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Hyperfine Structure for Hypogen

  1. May 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Hyperfine structure is the splitting of energy levels due to the coupling of the magnetic moment of the atomic nucleus and the electron’s total angular momentum.
    In a hypothetical atom called “hypogen” the electron is replaced by a negatively
    charged pion (π-) which has a rest mass of mπ=140 MeV/c2 and a spin of s=0. The
    nucleus has a total spin of I=1 and a nuclear g-factor of gN=-4.5. You can assume that the mass of the nucleus is much larger than the pion mass.

    1. Derive the hyperfine Hamiltonian

    2. Derive an equation of the energy shifts for all possible states with n=2, l=1 and
    calculate the energy shifts in eV. Use the Hamiltonian of question 1.

    3. Explain why the level structure of real hydrogen is more complicated than
    that of hypogen (proton spin I=1/2). Use the n=2 state for your explanation
    (not more than 50 words!). (Hint: take the fine structure into account)
    [20]


    2. Relevant equations

    Hamiltonian of the energy shift,
    [tex] H=-\widehat{\mu_{N}}\bullet \widehat{B_{l}}[/tex]

    Magnetic field produces by the pion,
    [tex]\widehat{B_{l}}=\frac{-\mu_{0} e}{4\pi r^{3}}\widehat{r}\times\widehat{v}[/tex]

    Angular momentum,
    [tex] \widehat{L}=\widehat{r}\times\widehat{v}m_{\pi}[/tex]

    Nuclear magnetic moment,
    [tex]\frac{g_{N}\mu_{N}}{\frac{h}{2\pi}}\widehat{I}[/tex]

    Use first order perturbation theory to evaluate the energy shifts.

    3. The attempt at a solution

    Hello!

    So I have done part one I think. The method I used was to multiply and divide the magnetic term by the pion mass so that I could make the r cross v into the angular momentum. Here is the major steps in my calculation;

    [tex] \widehat{B_{l}}=\frac{-\mu_{0} e}{4\pi r^{3}}\widehat{r}\times\widehat{v}[/tex]

    [tex]\widehat{B_{l}}=\frac{-\mu_{0} e}{4\pi r^{3}m_{\pi}}\widehat{r}\times\widehat{v}m_{\pi}[/tex]

    [tex]\widehat{B_{l}}=\frac{-\mu_{0} e}{4\pi r^{3} m_{\pi}}\widehat{L}[/tex]

    Substuting the nuclear magnetic moment into the hamiltonian gives,

    [tex]H=\frac{g_{N}\mu_{N}\mu_{0} e}{2 \m_{\pi}r^{3}h}\widehat{I}\bullet\widehat{L}[/tex]

    Hopefully I have made no errors!

    For part two I get down to a final equation. However my answer depends upon the mass of the nucleus, which is not given. Therefore I was wondering if there is a way to calculate the energy without the mass? My Professor insited that there was a way, however I did not understand his explanation. Please could someone help?

    Here is my equation,

    [tex]\frac{g_{N}\mu_{0}e^{2}h(bar)^{2}}{192 a_{B}^{3}\pi m_{N}m_{\pi}}[/tex]

    If you want me to show the rest of my working just say, I will add some of it later, i have to go out for a while so don' have time to perfect this post right now (plus it's due on friday so I thought I'd better get this out there).
     
  2. jcsd
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