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Energy level of a 'protonium atom' (answer Check)

  1. May 9, 2009 #1

    TFM

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

    In 1989, a new type of exotic atom called protonium was produced. Its
    structure is hydrogen-like, with the electron replaced by an antiproton. As
    the antiparticle of the proton, an antiproton has the same mass but
    opposite charge. In this problem, you should consider only the
    electromagnetic interaction between the antiproton and the proton, by
    analogy with ordinary hydrogen. The mass of the proton is 1836 times
    larger than the mass of an electron.

    2)

    What is the energy in eV of a photon emitted when protonium
    undergoes a transition from the state with principal quantum
    number n = 4 to the state with n = 2?

    2. Relevant equations

    [tex] E_n = \frac{E_r}{n} [/tex]

    [tex] E_r = -\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2 [/tex]

    3. The attempt at a solution

    Okay, so I have done this question, but my answer seems really large...

    using:

    [tex] \Delta E_{4 > 2} = \frac{E_r}{4^2} - \frac{E_r}{2^2} [/tex]

    using the constants:

    m = 1.67 e-27
    hbar = 1.054e-34
    e = -1.6e-19
    epsilon-naught = 8.85e-12

    I have gotten the Rydberg energy, [tex] E_r = 1.73 * 10^{13} [/tex]

    I have thus got a value for the energy shift of -3.24 e12 Joules, which is [tex]2 *10^{31}[/tex] eV. This seems to big. does this answer seem reasonable...?
     
  2. jcsd
  3. May 9, 2009 #2

    diazona

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    Homework Helper

    Well, it's going to be large, because [tex]E_r[/tex] is proportional to mass and there's a lot more of it in this "protonium" than in hydrogen. But I do see a couple of things to fix: I think you may have done your calculation of [tex]E_r[/tex] incorrectly - I get a value on the order of 10^4 eV. Go back and check that.

    Also, when you have a two-particle system, you use the reduced mass in place of m:
    [tex]\mu = \frac{m_1 m_2}{m_1 + m_2}[/tex]
     
  4. May 10, 2009 #3

    TFM

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    I expected it to be large, just not that large.

    Okay, so the reduced mass is [tex] 8.35 * 10^{-28}kg [/tex] from previous question. :redface:

    Okay so if I work through this...

    [tex] E_r = -\frac{\mu}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0} \right)^2 [/tex]

    [tex] -\frac{\mu}{2\hbar^2} = -\frac{8.35*10^{-28}}{2*(1.054*10^{-34})^2} = 3.76 * 10^{40} [/tex]

    [tex] \frac{e^2}{4\pi\epsilon_0} = \frac{(1.6*10^{-19})^2}{4\pi*(8.85*10^{-12})} = 2.3 * 10^{-28}[/tex]

    square this:

    [tex] 5.3 * 10^{-56} [/tex]

    put together gives:

    [tex] E_r = 1.99*10^{-15} [/tex]

    ???
     
  5. May 10, 2009 #4

    diazona

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    Homework Helper

    What are the units of that?

    Always keep track of units - if not explicitly, at least make sure you know what units your final answer is going to be in, and write them down! You can't get a correct answer without the correct units.
     
  6. May 10, 2009 #5

    TFM

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    The units would be Joules since:

    [tex]

    \mu = kg

    \hbar = Js

    e = C

    \epsilon_0 = C^2/nm^2

    [/tex]

    also this does convert to [tex] 1.2 x 10^4 [/tex] eV.

    And this new value gives me a change of energy of [tex] -2.25 * 10^3 [/tex] eV, which seems more reasonable.
     
  7. May 10, 2009 #6

    diazona

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    Homework Helper

    That agrees with my calculations :smile:
     
  8. May 10, 2009 #7

    TFM

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    Excellent. :smile:

    Many Thanks.
     
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