it in the following way:
Let m(l) represent the magnetic quantum number. Phi is the angle between L and z axis. For l=3, m(l)=-3,-2,-1,0,1,2,3
m(l)x(h/2(pi)) = L[cos(phi)]{[lx(l+1)]^(1/2)}(h/2(pi))
i.e. cos(phi) = m(l)/{[lx(l+1)]^(1/2)}
For m(l) = 3,
Cos(phi) = 3/(12^(1/2))
phi = 30 degrees
For m(l) = 2,
Cos(phi) = 2/(12^(1/2))
phi = 54.7 degrees

For m(l) = 1,
Cos(phi) = 1/(12^(1/2))
phi = 73.2 degrees
But the answer given in my book is 60, 35.3 and 16.8 degrees. I would get this answer if I take sine instead of cosine.

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2. Mar 21, 2007

hage567

I think your answers are correct. If I draw it out to scale, I get your values for theta. Either they made a mistake by using sin, or are for some reason defining the angle in a different place (not between L and Lz). So personally I think the book is wrong.

3. Mar 22, 2007

Thanx buddy.