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de Broglie Wave and Bohr Radius
Name: J. Alfredo G.
Status: educator
Age: 40s
Location: N/A
Country: N/A
Date: 2/26/2003
Question:
By using de Broglie's equation one find that lambda=3.35A
for the electron in the hydrogen atom (speed of electron = 2.187 E6 m/s,
according to Bohr's atom model). How can this wave exists in a space of
only 0.529A (Bohr radius)?
Replies:
Your numbers explain it! According to Bohr, for a stable orbit, exactly one
wavelength must fit on the path of the orbit. And sure enough!
2 * pi * 0.529 = 3.32A, awfully close to 3.35A. I suspect a more careful
calculation would produce perfect agreement.
Your question, though, is a very good one, since the electron in the lowest
Bohr orbit is not really a tiny planet circling a sun-like proton.
Nonetheless (luckily for the progress of physics), this simple model works
beautifully, which has always amazed me (and many others.)
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
The circumference corresponding with the Borh radius is 3.32A, so this
does not
seem inconsistent to me, given the approximate nature of the model.
Tim Mooney
The "wave" follows the orbit in the Bohr model. The path length for the
orbit = 2 * Pi * Bohr radius.
Greg Bradburn
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