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The Bohr Radius and reduced mass

  1. May 3, 2015 #1

    squelch

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    1. The problem statement, all variables and given/known data
    A muon is a particle with a charge equal to that of an electron and a mass equal to 207 times the mass of an electron. What is the radius and energy of the ground state of hydrogen if the electron is replaced with a muon? Don't forget to use the concept of reduced mass.

    2. Relevant equations

    [tex]\begin{array}{l}
    r = \frac{{4\pi {\varepsilon _0}{\hbar ^2}{n^2}}}{{m{e^2}}}\\
    {E_n} = - \frac{{m{e^4}}}{{2{{(4\pi {\varepsilon _0})}^2}{\hbar ^2}{n^2}}}\\
    \mu = \frac{1}{{\frac{1}{{{m_1}}} + \frac{1}{{{m_2}}}}} = \frac{{{m_1}{m_2}}}{{{m_1} + {m_2}}}
    \end{array}[/tex]

    3. The attempt at a solution

    For the ground state, n=1. To find this hypothetical radius, it seems we simply substitute the mass of the muon in for the mass of the electron, which yields r = 2.53528x10^-13 meters, or 0.0048 the Bohr radius. Similarly for the energy.

    However, I'm not precisely sure what to do with the reduced mass. The reduced mass seems to be a product if the mass of both a proton and electron -- is it more appropriate to substitute this reduced mass in for the mass of the electron given the parameters of the problem?
     
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  3. May 3, 2015 #2

    DrClaude

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    In the "correct" derivation of the Bohr model, you have to consider that both the nucleus and the electron can move. Just as in classical physics, you then find that the motion can be separated into two parts, the center-of-mass motion and the relative motion of the two particles. It is the latter that give rise to the equations for the Bohr orbits. What enters in these equations is the reduced mass of the system proton + electron. However, since the mass of the proton is much greater than that of the electron, you can approximate
    $$
    \mu = \frac{m_p m_e}{m_p + m_e} \approx \frac{m_p m_e}{m_p} = m_e
    $$
    So you will find in all equations the mass of the electron, which is the same result you get if you start the derivation by assuming that the nucleus is fixed.

    You can check for yourself how good this approximation is if you replace the mass of the electron by that of the muon.
     
  4. May 3, 2015 #3

    squelch

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    So is it appropriate to simply take the reduced mass of a muon and use that in place of the mass of the electron in both equations?
     
  5. May 3, 2015 #4

    DrClaude

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    I don't understand what you mean, as there is no such thing as "the reduced mass of a muon." The reduced mass requires two particles.
     
  6. May 3, 2015 #5

    squelch

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    Well, that makes a bit more sense. I suppose I should ask if I simply substitute the reduced mass [tex]\mu = \frac{{{m_{{\mu ^ - }}}{m_p}}}{{{m_p} + {m_{{\mu ^ - }}}}}[/tex]

    ...in place of m in the equations for the Bohr radius and energy.
     
  7. May 3, 2015 #6

    DrClaude

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    Yes, that's what you should do.
     
  8. May 3, 2015 #7

    squelch

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    Alrighty.

    Calculating using the reduced mass, I find:
    [tex]\begin{array}{l}
    r = 2.8473 \times {10^{ - 13}}m\\
    E = - 4.05134 \times {10^{ - 16}}J
    \end{array}[/tex]

    Thanks for the help.
     
  9. May 3, 2015 #8

    mfb

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    If you care about precision, you have to do the same for the electron, but there the difference between particle mass and reduced mass is much smaller (about 1/2000, while it is about 1/9 for the muon).
     
  10. May 3, 2015 #9

    squelch

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    I'm not sure precisely what you mean. There's only one parameter in each equation for mass that I can see.
     
  11. May 3, 2015 #10

    mfb

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    The difference between electron mass and the reduced mass for the electron/proton system is small.
    The difference between muon mass and the reduced mass for the muon/proton system is significant.
     
  12. May 3, 2015 #11

    squelch

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    Oh. Are you just saying that were I to calculate the Bohr radius for an actual hydrogen atom I should use the reduced mass of the electron/proton system to get a higher degree of precision, and that failing to use the reduced mass in the muon/proton system introduces a greater degree of imprecision than it might in the electron/proton system?

    I'm sorry, I just wanted to sort of restate it in words I understand to make sure I'm getting my mind around what you're explaining.
     
  13. May 3, 2015 #12

    mfb

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    Right.
     
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