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Orbital Angular Momentum Question

  • 27-04-2010 7:37pm
    #1
    Closed Accounts Posts: 206 ✭✭


    Hi there,

    Im trying at the minute ( and Failing ) to do my solid state physics homework, so i was wondering if anyone could point me in the right direction, because ive been googling for a while to no avail.

    The question asks to show that for a fully filled subshell, the total spin S and orbital angular momentum L are both zero.

    I understand the point about S, since the pairs of electrons in a full shell will add up to 0 net.

    But with the orbital momentum im getting very bogged down, I assume it means the sum of each separate electrons momentum. this should be

    sqrt(l*l+1) * h/2pi

    which i understand but for this will only equal zero if l = 0.

    I also understand that z component L = magnetic quantum number * bar h

    I can see how the z components would cancel to 0 when summed since
    m<= +-l, each plus and minus pair will cancel.

    So am I missing something, should the question read that the z components should be zero or is there something I havent found.

    All help appreciated, thanks

    Thomas


Comments

  • Registered Users, Registered Users 2 Posts: 784 ✭✭✭thecornflake


    hey tom,

    page 7 and 8 in the 1st set of magnetic notes explains it. I think it really is just what you said, the angular momentums cancel out on the sum up example on page 9. Its all due to the -1,0,1 like you said, you assign a value to each pair of electrons and they cancel as you add them all.

    conor AP2


  • Closed Accounts Posts: 206 ✭✭eagleye7


    ha conor,

    ypou have no idea how long ive spent looking for that, 3 different books and an hour on google.

    I still think thats not what the question really means but sure if thats whats in the notes ill take it:D


  • Registered Users, Registered Users 2 Posts: 784 ✭✭✭thecornflake


    put it this way , they grade them very easily.

    I got 7/10 on a question where i calculated the velocity of an electron to be on the order of 10 to the 27 m/s. So i pointed out that it was wrong, why it was wrong and a possible solution so they gave me most of the marks. I think it's very easy to over analyse these questions thats usually what i get stuck on.


  • Closed Accounts Posts: 183 ✭✭pvt6zh395dqbrj


    Hey,

    so L is not really a quantity itself but rather an operator on the wavefunction.

    The Heisenberg uncertainty principle says that the only thing we can measure at the same time is the magnitude of the angular momentum and its component along a paritcular axis. We generally call it the z-direction for the laugh (also because we can show that it doesn't matter what direction we pick).

    So we have the two equations then

    MagnitudeL = hbar*sqrt(l(l+1))

    Lz = hbar*ml

    Now, those two are the eigenvalues of the momentum operator. Not only are they the only eigenvalues but they only exist for when l is an integer, when l is between 0 and n-1 and when ml ranges from -l to l.

    So if you take helium which has two electrons in the n=1 state, then then can only have l=0. Meaning they can only have ml=0;

    If you take Neon which has ten electrons in the n=2 state. Two of those electrons will have l=0 and six will have l=1. So the ml value can range from -1 to 1. In the l=0 subshell there will be two electrons (up and down). In the l=1 shell there will be six electrons (-1 up, down; 0 up,down ; 1 up, down).

    If you add all the ml's together you get zero.

    So the total L which is the sum of the ml's is zero for any full subshell.

    With the spins, you pretty much have it right.


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