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 Click here to return to the Physics Archives

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