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Distance of electron from nucleus

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    a) Consider a hydrogenic atom consisting of a carbon-12
    nucleus and a single electron. What is the wavelength of the
    photon emitted when this atom drops from n = 5 to the n = 2
    state?

    b) According to the Bohr theory, what is the distance
    between the electron and the nucleus in the n = 2 state of the
    hydrogenic carbon-12 atom?

    2. Relevant equations

    [itex]r_n = n^2 \times a_0[/itex]

    where [itex]a_0[/itex] is the bohr radius

    3. The attempt at a solution

    I found the answer to part A no problem, but b is giving me trouble.

    I know that the equation for the radius for excited states is for a hydrogen atom. WIth a carbon nucleus I expect it to be much smaller that the bohr radius, but I can't figure out how to account for the change in nucleus size.

    Any ideas?
     
  2. jcsd
  3. Apr 18, 2013 #2
    What are the postulates of Bohr's model?
     
  4. Apr 18, 2013 #3
    Bohr's Model says that angular momentum is quantized, or only has certain values, so

    [itex]\vec{l} = \vec{r} \times \vec{p} [/itex]

    simplified, it is

    [itex]l = rmv = n\hbar[/itex]

    Right direction?
     
  5. Apr 18, 2013 #4
    That is not incorrect, but that does not seem to help much. What is the shape of orbits in Bohr's model, and what holds electrons in place?
     
  6. Apr 18, 2013 #5
    Ok so the electron has energy of -122.4 eV = -1.96*10-17J

    The total energy of the electron is:
    [itex] E = KE + U

    E = \frac{1}{2} m_e v^{2} + \frac{k q^2}{r}[/itex]

    And by F = ma

    [itex] \frac{m_{e} v^{2}}{r} = \frac{k q^{2}}{r^2} [/itex]

    I solved for v2

    and found this expression for r:

    [itex] r = \frac{2 k q^{2}}{k q^{2}-2 E} [/itex]

    However I found this radius to be 1.18 * 10-11m which is wrong
     
  7. Apr 18, 2013 #6
    The equation you termed F = ma is not entirely correct, as you neglected the charge of the nucleus.

    Then you could use the angular momentum rule.
     
  8. Apr 18, 2013 #7
    Ah I see.

    so the force equation would be

    [itex]\frac{m_{e} v^{2}}{r} = \frac{k 6 q^{2}}{r^2}[/itex]?
     
  9. Apr 18, 2013 #8
    Yes.
     
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