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Electric Current in a Hydrogen Atom

  1. May 4, 2005 #1
    This selection comes from the Serway . Beichner, Physics for Scientists and Engineers text that many undergrads are (I'm sure) familiar with!

    I've looked at this problem on 3 different casual occasions and on one test and still have yet to arrange the concept in my mind...

    "In the Bohr modle of the hydrogen atom, an electron in the lowest energy state follows a circular path at a distance of 5.29 x 10^-11m from the proton. (a) Show that the speed of the electron is 2.19 x 10^6 m/s (b) What is the effective current associated with this orbiting electron?"

    I've been considering the equation [tex] I_{av} = nqv_{d}A [/tex], which makes the answer to part (b) a gradeschool word problem. BUT, the answer probably involves implementing some equation derived four chapters ago. :smile:

    Any hints would be greatly appreciated! Thanks.

  2. jcsd
  3. May 4, 2005 #2

    Doc Al

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    Staff: Mentor

    The equation you need for part (b) is even simpler than that one, since you just have a single charge (an electron) moving at the given speed as it orbits the nucleus.
  4. May 4, 2005 #3
    So, I have a charge 1.602 x 10^-19 C, a distance 5.29 x 10^-11 m, and a value n = 1. I suppose what has me confused is the cross-sectional area.
  5. May 4, 2005 #4

    Doc Al

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    That formula doesn't apply here. It's meant for situations where you have a charge density with a drift speed. This situation is simpler. How long does it take the electron to make one revolution about the nucleus? I = Q/T.
  6. May 4, 2005 #5
    If I use the velocity given, I calculate 1.52 x 10^-16 sec for one rev. But I 'have' neither the velocity nor the current in part (a).
  7. May 4, 2005 #6

    Doc Al

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    To solve part (a) you need to review the Bohr model. Given the answer from part (a), you can find the effective current (part (b)). The current (Q/T) is one electron charge per period.
  8. May 4, 2005 #7
    For (b) you derived T, and know Q, so some algebra will give you I.

    For (a) my guess would be equate coulombs law to the centripetal force and solve for the velocity.
  9. May 4, 2005 #8
    Ah, so I guess it would have helped to have been taught the Bohr model!

    [tex]F_{e} = k_{e}\frac{q^2}{r^2} = m_{e}\frac{v^2}{r}[/tex]

    [tex] v = [k_{e}\frac{q^2}{rm_{e}}]^\frac{1}{2} [/tex]
    [tex] v = [(8.99 x 10^7)\frac{(1.602 x 10e-19)^2}{(5.29 x 10e-11)(9.10 x 10e-31)}]^\frac{1}{2}[/tex]
    [tex] v = 2,189,240 \frac{m}{s} = 2.19 x 10e6 \frac{m}{s}[/tex]

    Golly, that feels gooood! Thanks guys, funny how equations from forever ago come back at ya.. Thanks again! :smile:

  10. May 5, 2005 #9


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    Don't use "x" for multiplication,but either \cdot or \times...

  11. May 5, 2005 #10
    Thanks, I did my best with the scientific notation there. Apparently the ^2 doesn't work when superscripting an integer.
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