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Angles made by angular momentum vector with magnetic field

  1. Mar 21, 2007 #1
    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.
     

    Attached Files:

  2. jcsd
  3. Mar 21, 2007 #2

    hage567

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    Homework Helper

    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.
     
  4. Mar 22, 2007 #3
    Thanx buddy.
     
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